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Praxis Core Math

Course: praxis core math   >   unit 1.

  • Algebraic properties | Lesson
  • Algebraic properties | Worked example
  • Solution procedures | Lesson
  • Solution procedures | Worked example
  • Equivalent expressions | Lesson
  • Equivalent expressions | Worked example
  • Creating expressions and equations | Lesson
  • Creating expressions and equations | Worked example

Algebraic word problems | Lesson

  • Algebraic word problems | Worked example
  • Linear equations | Lesson
  • Linear equations | Worked example
  • Quadratic equations | Lesson
  • Quadratic equations | Worked example

What are algebraic word problems?

What skills are needed.

  • Translating sentences to equations
  • Solving linear equations with one variable
  • Evaluating algebraic expressions
  • Solving problems using Venn diagrams

How do we solve algebraic word problems?

  • Define a variable.
  • Write an equation using the variable.
  • Solve the equation.
  • If the variable is not the answer to the word problem, use the variable to calculate the answer.

What's a Venn diagram?

  • 7 + 10 − 13 = 4 ‍   brought both food and drinks.
  • 7 − 4 = 3 ‍   brought only food.
  • 10 − 4 = 6 ‍   brought only drinks.
  • Your answer should be
  • an integer, like 6 ‍  
  • a simplified proper fraction, like 3 / 5 ‍  
  • a simplified improper fraction, like 7 / 4 ‍  
  • a mixed number, like 1   3 / 4 ‍  
  • an exact decimal, like 0.75 ‍  
  • a multiple of pi, like 12   pi ‍   or 2 / 3   pi ‍  
  • (Choice A)   $ 4 ‍   A $ 4 ‍  
  • (Choice B)   $ 5 ‍   B $ 5 ‍  
  • (Choice C)   $ 9 ‍   C $ 9 ‍  
  • (Choice D)   $ 14 ‍   D $ 14 ‍  
  • (Choice E)   $ 20 ‍   E $ 20 ‍  
  • (Choice A)   10 ‍   A 10 ‍  
  • (Choice B)   12 ‍   B 12 ‍  
  • (Choice C)   24 ‍   C 24 ‍  
  • (Choice D)   30 ‍   D 30 ‍  
  • (Choice E)   32 ‍   E 32 ‍  
  • (Choice A)   4 ‍   A 4 ‍  
  • (Choice B)   10 ‍   B 10 ‍  
  • (Choice C)   14 ‍   C 14 ‍  
  • (Choice D)   18 ‍   D 18 ‍  
  • (Choice E)   22 ‍   E 22 ‍  

Things to remember

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How to Solve an Algebraic Expression

Last Updated: October 27, 2023 Fact Checked

This article was co-authored by David Jia . David Jia is an Academic Tutor and the Founder of LA Math Tutoring, a private tutoring company based in Los Angeles, California. With over 10 years of teaching experience, David works with students of all ages and grades in various subjects, as well as college admissions counseling and test preparation for the SAT, ACT, ISEE, and more. After attaining a perfect 800 math score and a 690 English score on the SAT, David was awarded the Dickinson Scholarship from the University of Miami, where he graduated with a Bachelor’s degree in Business Administration. Additionally, David has worked as an instructor for online videos for textbook companies such as Larson Texts, Big Ideas Learning, and Big Ideas Math. There are 10 references cited in this article, which can be found at the bottom of the page. This article has been fact-checked, ensuring the accuracy of any cited facts and confirming the authority of its sources. This article has been viewed 485,677 times.

An algebraic expression is a mathematical phrase that contains numbers and/or variables. Though it cannot be solved because it does not contain an equals sign (=), it can be simplified. You can, however, solve algebraic equations , which contain algebraic expressions separated by an equals sign. If you want to know how to master this mathematical concept, then see Step 1 to get started.

Understanding the Basics

Step 1 Understand the difference between an algebraic expression and an algebraic equation.

  • Algebraic expression : 4x + 2
  • Algebraic equation : 4x + 2 = 100

Step 2 Know how to combine like terms.

  • 3x 2 + 5 + 4x 3 - x 2 + 2x 3 + 9 =
  • 3x 2 - x 2 + 4x 3 + 2x 3 + 5 + 9 =
  • 2x 2 + 6x 3 + 14

Step 3 Know how to factor a number.

  • You can see that each coefficient can be divisible by 3. Just "factor out" the number 3 by dividing each term by 3 to get your simplified equation.
  • 3x/3 + 15/3 = 9x/3 + 30/3 =
  • x + 5 = 3x + 10

Step 4 Know the order of operations.

  • (3 + 5) 2 x 10 + 4
  • First, follow P, the operation in the parentheses:
  • = (8) 2 x 10 + 4
  • Then, follow E, the operation of the exponent:
  • = 64 x 10 + 4
  • Next, do multiplication:
  • And last, do addition:

Step 5 Learn how to isolate a variable.

  • 5x + 15 = 65 =
  • 5x/5 + 15/5 = 65/5 =
  • x + 3 = 13 =

Solve an Algebraic Equation

Step 1 Solve a basic linear algebraic equation.

  • 4x + 16 = 25 -3x =
  • 4x = 25 -16 - 3x
  • 4x + 3x = 25 -16 =
  • 7x/7 = 9/7 =

Step 2 Solve an algebraic equation with exponents.

  • First, subtract 12 from both sides.
  • 2x 2 + 12 -12 = 44 -12 =
  • Next, divide both sides by 2.
  • 2x 2 /2 = 32/2 =
  • Solve by taking the square root of both sides, since that will turn x 2 into x.
  • √x 2 = √16 =
  • State both answers:x = 4, -4

Step 3 Solve an algebraic expression with fractions.

  • First, cross multiply to get rid of the fraction. You have to multiply the numerator of one fraction by the denominator of the other.
  • (x + 3) x 3 = 2 x 6 =
  • Now, combine like terms. Combine the constant terms, 9 and 12, by subtracting 9 from both sides.
  • 3x + 9 - 9 = 12 - 9 =
  • Isolate the variable, x, by dividing both sides by 3 and you've got your answer.
  • 3x/3 = 3/3 =

Step 4 Solve an algebraic expression with radical signs.

  • First, move everything that isn't under the radical sign to the other side of the equation:
  • √(2x+9) = 5
  • Then, square both sides to remove the radical:
  • (√(2x+9)) 2 = 5 2 =
  • Now, solve the equation as you normally would by combining the constants and isolating the variable:
  • 2x = 25 - 9 =

Step 5 Solve an algebraic expression that contains absolute value.

  • |4x +2| - 6 = 8 =
  • |4x +2| = 8 + 6 =
  • |4x +2| = 14 =
  • 4x + 2 = 14 =
  • Now, solve again by flipping the sign of the term on the other side of the equation after you've isolated the absolute value:
  • 4x + 2 = -14
  • 4x = -14 -2
  • 4x/4 = -16/4 =
  • Now, just state both answers: x = -4, 3

Community Q&A

Donagan

  • The degree of a polynomial is the highest power within the terms. Thanks Helpful 9 Not Helpful 1
  • Once you're done, replace the variable with the answer, and solve the sum to see if it makes sense. If it does, then, congratulations! You just solved an algebraic equation! Thanks Helpful 7 Not Helpful 3
  • To cross-check your answer, visit wolfram-alpha.com. They give the answer and often the two steps. Thanks Helpful 8 Not Helpful 5

algebraic expression problem solving with solution

You Might Also Like

Evaluate an Algebraic Expression

  • ↑ https://www.math4texas.org/Page/527
  • ↑ https://www.khanacademy.org/math/cc-sixth-grade-math/cc-6th-expressions-and-variables/cc-6th-combining-like-terms/v/combining-like-terms-2
  • ↑ https://www.mathsisfun.com/algebra/factoring.html
  • ↑ https://www.mathsisfun.com/operation-order-pemdas.html
  • ↑ https://sciencing.com/tips-for-solving-algebraic-equations-13712207.html
  • ↑ https://www.mathsisfun.com/algebra/equations-solving.html
  • ↑ https://tutorial.math.lamar.edu/Classes/Alg/SolveExpEqns.aspx
  • ↑ https://www.mathsisfun.com/algebra/fractions-algebra.html
  • ↑ https://math.libretexts.org/Courses/Coastline_College/Math_C045%3A_Beginning_and_Intermediate_Algebra_(Chau_Duc_Tran)/10%3A_Roots_and_Radicals/10.07%3A_Solve_Radical_Equations
  • ↑ https://www.mathplanet.com/education/algebra-1/linear-inequalitites/solving-absolute-value-equations-and-inequalities

About This Article

David Jia

If you want to solve an algebraic expression, first understand that expressions, unlike equations, are mathematical phrase that can contain numbers and/or variables but cannot be solved. For example, 4x + 2 is an expression. To reduce the expression, combine like terms, for example everything with the same variable. After you've done that, factor numbers by finding the lowest common denominator. Then, use the order of operations, which is known by the acronym PEMDAS, to reduce or solve the problem. To learn how to solve algebraic equations, keep scrolling! Did this summary help you? Yes No

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Simple Algebra Problems – Easy Exercises with Solutions for Beginners

JUMP TO TOPIC

Understanding Algebraic Expressions

Breaking down algebra problems, solving algebraic equations, tackling algebra word problems, types of algebraic equations, algebra for different grades.

Simple Algebra Problems Easy Exercises with Solutions for Beginners

For instance, solving the equation (3x = 7) for (x) helps us understand how to isolate the variable to find its value.

Illustration of Simple Algebra Problems

I always find it fascinating how algebra serves as the foundation for more advanced topics in mathematics and science. Starting with basic problems such as ( $(x-1)^2 = [4\sqrt{(x-4)}]^2$ ) allows us to grasp key concepts and build the skills necessary for tackling more complex challenges.

So whether you’re refreshing your algebra skills or just beginning to explore this mathematical language, let’s dive into some examples and solutions to demystify the subject. Trust me, with a bit of practice, you’ll see algebra not just as a series of problems, but as a powerful tool that helps us solve everyday puzzles.

Simple Algebra Problems and Strategies

When I approach simple algebra problems, one of the first things I do is identify the variable.

The variable is like a placeholder for a number that I’m trying to find—a mystery I’m keen to solve. Typically represented by letters like ( x ) or ( y ), variables allow me to translate real-world situations into algebraic expressions and equations.

An algebraic expression is a mathematical phrase that can contain ordinary numbers, variables (like ( x ) or ( y )), and operators (like add, subtract, multiply, and divide). For example, ( 4x + 7 ) is an algebraic expression where ( x ) is the variable and the numbers ( 4 ) and ( 7 ) are terms. It’s important to manipulate these properly to maintain the equation’s balance.

Solving algebra problems often starts with simplifying expressions. Here’s a simple method to follow:

  • Combine like terms : Terms that have the same variable can be combined. For instance, ( 3x + 4x = 7x ).
  • Isolate the variable : Move the variable to one side of the equation. If the equation is ( 2x + 5 = 13 ), my job is to get ( x ) by itself by subtracting ( 5 ) from both sides, giving me ( 2x = 8 ).

With algebraic equations, the goal is to solve for the variable by performing the same operation on both sides. Here’s a table with an example:

Algebra word problems require translating sentences into equations. If a word problem says “I have six less than twice the number of apples than Bob,” and Bob has ( b ) apples, then I’d write the expression as ( 2b – 6 ).

Understanding these strategies helps me tackle basic algebra problems efficiently. Remember, practice makes perfect, and each problem is an opportunity to improve.

In algebra, we encounter a variety of equation types and each serves a unique role in problem-solving. Here, I’ll brief you about some typical forms.

Linear Equations : These are the simplest form, where the highest power of the variable is one. They take the general form ( ax + b = 0 ), where ( a ) and ( b ) are constants, and ( x ) is the variable. For example, ( 2x + 3 = 0 ) is a linear equation.

Polynomial Equations : Unlike for linear equations, polynomial equations can have variables raised to higher powers. The general form of a polynomial equation is ( $a_nx^n + a_{n-1}x^{n-1} + … + a_2x^2 + a_1x + a_0 = 0$ ). In this equation, ( n ) is the highest power, and ( $a_n$ ), ( $a_{n-1} $), …, ( $a_0$ ) represent the coefficients which can be any real number.

  • Binomial Equations : They are a specific type of polynomial where there are exactly two terms. Like ($ x^2 – 4 $), which is also the difference of squares, a common format encountered in factoring.

To understand how equations can be solved by factoring, consider the quadratic equation ( $x^2$ – 5x + 6 = 0 ). I can factor this into ( (x-2)(x-3) = 0 ), which allows me to find the roots of the equation.

Here’s how some equations look when classified by degree:

Remember, identification and proper handling of these equations are essential in algebra as they form the basis for complex problem-solving.

In my experience with algebra, I’ve found that the journey begins as early as the 6th grade, where students get their first taste of this fascinating subject with the introduction of variables representing an unknown quantity.

I’ve created worksheets and activities aimed specifically at making this early transition engaging and educational.

6th Grade :

Moving forward, the complexity of algebraic problems increases:

7th and 8th Grades :

  • Mastery of negative numbers: students practice operations like ( -3 – 4 ) or ( -5 $\times$ 2 ).
  • Exploring the rules of basic arithmetic operations with negative numbers.
  • Worksheets often contain numeric and literal expressions that help solidify their concepts.

Advanced topics like linear algebra are typically reserved for higher education. However, the solid foundation set in these early grades is crucial. I’ve developed materials to encourage students to understand and enjoy algebra’s logic and structure.

Remember, algebra is a tool that helps us quantify and solve problems, both numerical and abstract. My goal is to make learning these concepts, from numbers to numeric operations, as accessible as possible, while always maintaining a friendly approach to education.

I’ve walked through various simple algebra problems to help establish a foundational understanding of algebraic concepts. Through practice, you’ll find that these problems become more intuitive, allowing you to tackle more complex equations with confidence.

Remember, the key steps in solving any algebra problem include:

  • Identifying variables and what they represent.
  • Setting up the equation that reflects the problem statement.
  • Applying algebraic rules such as the distributive property ($a(b + c) = ab + ac$), combining like terms, and inverse operations.
  • Checking your solutions by substituting them back into the original equations to ensure they work.

As you continue to engage with algebra, consistently revisiting these steps will deepen your understanding and increase your proficiency. Don’t get discouraged by mistakes; they’re an important part of the learning process.

I hope that the straightforward problems I’ve presented have made algebra feel more manageable and a little less daunting. Happy solving!

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6.1: Evaluating Algebraic Expressions

  • Last updated
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  • Page ID 57572

  • David Arnold
  • College of the Redwoods

In this section we will evaluate algebraic expressions for given values of the variables contained in the expressions. Here are some simple tips to help you be successful.

Tips for Evaluating Algebraic Expressions

  • Replace all occurrences of variables in the expression with open parentheses. Leave room between the parentheses to substitute the given value of the variable.
  • Substitute the given values of variables in the open parentheses prepared in the first step.
  • Evaluate the resulting expression according to the Rules Guiding Order of Operations.

Let's begin with an example.

Evaluate the expression \(x^2 − 2xy + y^2\) at \(x = −3\) and \(y = 2\).

Following “Tips for Evaluating Algebraic Expressions,” first replace all occurrences of variables in the expression x 2 − 2 xy + y 2 with open parentheses.

\[ x^2 -2xy + y^2 = ( ~ )^2 -2(~)(~) + ( ~ )^2 \nonumber\nonumber \]

Secondly, replace each variable with its given value, and thirdly, follow the “Rules Guiding Order of Operations” to evaluate the resulting expression.

\[ \begin{aligned} x^2 -2xy + y^2 ~ & \textcolor{red}{ \text{ Original expression.}} \\ =( \textcolor{red}{-3} )^2 -2 ( \textcolor{red}{-3})( \textcolor{red}{2}) + (\textcolor{red}{2})^2 ~ & \textcolor{red}{ \text{ Substitute } -3 \text{ for } x \text{and 2 for }y.} \\ =9-2(-3)(2)+4 ~ & \textcolor{red}{ \text{ Evaluate exponents first.}} \\ = 9-(-6)(2)+4 ~ & \textcolor{red}{ \text{ Left to right, multiply } 2(-3)=-6.} \\ =9-(-12)+4 ~ & \textcolor{red}{ \text{ Left to right, multiply: } (-6)(2) = -12.} \\ = 9 + 12 + 4 ~ & \textcolor{red}{ \text{ Add the opposite.}} \\ = 25 ~ & \textcolor{red}{ \text{ Add.}} \end{aligned}\nonumber \]

If x = −2 and y = −1, evaluate x 3 − y 3 .

Evaluate the expression ( a − b ) 2 If a = 3 and b = −5, at a = 3 and b = −5.

Following “Tips for Evaluating Algebraic Expressions,” first replace all occurrences of variables in the expression ( a − b ) 2 with open parentheses.

\[ (a-b)^2 = (()-())^2\nonumber \]

\[ \begin{aligned} (a-b)^2 = (( \textcolor{red}{3})-( \textcolor{red}{-5}))^2 ~ & \textcolor{red}{ \text{ Substitute 3 for } a \text{ and } -5 \text{ for } b.} \\ = (3+5)^2 ~ & \textcolor{red}{ \text{ Add the opposite: } (3)-(-5)=3+5} \\ = 8^2 ~ & \textcolor{red}{ \text{ Simplify inside parentheses: } 3+5 = 8} \\ =64 ~ & \textcolor{red}{ \text{ Evaluate exponent: } 8^2 = 64} \end{aligned}\nonumber \]

If a = 3 and b = −5, evaluate a 2 − b 2 .

Evaluate the expression |a|−|b| at a = 5 and b = −7.

Following “Tips for Evaluating Algebraic Expressions,” first replace all occurrences of variables in the expression |a|−|b| with open parentheses.

\[ |a| - |b| = |( ~ )| - |( ~ )|\nonumber \]

\[ \begin{aligned} |a| - |b| = |( \textcolor{red}{5} )| = |( \textcolor{red}{-7})| ~ & \textcolor{red}{ \text{ Substitute 5 for } a \text{ and } -7 \text{ for } b.} \\ = 5 - 7 ~ & \textcolor{red}{ \text{ Absolute values first: } |(5)| = 5 \text{ and } |(-7)|=7|} \\ =5+(-7) ~ & \textcolor{red}{ \text{ Add the opposites: } 5 - 7 = 5+(-7).} \\ =-2 ~ & \textcolor{red}{ \text{ Add: } 5+(-7)=-2.} \end{aligned}\nonumber \]

If a = 5 and b = −7, evaluate 2| a | − 3| b |.

Evaluate the expression | a − b | at a = 5 and b = −7.

Following “Tips for Evaluating Algebraic Expressions,” first replace all occurrences of variables in the expression | a − b | with open parentheses.

\[ |a-b| = |(~)-(~)|\nonumber \]

\[ \begin{aligned} |a-b| = |( \textcolor{red}{5})-( \textcolor{red}{-7})| ~ & \textcolor{red}{ \text{ Substitute 5 for } a \text{ and } -7 \text{ for } b.} \\ = |5+7| ~ & \textcolor{red}{ \text{ Add the opposite: } 5-(-7)=5+7.} \\ =|12| ~ & \textcolor{red}{ \text{ Add: } 5+7=12.} \\ =12 ~ & \textcolor{red}{ \text{ Take the absolute value: } |12| = 12.} \end{aligned}\nonumber \]

If a = 5 and b = −7, evaluate |2 a − 3 b |.

Evaluate the expression

\[ \frac{ad-bc}{a+b}\nonumber \]

at a = 5, b = −3, c = 2, and d = −4.

Following “Tips for Evaluating Algebraic Expressions,” first replace all occurrences of variables in the expression with open parentheses.

\[ \frac{ad-bc}{a+b} = \frac{(~)(~)-(~)(~)}{(~)+(~)}\nonumber \]

\[ \begin{aligned} \frac{ad-bc}{a+b} = \frac{( \textcolor{red}{5}) -( \textcolor{red}{-3}) ( \textcolor{red}{2})}{( \textcolor{red}{5}) + ( \textcolor{red}{-3})} ~ & \textcolor{red}{ \text{ Substitute: } 5 \text{ for } a,~ -3 \text{ for } b,~ 2 \text{ for } c,~ -4 \text{ for } d.} \\ = \frac{-20-(-6)}{2} ~ & \begin{aligned} \textcolor{red}{ \text{ Numerator: } (5)(=4)=-20,~ (-3)(2) = -6.} \\ \textcolor{red}{ \text{ Denominator: } 5+(-3)=2.} \end{aligned} \\ = \frac{-20+6}{2} ~ & \textcolor{red}{ \text{ Numerator: Add the opposite.}} \\ = \frac{-14}{2} ~ & \textcolor{red}{ \text{ Numerator: } -20+6=-14.} \\ = -7 ~ & \textcolor{red}{ \text{Divide.}} \end{aligned}\nonumber \]

If a = −7, b = −3, c = −15, 15, and d = −14, evaluate:

\[\frac{a^2+b^2}{c+d}\nonumber \]

Pictured below is a rectangular prism.

Screen Shot 2019-08-19 at 6.09.16 PM.png

The volume of the rectangular prism is given by the formula

\[V=LWH,\nonumber \]

where L is the length, W is the width, and H is the height of the rectangular prism. Find the volume of a rectangular prism having length 12 feet, width 4 feet, and height 6 feet.

Following “Tips for Evaluating Algebraic Expressions,” first replace all occurrences of of L, W, and H in the formula

\[ V = LWH\nonumber \]

with open parentheses.

\[V = (~)(~)(~)\nonumber \]

Next, substitute 12 ft for L , 4 ft for W , and 6 ft for H and simplify.

\[ \begin{aligned} V = (12 \text{ft})(4 \text{ft})(6 \text{ft}) \\ = 288 \text{ft}^3 \end{aligned}\nonumber \]

Hence, the volume of the rectangular prism is 288 cubic feet.

The surface area of the prism pictured in this example is given by the following formula:

\[S = 2(W H + LH + LW) \nonumber \]

If L = 12, W = 4, and H = 6 feet, respectively, calculate the surface area.

288 square feet

In Exercises 1-12, evaluate the expression at the given value of x.

1. −3x 2 − 6x + 3 at x = 7

2. 7x 2 − 7x + 1 at x = −8

3. −6x − 6 at x = 3

4. 6x − 1 at x = −10

5. 5x 2 + 2x + 4 at x = −1

6. 4x 2 − 9x + 4 at x = −3

7. −9x − 5 at x = −2

8. −9x + 12 at x = 5

9. 4x 2 + 2x + 6 at x = −6

10. −3x 2 + 7x + 4 at x = −7

11. 12x + 10 at x = −12

12. −6x + 7 at x = 11

In Exercises 13-28, evaluate the expression at the given values of x and y.

13. |x|−|y| at x = −5 and y = 4

14. |x|−|y| at x = −1 and y = −2

15. −5x 2 + 2y 2 at x = 4 and y = 2

16. −5x 2 − 4y 2 at x = −2 and y = −5

17. |x|−|y| at x = 0 and y = 2

18. |x|−|y| at x = −2 and y = 0

19. |x − y| at x = 4 and y = 5

20. |x − y| at x = −1 and y = −4

21. 5x 2 − 4xy + 3y 2 at x = 1 and y = −4

22. 3x 2 + 5xy + 3y 2 at x = 2 and y = −1

23. |x − y| at x = 4 and y = 4

24. |x − y| at x = 3 and y = −5

25. −5x 2 − 3xy + 5y 2 at x = −1 and y = −2

26. 3x 2 − 2xy − 5y 2 at x = 2 and y = 5

27. 5x 2 + 4y 2 at x = −2 and y = −2

28. −4x 2 + 2y 2 at x = 4 and y = −5

In Exercises 29-40, evaluate the expression at the given value of x.

29. \( \frac{9+9x}{−x}\) at x = −3

30. \( \frac{9 − 2x}{−x}\) at x = −1

31. \(\frac{−8x + 9}{−9 + x}\) at x = 10

32. \(\frac{2x + 4}{1 + x}\) at x = 0

33. \(\frac{−4+9x}{7x}\) at x = 2

34. \(\frac{−1 − 9x}{x}\) at x = −1

35. \(\frac{−12 − 7x}{x}\) at x = −1

36. \(\frac{12 + 11x}{3x}\) at x = −6

37. \(\frac{6x − 10}{5}\) + x at x = −6

38. \(\frac{11x + 11}{−4}\) + x at x = 5

39. \(\frac{10x + 11}{5}\) + x at x = −4

40. \(\frac{6x + 12}{−3}\) + x at x = 2

41. The formula

\[d=16t^2\nonumber \]

gives the distance (in feet) that an object falls from rest in terms of the time t that has elapsed since its release. Find the distance d (in feet) that an object falls in t = 4 seconds.

42. The formula

\[d = 16t^2\nonumber \]

gives the distance (in feet) that an object falls from rest in terms of the time t that has elapsed since its release. Find the distance d (in feet) that an object falls in t = 24 seconds.

43. The formula

\[C = \frac{5(F − 32)}{9}\nonumber \]

gives the Celcius temperature C in terms of the Fahrenheit temperature F. Use the formula to find the Celsius temperature (◦ C) if the Fahrenheit temperature is F = 230◦ F.

44. The formula

gives the Celcius temperature C in terms of the Fahrenheit temperature F. Use the formula to find the Celsius temperature ( ◦ C) if the Fahrenheit temperature is F = 95 ◦ F.

45. The Kelvin scale of temperature is used in chemistry and physics. Absolute zero occurs at 0 ◦ K, the temperature at which molecules have zero kinetic energy. Water freezes at 273 ◦ K and boils at K = 373 ◦ K. To change Kelvin temperature to Fahrenheit temperature, we use the formula

\[F = \frac{9(K − 273)}{5} + 32.\nonumber \]

Use the formula to change 28 ◦ K to Fahrenheit.

46. The Kelvin scale of temperature is used in chemistry and physics. Absolute zero occurs at 0 ◦ K, the temperature at which molecules have zero kinetic energy. Water freezes at 273 ◦ K and boils at K = 373 ◦ K. To change Kelvin temperature to Fahrenheit temperature, we use the formula

Use the formula to change 248 ◦ K to Fahrenheit.

47. A ball is thrown vertically upward. Its velocity t seconds after its release is given by the formula

\[v = v0 − gt,\nonumber \]

where v 0 is its initial velocity, g is the acceleration due to gravity, and v is the velocity of the ball at time t. The acceleration due to gravity is g = 32 feet per second per second. If the initial velocity of the ball is v 0 = 272 feet per second, find the speed of the ball after t = 6 seconds.

48. A ball is thrown vertically upward. Its velocity t seconds after its release is given by the formula

\[v = v_0 − gt,\nonumber \]

where v 0 is its initial velocity, g is the acceleration due to gravity, and v is the velocity of the ball at time t. The acceleration due to gravity is g = 32 feet per second per second. If the initial velocity of the ball is v 0 = 470 feet per second, find the speed of the ball after t = 4 seconds.

49. Even numbers . Evaluate the expression 2n for the following values:

iv) n = −4

v) n = −5

vi) Is the result always an even number? Explain.

50. Odd numbers . Evaluate the expression 2n + 1 for the following values:

vi) Is the result always an odd number? Explain.

1. −186

3. −24

11. −134

15. −72

17. −2

29. −6

31. −71

39. −29

41. 256 feet

43. 110 degrees

45. −409 ◦ F

47. 80 feet per second

iv) −8

v) −10

vi) Yes, the result will always be an even number because 2 will always be a factor of the product 2n.

Solving Equations

What is an equation.

An equation says that two things are equal. It will have an equals sign "=" like this:

That equations says:

what is on the left (x − 2)  equals  what is on the right (4)

So an equation is like a statement " this equals that "

What is a Solution?

A Solution is a value we can put in place of a variable (such as x ) that makes the equation true .

Example: x − 2 = 4

When we put 6 in place of x we get:

which is true

So x = 6 is a solution.

How about other values for x ?

  • For x=5 we get "5−2=4" which is not true , so x=5 is not a solution .
  • For x=9 we get "9−2=4" which is not true , so x=9 is not a solution .

In this case x = 6 is the only solution.

You might like to practice solving some animated equations .

More Than One Solution

There can be more than one solution.

Example: (x−3)(x−2) = 0

When x is 3 we get:

(3−3)(3−2) = 0 × 1 = 0

And when x is 2 we get:

(2−3)(2−2) = (−1) × 0 = 0

which is also true

So the solutions are:

x = 3 , or x = 2

When we gather all solutions together it is called a Solution Set

The above solution set is: {2, 3}

Solutions Everywhere!

Some equations are true for all allowed values and are then called Identities

Example: sin(−θ) = −sin(θ) is one of the Trigonometric Identities

Let's try θ = 30°:

sin(−30°) = −0.5 and

−sin(30°) = −0.5

So it is true for θ = 30°

Let's try θ = 90°:

sin(−90°) = −1 and

−sin(90°) = −1

So it is also true for θ = 90°

Is it true for all values of θ ? Try some values for yourself!

How to Solve an Equation

There is no "one perfect way" to solve all equations.

A Useful Goal

But we often get success when our goal is to end up with:

x = something

In other words, we want to move everything except "x" (or whatever name the variable has) over to the right hand side.

Example: Solve 3x−6 = 9

Now we have x = something ,

and a short calculation reveals that x = 5

Like a Puzzle

In fact, solving an equation is just like solving a puzzle. And like puzzles, there are things we can (and cannot) do.

Here are some things we can do:

  • Add or Subtract the same value from both sides
  • Clear out any fractions by Multiplying every term by the bottom parts
  • Divide every term by the same nonzero value
  • Combine Like Terms
  • Expanding (the opposite of factoring) may also help
  • Recognizing a pattern, such as the difference of squares
  • Sometimes we can apply a function to both sides (e.g. square both sides)

Example: Solve √(x/2) = 3

And the more "tricks" and techniques you learn the better you will get.

Special Equations

There are special ways of solving some types of equations. Learn how to ...

  • solve Quadratic Equations
  • solve Radical Equations
  • solve Equations with Sine, Cosine and Tangent

Check Your Solutions

You should always check that your "solution" really is a solution.

How To Check

Take the solution(s) and put them in the original equation to see if they really work.

Example: solve for x:

2x x − 3 + 3 = 6 x − 3     (x≠3)

We have said x≠3 to avoid a division by zero.

Let's multiply through by (x − 3) :

2x + 3(x−3) = 6

Bring the 6 to the left:

2x + 3(x−3) − 6 = 0

Expand and solve:

2x + 3x − 9 − 6 = 0

5x − 15 = 0

5(x − 3) = 0

Which can be solved by having x=3

Let us check x=3 using the original question:

2 × 3 3 − 3 + 3  =   6 3 − 3

Hang On: 3 − 3 = 0 That means dividing by Zero!

And anyway, we said at the top that x≠3 , so ...

x = 3 does not actually work, and so:

There is No Solution!

That was interesting ... we thought we had found a solution, but when we looked back at the question we found it wasn't allowed!

This gives us a moral lesson:

"Solving" only gives us possible solutions, they need to be checked!

  • Note down where an expression is not defined (due to a division by zero, the square root of a negative number, or some other reason)
  • Show all the steps , so it can be checked later (by you or someone else)

Logo image

Modeling, Functions, and Graphs

Katherine Yoshiwara

Section A.3 Algebraic Expressions and Problem Solving

Example a.17 ..

  • Choose a variable for the number of hours Loren works per week.
  • Write an algebraic expression for the amount of Loren’s weekly earnings.
  • Let \(h\) stand for the number of hours Loren works per week.
  • The amount Loren earns is given by \begin{equation*} \blert{6\times (\text{number of hours Loren worked})} \end{equation*} or \(6\cdot h\text{.}\) Loren’s weekly earnings can be expressed as \(6h\text{.}\)

Example A.18 .

Example a.19 ..

  • Choose variables to represent the unknown quantities and write an algebraic expression for April’s weekly income in terms of her sales.
  • Find April’s income for a week in which she sells $ \(350\) worth of cleaning products.
  • Let \(I\) represent April’s total income for the week, and let \(S\) represent the total amount of her sales. We translate the information from the problem into mathematical language as follows: \begin{gather*} \blert{\text{Her income consists of }\$200 . . .\text{ plus }. . . 9\% \text{ of her sales}} \\ I \hphantom{consists of}= \hphantom{of}200 \hphantom{plus+}+ \hphantom{....}0.09 \hphantom{of her}S \end{gather*} Thus, \(I = 200 + 0.09S\text{.}\)
  • We want to evaluate our expression from part (a) with \(S = 350\text{.}\) We substitute \(\alert{350}\) for \(S\) to find \begin{equation*} I = 200 + 0.09(\alert{350}) \end{equation*} Following the order of operations, we perform the multiplication before the addition. Thus, we begin by computing \(0.09(350)\text{.}\) \begin{align*} I \amp = 200 + 0.09(350)\amp\amp\blert{\text{Multiply }0.09 (350) \text{ first.}}\\ \amp = 200 + 31.5\\ \amp = 231.50 \end{align*} April’s income for the week is $ \(231.50\text{.}\)

Remark A.20 . Calculator Tip.

Example a.21 ..

  • Choose variables to represent the unknown quantities and write an expression for the cost of shipping Andrew’s painting.
  • Find the shipping cost if Andrew uses \(2.9\) pounds of packing material.
  • Let \(C\) stand for the shipping cost and let \(w\) stand for the weight of the packing material. Andrew must find the total weight of his package first, then multiply by the shipping charge. The total weight of the package is \(8.3 + w\) pounds. We use parentheses around this expression to show that it should be computed first, and the sum should be multiplied by the shipping charge of $ \(2.80\) per pound. Thus, \begin{equation*} C = 2.80(8.3 + w) \end{equation*}
  • Evaluate the formula from part (a) with \(w = \alert{2.9}\text{.}\) \begin{align*} C \amp = 2.80(8.3 + \alert{2.9})\amp\amp\blert{\text{Add inside parentheses.}}\\ \amp = 2.80(11.2)\amp\amp\blert{\text{Multiply.}}\\ \amp = 31.36 \end{align*} The cost of shipping the painting is $ \(31.36\text{.}\)

Remark A.22 . Calculator Tip.

Caution a.23 ., subsection problem solving, guidelines for problem solving..

  • Identify the unknown quantity and assign a variable to represent it.
  • Find some quantity that can be expressed in two different ways and write an equation.
  • Solve the equation.
  • Interpret your solution to answer the question in the problem.

Subsection Supply and Demand

Example a.24 ..

  • We are looking for the equilibrium price, \(p\text{.}\)
  • The Coffee Connection would like the demand for its coffee to equal its supply. We equate the expressions for supply and for demand to obtain the equation \begin{equation*} 800 - 60p = 175 + 40p \end{equation*}
  • Solve the equation. To get all terms containing the variable, \(p\text{,}\) on one side of the equation, we add \(60p\) to both sides and subtract \(175\) from both sides to obtain \begin{align*} 800 - 60p + \alert{60p - 175} \amp= 175 + 40p + \alert{60p - 175}\\ 625 \amp = 100p \amp\amp\blert{\text{Divide both sides by }100.}\\ 6.25 \amp = p \end{align*}
  • The Coffee Connection should charge $ \(6.25\) per pound for its coffee.

Subsection Percent Problems

Percent formula., example a.25 ..

  • Let \(c\) represent the cost of the house last year.
  • Express the current price of the house in two different ways. During the past year, the price of the house increased by \(4\%\text{,}\) or \(0.04c\text{.}\) Its current price is thus \begin{equation*} \stackrel{\text{Original cost}}{(1)c} + \stackrel{\text{Price increase}}{0.04c} = c(1 + 0.04) = 1.04c \end{equation*} This expression is equal to the value given for current price of the house: \begin{equation*} 1.04c = 100,000 \end{equation*}
  • To solve this equation, we divide both sides by \(1.04\) to find \begin{equation*} c = \frac{100,000}{1.04}= 96,153.846 \end{equation*}
  • To the nearest cent, the cost of the house last year was $ \(96,153.85\text{.}\)

Caution A.26 .

Subsection weighted averages, example a.27 ..

  • Let \(x\) represent the final exam score Kwan needs.
  • Kwan’s grade is the weighted average of his test, homework, and final exam scores. \begin{equation*} \frac{\alert{0.50}(84) + \alert{0.20}(92) + \alert{0.30}x}{1.00}=90 \end{equation*} (The sum of the weights is 1.00, or 100% of Kwan’s grade.) Multiply both sides of the equation by \(1.00\) to get \begin{equation*} 0.50(84) + 0.20(92) + 0.30x = 1.00(90) \end{equation*}
  • Solve the equation. Simplify the left side first. \begin{align*} 60.4 + 0.30x \amp = 90\amp\amp\blert{\text{Subtract 60.4 from both sides.}}\\ 0.30x \amp = 29.6\amp\amp\blert{\text{Divide both sides by 0.30.}}\\ x \amp = 98.7 \end{align*}
  • Kwan needs a score of \(98.7\) on the final exam to earn a grade of \(90\text{.}\)

Weighted Average.

Example a.28 ..

  • Let \(p\) represent the number of pounds of LeanMeal needed.
  • Simplify each side of the equation, using the distributive law on the right side, then solve. \begin{align*} 7.5 + 0.05p \amp = 4 + 0.08p \amp\amp\blert{\text{Subtract}~4+0.05p~\text{from both sides.}}\\ 3.5 \amp = 0.03p \amp\amp \blert{\text{Divid both sides by}~ 0.03.}\\ p \amp = 116.\overline{6} \end{align*}
  • Delbert should mix \(116\frac{2}{3}\) pounds of LeanMeal with \(50\) pounds of JuicyBits to make a mixture that is \(8\%\) fat.

Subsection Section Summary

Subsubsection vocabulary.

  • Weighted average
  • Equilibrium price
  • Algebraic expression
  • Evaluate an expression

Subsubsection SKILLS

  • Write an algebraic expression: #1–12
  • Evaluate an algebraic expression: #1–12
  • Write and solve an equation to solve a problem: #13–28

Exercises Exercises A.3

Exercise group..

  • Write an expression for Jim’s age in terms of Ana’s age.
  • Use your expression to find Jim’s age when ana is 22 years old.
  • Write an expression for the total cost of new wheels in terms of the number of wheels Rani must replace.
  • Use your expression to find the total cost if Rani must replace 8 wheels.
  • Write an expression for the total number of hours Helen must drive in terms of her average driving speed.
  • Use your expression to find how long Helen must drive if she averages 45 miles per hour.
  • Write an expression for the number of years before Ben gets his inheritance in terms of his present age.
  • Use your expression to find how many more years Ben must wait after he turns 13 years old.
  • Write an expression for the area of a circle in terms of its radius.
  • Find the area of a circle whose radius is 5 centimeters.
  • Write an expression for the volume of a sphere in terms of its radius.
  • Find the volume of a sphere whose radius is 5 centimeters.
  • Write an expression for the total bill for an item (price plus tax) in terms of the price of the item.
  • Find the total bill for an item whose price is $490.
  • Write an expression for the balance (initial deposit plus interest) in the account after one year in terms of the amount deposited.
  • Find the total amount in the account after one year if $350 was deposited.
  • Write an expression for the cost of a long-distance phone call in terms of the number of minutes of the call.
  • Find the cost of a 27-minute phone call.
  • Write an expression for the cost of the flight in terms of Mr. Owlsley’s weight.
  • Find the cost if Mr. Owsley weights 162 pounds.
  • Write an expression for the amount of rice Juan has consumed in terms of the number of weeks since he bought the bag.
  • Write an expression for the amount of rice Juan has left in terms of the number of weeks since he bought the bag.
  • Find the amount of rice Juan has left after 6 weeks.
  • Write an expression for the elevation that Trinh has lost in terms of the distance she has cycled.
  • Write an expression for Trinh’s elevation in terms of the number of miles she has cycled.
  • Find Trinh’s elevation after she has cycled 9 miles.
  • What are we asked to find in this problem? Assign a variable to represent it.
  • Write an expression in terms of your variable for the distance Roger’s wife drives.
  • Write an expression in terms of your variable for the distance Roger has cycled.
  • Write an equation and solve it.
  • Write an expression in terms of your variable for the distance Kate and Julie sailed.
  • Write an expression in terms of your variable for the distance their father traveled.
  • Write an expression in terms of your variable for the total cost incurred by each machine.
  • Write an expression in terms of your variable for the total cost incurred by each refrigerator.
  • Assuming the same rate of growth, what do you predict for the population of Midland next year?
  • What was the population of Midland last year?
  • Assuming the same rate of inflation, what do you predict for the price of a steak dinner next year?
  • What did a steak dinner cost last year?
  • Write algebraic expressions in terms of your variable for the amounts of each fertilizer the horticulturist uses. Use the table.
  • Write expressions for the amount of potash in each batch of fertilizer.
  • Write two different expressions for the amount of potash in the mixture. Now write an equation and solve it.
  • Write algebraic expressions in terms of your variable for the amounts of each alloy the sculptor uses. Use the table.
  • Write expressions for the amount of copper in each batch of alloy.
  • Write two different expressions for the amount of copper in the mixture. Now write an equation and solve it.
  • Write algebraic expressions for the total amounts Lacy’s pays its managers, its department heads, and its clerks.
  • Write two different expressions for the total amount Lacy’s pays in salaries each year.
  • Write algebraic expressions for the number of trucks and the number of cars that will meet emission standards.
  • Write two different expressions for the total number of vehicles that will meet the standards.

algebraic expression problem solving with solution

Algebraic Expressions and Word Problems

Related Topics: More Lessons for Grade 7 Math Worksheets

Examples, solutions, videos, worksheets, games and activities to help Algebra 1 or grade 7 students learn how to write algebraic expressions from word problems.

Beginning Algebra & Word Problem Steps

  • Name what x is.
  • Define everything in the problem in terms of x.
  • Write the equation.
  • Solve the equation.
  • Kevin’s age is 3 years more than twice Jane’s age. The sum of their ages is 39. How old is Kevin and Jane?
  • The difference between two numbers is 7. Find the two numbers if the larger number is three times the smaller.
  • Mary and Jim collect baseball cards, Mary has 5 more than 3 times as many cards as Jim. The total number of cards they both have is 253. How many cards does Mary have?

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Chapter 1: Solving Equations and Inequalities

Problem solving, learning objectives.

  • Translate words into algebraic expressions and equations
  • Define a process for solving word problems
  • Apply the steps for solving word problems to distance, rate, and time problems
  • Apply the steps for solving word problems to interest rate problems
  • Evaluate a formula using substitution
  • Rearrange formulas to isolate specific variables
  • Identify an unknown given a formula
  • Apply the steps for solving word problems to geometry problems
  • Use the formula for converting between Fahrenheit and Celsius

Define a Process for Problem Solving

Word problems can be tricky. Often it takes a bit of practice to convert an English sentence into a mathematical sentence, which is one of the first steps to solving word problems. In the table below, words or phrases commonly associated with mathematical operators are categorized. Word problems often contain these or similar words, so it’s good to see what mathematical operators are associated with them.

Some examples follow:

  • [latex]x\text{ is }5[/latex]  becomes [latex]x=5[/latex]
  • Three more than a number becomes [latex]x+3[/latex]
  • Four less than a number becomes [latex]x-4[/latex]
  • Double the cost becomes [latex]2\cdot\text{ cost }[/latex]
  • Groceries and gas together for the week cost $250 means [latex]\text{ groceries }+\text{ gas }=250[/latex]
  • The difference of 9 and a number becomes [latex]9-x[/latex]. Notice how 9 is first in the sentence and the expression

Let’s practice translating a few more English phrases into algebraic expressions.

Translate the table into algebraic expressions:

In this example video, we show how to translate more words into mathematical expressions.

The power of algebra is how it can help you model real situations in order to answer questions about them.

Here are some steps to translate problem situations into algebraic equations you can solve. Not every word problem fits perfectly into these steps, but they will help you get started.

  • Read and understand the problem.
  • Determine the constants and variables in the problem.
  • Translate words into algebraic expressions and equations.
  • Write an equation to represent the problem.
  • Solve the equation.
  • Check and interpret your answer. Sometimes writing a sentence helps.

Twenty-eight less than five times a certain number is 232. What is the number?

Following the steps provided:

  • Read and understand: we are looking for a number.
  • Constants and variables: 28 and 232 are constants, “a certain number” is our variable because we don’t know its value, and we are asked to find it. We will call it x.
  • Translate:  five times a certain number translates to [latex]5x[/latex] Twenty-eight less than five times a certain number translates to [latex]5x-28[/latex] because subtraction is built backward. is 232 translates to [latex]=232[/latex] because “is” is associated with equals.
  • Write an equation:  [latex]5x-28=232[/latex]

[latex]\begin{array}{r}5x-28=232\\5x=260\\x=52\,\,\,\end{array}[/latex]

[latex]\begin{array}{r}5\left(52\right)-28=232\\5\left(52\right)=260\\260=260\end{array}[/latex].

In the video that follows, we show another example of how to translate a sentence into a mathematical expression using a problem solving method.

Another type of number problem involves consecutive numbers. Consecutive numbers are numbers that come one after the other, such as 3, 4, 5. If we are looking for several consecutive numbers it is important to first identify what they look like with variables before we set up the equation.

For example, let’s say I want to know the next consecutive integer after 4. In mathematical terms, we would add 1 to 4 to get 5. We can generalize this idea as follows: the consecutive integer of any number, x , is [latex]x+1[/latex]. If we continue this pattern we can define any number of consecutive integers from any starting point. The following table shows how to describe four consecutive integers using algebraic notation.

We apply the idea of consecutive integers to solving a word problem in the following example.

The sum of three consecutive integers is 93. What are the integers?

  • Read and understand:  We are looking for three numbers, and we know they are consecutive integers.
  • Constants and Variables:  93 is a constant. The first integer we will call x . Second: [latex]x+1[/latex] Third: [latex]x+2[/latex]
  • Translate:  The sum of three consecutive integers translates to [latex]x+\left(x+1\right)+\left(x+2\right)[/latex], based on how we defined the first, second, and third integers. Notice how we placed parentheses around the second and third integers. This is just to make each integer more distinct. is 93 translates to [latex]=93[/latex] because is is associated with equals.
  • Write an equation:  [latex]x+\left(x+1\right)+\left(x+2\right)=93[/latex]

[latex]x+x+1+x+2=93[/latex]

Combine like terms, simplify, and solve.

[latex]\begin{array}{r}x+x+1+x+2=93\\3x+3 = 93\\\underline{-3\,\,\,\,\,-3}\\3x=90\\\frac{3x}{3}=\frac{90}{3}\\x=30\end{array}[/latex]

  • Check and Interpret: Okay, we have found a value for x . We were asked to find the value of three consecutive integers, so we need to do a couple more steps. Remember how we defined our variables: The first integer we will call [latex]x[/latex], [latex]x=30[/latex] Second: [latex]x+1[/latex] so [latex]30+1=31[/latex] Third: [latex]x+2[/latex] so [latex]30+2=32[/latex] The three consecutive integers whose sum is [latex]93[/latex] are [latex]30\text{, }31\text{, and }32[/latex]

There is often a well-known formula or relationship that applies to a word problem. For example, if you were to plan a road trip, you would want to know how long it would take you to reach your destination. [latex]d=rt[/latex] is a well-known relationship that associates distance traveled, the rate at which you travel, and how long the travel takes.

Distance, Rate, and Time

If you know two of the quantities in the relationship [latex]d=rt[/latex], you can easily find the third using methods for solving linear equations. For example, if you know that you will be traveling on a road with a speed limit of [latex]30\frac{\text{ miles }}{\text{ hour }}[/latex] for 2 hours, you can find the distance you would travel by multiplying rate times time or [latex]\left(30\frac{\text{ miles }}{\text{ hour }}\right)\left(2\text{ hours }\right)=60\text{ miles }[/latex].

We can generalize this idea depending on what information we are given and what we are looking for. For example, if we need to find time, we could solve the [latex]d=rt[/latex] equation for t using division:

[latex]d=rt\\\frac{d}{r}=t[/latex]

Likewise, if we want to find rate, we can isolate r using division:

[latex]d=rt\\\frac{d}{t}=r[/latex]

In the following examples you will see how this formula is applied to answer questions about ultra marathon running.

Ann Trason

Ultra marathon running (defined as anything longer than 26.2 miles) is becoming very popular among women even though it remains a male-dominated niche sport. Ann Trason has broken twenty world records in her career. One such record was the American River 50-mile Endurance Run which begins in Sacramento, California, and ends in Auburn, California. [1] In 1993 Trason finished the run with a time of 6:09:08.  The men’s record for the same course was set in 1994 by Tom Johnson who finished the course with a time of 5:33:21. [2]

In the next examples we will use the [latex]d=rt[/latex] formula to answer the following questions about the two runners.

What was each runner’s rate for their record-setting runs?

By the time Johnson had finished, how many more miles did Trason have to run?

How much further could Johnson have run if he had run as long as Trason?

What was each runner’s time for running one mile?

To make answering the questions easier, we will round the two runners’ times to 6 hours and 5.5 hours.

Read and Understand:  We are looking for rate and we know distance and time, so we can use the idea: [latex]d=rt\\\frac{d}{t}=r[/latex]

Define and Translate: Because there are two runners, making a table to organize this information helps. Note how we keep units to help us keep track of what how all the terms are related to each other.

Write and Solve:

Trason’s rate:

[latex]\begin{array}{c}d=rt\\\\50\text{ miles }=\text{r}\left(6\text{ hours }\right)\\\frac{50\text{ miles }}{6\text{ hours }}=\frac{8.33\text{ miles }}{\text{ hour }}\end{array}[/latex].

(rounded to two decimal places)

Johnson’s rate:

[latex]\begin{array}{c}d=rt\\\\,\,\,\,\,\,\,50\text{ miles }=\text{r}\left(5.5\text{ hours }\right)\\\frac{50\text{ miles }}{6\text{ hours }}=\frac{9.1\text{ miles }}{\text{ hour }}\end{array}[/latex]

Check and Interpret:

We can fill in our table with this information.

Now that we know each runner’s rate we can answer the second question.

Here is the table we created for reference:

Read and Understand:  We are looking for how many miles Trason still had on the trail when Johnson had finished after 5.5 hours. This is a distance, and we know rate and time.

Define and Translate:  We can use the formula [latex]d=rt[/latex] again. This time the unknown is d , and the time Trason had run is 5.5 hours.

[latex]\begin{array}{l}d=rt\\\\d=8.33\frac{\text{ miles }}{\text{ hour }}\left(5.5\text{ hours }\right)\\\\d=45.82\text{ miles }\end{array}[/latex].

Have we answered the question? We were asked to find how many more miles she had to run after 5.5 hours.  What we have found is how long she had run after 5.5 hours. We need to subtract [latex]d=45.82\text{ miles }[/latex] from the total distance of the course.

[latex]50\text{ miles }-45.82\text{ miles }=1.48\text{ miles }[/latex]

The third question is similar to the second. Now that we know each runner’s rate, we can answer questions about individual distances or times.

Read and Understand:  The word further implies we are looking for a distance.

Define and Translate:  We can use the formula [latex]d=rt[/latex] again. This time the unknown is d , the time is 6 hours, and Johnson’s rate is [latex]9.1\frac{\text{ miles }}{\text{ hour }}[/latex]

[latex]\begin{array}{l}d=rt\\\\d=9.1\frac{\text{ miles }}{\text{ hour }}\left(6\text{ hours }\right)\\\\d=54.6\text{ miles }\end{array}[/latex].

Have we answered the question? We were asked to find how many more miles Johnson would have run if he had run at his rate of [latex]9.1\frac{\text{ miles }}{\text{ hour }}[/latex] for 6 hours.

Johnson would have run 54.6 miles, so that’s 4.6 more miles than than he ran for the race.

Now we will tackle the last question where we are asked to find a time for each runner.

Read and Understand:  we are looking for time, and this time our distance has changed from 50 miles to 1 mile, so we can use

Define and Translate: we can use the formula [latex]d=rt[/latex] again. This time the unknown is t , the distance is 1 mile, and we know each runner’s rate. It may help to create a new table:

We will need to divide to isolate time.

[latex]\begin{array}{c}d=rt\\\\1\text{ mile }=8.33\frac{\text{ miles }}{\text{ hour }}\left(t\text{ hours }\right)\\\\\frac{1\text{ mile }}{\frac{8.33\text{ miles }}{\text{ hour }}}=t\text{ hours }\\\\0.12\text{ hours }=t\end{array}[/latex].

0.12 hours is about 7.2 minutes, so Trason’s time for running one mile was about 7.2 minutes. WOW! She did that for 6 hours!

[latex]\begin{array}{c}d=rt\\\\1\text{ mile }=9.1\frac{\text{ miles }}{\text{ hour }}\left(t\text{ hours }\right)\\\\\frac{1\text{ mile }}{\frac{9.1\text{ miles }}{\text{ hour }}}=t\text{ hours }\\\\0.11\text{ hours }=t\end{array}[/latex].

0.11 hours is about 6.6 minutes, so Johnson’s time for running one mile was about 6.6 minutes. WOW! He did that for 5.5 hours!

Have we answered the question? We were asked to find how long it took each runner to run one mile given the rate at which they ran the whole 50-mile course.  Yes, we answered our question.

Trason’s mile time was [latex]7.2\frac{\text{minutes}}{\text{mile}}[/latex] and Johnsons’ mile time was [latex]6.6\frac{\text{minutes}}{\text{mile}}[/latex]

In the following video, we show another example of answering many rate questions given distance and time.

Simple Interest

In order to entice customers to invest their money, many banks will offer interest-bearing accounts. The accounts work like this: a customer deposits a certain amount of money (called the Principal, or P ), which then grows slowly according to the interest rate ( R , measured in percent) and the length of time ( T , usually measured in months) that the money stays in the account. The amount earned over time is called the interest ( I ), which is then given to the customer.

Caution

The simplest way to calculate interest earned on an account is through the formula [latex]\displaystyle I=P\,\cdot \,R\,\cdot \,T[/latex].

If we know any of the three amounts related to this equation, we can find the fourth. For example, if we want to find the time it will take to accrue a specific amount of interest, we can solve for T using division:

[latex]\displaystyle\begin{array}{l}\,\,\,\,\,\,\,\,\,\,\,\,I=P\,\cdot \,R\,\cdot \,T\\\\ \frac{I}{{P}\,\cdot \,R}=\frac{P\cdot\,R\,\cdot \,T}{\,P\,\cdot \,R}\\\\\,\,\,\,\,\,\,\,\,\,\,{T}=\frac{I}{\,R\,\cdot \,T}\end{array}[/latex]

Below is a table showing the result of solving for each individual variable in the formula.

In the next examples, we will show how to substitute given values into the simple interest formula, and decipher which variable to solve for.

If a customer deposits a principal of $2000 at a monthly rate of 0.7%, what is the total amount that she has after 24 months?

Substitute in the values given for the Principal, Rate, and Time.

[latex]\displaystyle\begin{array}{l}I=P\,\cdot \,R\,\cdot \,T\\I=2000\cdot 0.7\%\cdot 24\end{array}[/latex]

Rewrite 0.7% as the decimal 0.007, then multiply.

[latex]\begin{array}{l}I=2000\cdot 0.007\cdot 24\\I=336\end{array}[/latex]

Add the interest and the original principal amount to get the total amount in her account.

[latex] \displaystyle 2000+336=2336[/latex]

She has $2336 after 24 months.

The following video shows another example of finding an account balance after a given amount of time, principal invested, and a rate.

In the following example you will see why it is important to make sure the units of the interest rate match the units of time when using the simple interest formula.

Alex invests $600 at 3.25% monthly interest for 3 years. What amount of interest has Alex earned?

Read and Understand: The question asks for an amount, so we can substitute what we are given into the simple interest formula [latex]I=P\,\cdot \,R\,\cdot \,T[/latex]

Define and Translate:  we know P, R, and T so we can use substitution. R  = 0.0325, P = $600, and T = 3 years. We have to be careful! R is in months, and T is in years.  We need to change T into months because we can’t change the rate—it is set by the bank.

[latex]{T}=3\text{ years }\cdot12\frac{\text{ months }}{ year }=36\text{ months }[/latex]

Substitute the given values into the formula.

[latex]\begin{array}{l} I=P\,\cdot \,R\,\cdot \,T\\\\I=600\,\cdot \,0.035\,\cdot \,36\\\\{I}=756\end{array}[/latex]

We were asked what amount Alex earned, which is the amount provided by the formula. In the previous example we were asked the total amount in the account, which included the principal and interest earned.

Alex has earned $756.

After 10 years, Jodi’s account balance has earned $1080 in interest. The rate on the account is 0.09% monthly. What was the original amount she invested in the account?

Read and Understand: The question asks for the original amount invested, the principal. We are given a length of time in years, and an interest rate in months, and the amount of interest earned.

Define and Translate:  we know I = $1080, R = 0.009, and T = 10 years so we can use [latex]{P}=\frac{I}{{R}\,\cdot \,T}[/latex]

We also need to make sure the units on the interest rate and the length of time match, and they do not. We need to change time into months again.

[latex]{T}=10\text{ years }\cdot12\frac{\text{ months }}{ year }=120\text{ months }[/latex]

Substitute the given values into the formula

[latex]\begin{array}{l}{P}=\frac{I}{{R}\,\cdot \,T}\\\\{P}=\frac{1080}{{0.009}\,\cdot \,120}\\\\{P}=\frac{1080}{1.08}=1000\end{array}[/latex]

We were asked to find the principal given the amount of interest earned on an account.  If we substitute P = $1000 into the formula [latex]I=P\,\cdot \,R\,\cdot \,T[/latex] we get

[latex]I=1000\,\cdot \,0.009\,\cdot \,120\\I=1080[/latex]

Our solution checks out. Jodi invested $1000.

Further Applications of Linear Equations

Formulas come up in many different areas of life. We have seen the formula that relates distance, rate, and time and the formula for simple interest on an investment. In this section we will look further at formulas and see examples of formulas for dimensions of geometric shapes as well as the formula for converting temperature between Fahrenheit and Celsius.

There are many geometric shapes that have been well studied over the years. We know quite a bit about circles, rectangles, and triangles. Mathematicians have proven many formulas that describe the dimensions of geometric shapes including area, perimeter, surface area, and volume.

Perimeter is the distance around an object. For example, consider a rectangle with a length of 8 and a width of 3. There are two lengths and two widths in a rectangle (opposite sides), so we add [latex]8+8+3+3=22[/latex]. Since there are two lengths and two widths in a rectangle, you may find the perimeter of a rectangle using the formula [latex]{P}=2\left({L}\right)+2\left({W}\right)[/latex] where

In the following example, we will use the problem-solving method we developed to find an unknown width using the formula for the perimeter of a rectangle. By substituting the dimensions we know into the formula, we will be able to isolate the unknown width and find our solution.

You want to make another garden box the same size as the one you already have. You write down the dimensions of the box and go to the lumber store to buy some boards. When you get there you realize you didn’t write down the width dimension—only the perimeter and length. You want the exact dimensions so you can have the store cut the lumber for you.

Here is what you have written down:

Perimeter = 16.4 feet Length = 4.7 feet

Can you find the dimensions you need to have your boards cut at the lumber store? If so, how many boards do you need and what lengths should they be?

Read and Understand:  We know perimeter = 16.4 feet and length = 4.7 feet, and we want to find width.

Define and Translate:

Define the known and unknown dimensions:

First we will substitute the dimensions we know into the formula for perimeter:

[latex]\begin{array}{l}\,\,\,\,\,P=2{W}+2{L}\\\\16.4=2\left(w\right)+2\left(4.7\right)\end{array}[/latex]

Then we will isolate w to find the unknown width.

[latex]\begin{array}{l}16.4=2\left(w\right)+2\left(4.7\right)\\16.4=2{w}+9.4\\\underline{-9.4\,\,\,\,\,\,\,\,\,\,\,\,\,-9.4}\\\,\,\,\,\,\,\,7=2\left(w\right)\\\,\,\,\,\,\,\,\frac{7}{2}=\frac{2\left(w\right)}{2}\\\,\,\,\,3.5=w\end{array}[/latex]

Write the width as a decimal to make cutting the boards easier and replace the units on the measurement, or you won’t get the right size of board!

If we replace the width we found, [latex]w=3.5\text{ feet }[/latex] into the formula for perimeter with the dimensions we wrote down, we can check our work:

[latex]\begin{array}{l}\,\,\,\,\,{P}=2\left({L}\right)+2\left({W}\right)\\\\{16.4}=2\left({4.7}\right)+2\left({3.5}\right)\\\\{16.4}=9.4+7\\\\{16.4}=16.4\end{array}[/latex]

Our calculation for width checks out. We need to ask for 2 boards cut to 3.5 feet and 2 boards cut to 4.7 feet so we can make the new garden box.

This video shows a similar garden box problem.

We could have isolated the w in the formula for perimeter before we solved the equation, and if we were going to use the formula many times, it could save a lot of time. The next example shows how to isolate a variable in a formula before substituting known dimensions or values into the formula.

Isolate the term containing the variable, w, from the formula for the perimeter of a rectangle :  

[latex]{P}=2\left({L}\right)+2\left({W}\right)[/latex].

First, isolate the term with  w by subtracting 2 l from both sides of the equation.

[latex] \displaystyle \begin{array}{l}\,\,\,\,\,\,\,\,\,\,p\,=\,\,\,\,2l+2w\\\underline{\,\,\,\,\,-2l\,\,\,\,\,-2l\,\,\,\,\,\,\,\,\,\,\,}\\p-2l=\,\,\,\,\,\,\,\,\,\,\,\,\,2w\end{array}[/latex]

Next, clear the coefficient of w by dividing both sides of the equation by 2.

[latex]\displaystyle \begin{array}{l}\underline{p-2l}=\underline{2w}\\\,\,\,\,\,\,2\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,2\\ \,\,\,\frac{p-2l}{2}\,\,=\,\,w\\\,\,\,\,\,\,\,\,\,\,\,w=\frac{p-2l}{2}\end{array}[/latex]

You can rewrite the equation so the isolated variable is on the left side.

[latex]w=\frac{p-2l}{2}[/latex]

The area of a triangle is given by [latex] A=\frac{1}{2}bh[/latex] where

A = area b = the length of the base h = the height of the triangle

Remember that when two variables or a number and a variable are sitting next to each other without a mathematical operator between them, you can assume they are being multiplied. This can seem frustrating, but you can think of it like mathematical slang. Over the years, people who use math frequently have just made that shortcut enough that it has been adopted as convention.

In the next example we will use the formula for area of a triangle to find a missing dimension, as well as use substitution to solve for the base of a triangle given the area and height.

Find the base ( b) of a triangle with an area ( A ) of 20 square feet and a height ( h) of 8 feet.

Use the formula for the area of a triangle, [latex] {A}=\frac{{1}}{{2}}{bh}[/latex] .

Substitute the given lengths into the formula and solve for  b.

[latex]\displaystyle \begin{array}{l}\,\,A=\frac{1}{2}bh\\\\20=\frac{1}{2}b\cdot 8\\\\20=\frac{8}{2}b\\\\20=4b\\\\\frac{20}{4}=\frac{4b}{4}\\\\ \,\,\,5=b\end{array}[/latex]

The base of the triangle measures 5 feet.

We can rewrite the formula in terms of b or h as we did with perimeter previously. This probably seems abstract, but it can help you develop your equation-solving skills, as well as help you get more comfortable with working with all kinds of variables, not just x .

Use the multiplication and division properties of equality to isolate the variable b .

[latex]\begin{array}{l}\,\,\,\,\,\,\,\,A=\frac{1}{2}bh\\\\\left(2\right)A=\left(2\right)\frac{1}{2}bh\\\\\,\,\,\,\,\,2A=bh\\\\\,\,\,\,\,\,\,\frac{2A}{h}=\frac{bh}{h}\\\\\,\,\,\,\,\,\,\,\frac{2A}{h}=\frac{b\cancel{h}}{\cancel{h}}\end{array}[/latex]

Write the equation with the desired variable on the left-hand side as a matter of convention:

Use the multiplication and division properties of equality to isolate the variable h .

[latex]\begin{array}{l}\,\,\,\,\,\,\,\,A=\frac{1}{2}bh\\\\\left(2\right)A=\left(2\right)\frac{1}{2}bh\\\\\,\,\,\,\,\,2A=bh\\\\\,\,\,\,\,\,\,\frac{2A}{b}=\frac{bh}{b}\\\\\,\,\,\,\,\,\,\,\frac{2A}{b}=\frac{h\cancel{b}}{\cancel{b}}\end{array}[/latex]

Temperature

Let’s look at another formula that includes parentheses and fractions, the formula for converting from the Fahrenheit temperature scale to the Celsius scale.

[latex]C=\left(F-32\right)\cdot \frac{5}{9}[/latex]

Given a temperature of [latex]12^{\circ}{C}[/latex], find the equivalent in [latex]{}^{\circ}{F}[/latex].

Substitute the given temperature in[latex]{}^{\circ}{C}[/latex] into the conversion formula:

[latex]12=\left(F-32\right)\cdot \frac{5}{9}[/latex]

Isolate the variable F to obtain the equivalent temperature.

[latex]\begin{array}{r}12=\left(F-32\right)\cdot \frac{5}{9}\\\\\left(\frac{9}{5}\right)12=F-32\,\,\,\,\,\,\,\,\,\,\,\,\,\\\\\left(\frac{108}{5}\right)12=F-32\,\,\,\,\,\,\,\,\,\,\,\,\,\\\\21.6=F-32\,\,\,\,\,\,\,\,\,\,\,\,\,\\\underline{+32\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,+32}\,\,\,\,\,\,\,\,\,\,\,\,\\\\53.6={}^{\circ}{F}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\end{array}[/latex]

As with the other formulas we have worked with, we could have isolated the variable F first, then substituted in the given temperature in Celsius.

Solve the formula shown below for converting from the Fahrenheit scale to the Celsius scale for F.

To isolate the variable F, it would be best to clear the fraction involving F first. Multiply both sides of the equation by [latex] \displaystyle \frac{9}{5}[/latex].

[latex]\begin{array}{l}\\\,\,\,\,\left(\frac{9}{5}\right)C=\left(F-32\right)\left(\frac{5}{9}\right)\left(\frac{9}{5}\right)\\\\\,\,\,\,\,\,\,\,\,\,\,\,\frac{9}{5}C=F-32\end{array}[/latex]

Add 32 to both sides.

[latex]\begin{array}{l}\frac{9}{5}\,C+32=F-32+32\\\\\frac{9}{5}\,C+32=F\end{array}[/latex]

[latex]F=\frac{9}{5}C+32[/latex]

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10 best AI math solver tools for math problem-solving

Homework AI makes it easier for students to learn difficult subjects. Boost your assignment and exam grades with these best AI homework helpers.

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  • March 18, 2024

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The traditional approach to learning involves acquiring knowledge through listening and observation.

However, professionally trained AI can better accommodate different learning styles and enhance comprehension by offering tailored, on-demand learning assistance, especially in challenging subjects like mathematics.

As a subject many students struggle with, having access to a reliable AI math solver is invaluable. Math AI solvers can provide students and other learners with instant homework help outside the classroom at any time when needed.

They can also help improve students’ math test scores and build their mathematical skills over time. Let’s look at some of the best math AI tools for mathematical problem-solving:

Ten best AI math solver tools

1. Mathful – Best AI math solver overall

2. HIX Tutor – Best AI math solver for rapid homework response

3. AI Math – Best AI math solver for increasing math test scores

4. HomeworkAI – Best AI math solver for 24/7 math homework help

5. GeniusTutor – Best AI math solver for high-level learning 

6. Mathway – Best AI math solver for solving algebra problems

7. Air Math – Best AI math solver for mobile uses

8. StudyMonkey – Best AI math solver for in-depth explanations 

9. Interactive Mathematics – Best AI math solver for comprehensive chat support

10. Smodin – Best AI math solver for step-by-step solutions

Mathful – Best AI math solver overall

Mathful ai math solver

Mathful is an AI-powered math homework solver that provides step-by-step answers to all types of math questions.

The math AI tool uses a large language model and advanced algorithms to help students improve their math grades and prepare for exams.

Mathful has proven to be one of the most accurate AI math solvers, boasting a remarkable 98% accuracy rate across various mathematical disciplines, such as calculus, algebra, and geometry.

Mathful can also help students of all levels, from elementary school to university and beyond.

Mathful can help students improve their math grades in school by enforcing core math concepts and providing detailed explanations that promote comprehension.

Students can start using Mathful for free; low-cost subscription plans are available after the initial trial. 

  • Able to provide highly accurate solutions and comprehensive explanations.
  • Can process text and image files.
  • Able to solve a variety of question types. 
  • Available to use 24/7. 
  • It cannot replace a real classroom education. 

Get instant answers to Math homework questions with Mathful AI math solver >>>

HIX Tutor – Best AI math solver for rapid homework response

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HIX Tutor is a powerful AI homework helper that provides comprehensive support in many subjects, such as chemistry, biology, and physics.

It also serves as a personal AI math tutor, helping students boost their math grades and overall academic success. 

To use HIX Tutor’s advanced math AI, type in a math problem or upload an image or document of the question.

The tool instantly generates a detailed explanation for each problem step, helping students understand the underlying math concepts. 

HIX Tutor’s AI math problem solver can help save users time spent struggling with complicated math assignments.

Try the AI math solver at no cost. Once you’ve reached your question limit, upgrade to an affordable monthly or annual plan.

  • Delivers step-by-step solutions to math questions.
  • Trained on a large math knowledge dataset. 
  • Reduces time spent on math homework. 
  • Requires payment after the initial trial. 
  • Some students may only use the tool to get answers without learning. 

Streamline the math learning experience with HIX Tutor’s math AI solver >>>

AI Math – Best AI math solver for increasing math test scores

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How you prepare for a math test can significantly impact your performance.

AI math solvers like AI Math help take the frustration out of studying by providing thorough explanations that teach students how to tackle similar math problems. 

The AI math problem solver generates answers to questions in under 10 seconds with a 99% accuracy rate.

AI Math supports over 30 languages so that students can get responses in their native language for better understanding.

Students who use AI Math to supplement their classroom education experience an increase in their math test scores of up to 35%. Starting with AI Math is free; subscriptions cost just a few dollars a month.

  • Covers most branches of math, such as arithmetic and trigonometry.
  • Walks students through the solution to facilitate understanding. 
  • Can solve simple to complex math problems.
  • Does not currently offer advanced math features. 

Choose AI Math and study for math tests in a smarter way >>>

HomeworkAI – Best AI math solver for 24/7 math homework help

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Students often need help with homework outside of traditional school hours. AI math solver tools like HomeworkAI allow students to get comprehensive support round-the-clock.

Much like a personal tutor, HomeworkAI focuses on teaching students how to solve homework problems instead of simply giving answers. 

HomeworkAI can handle math problems with multiple solution methods, meaning a primary solution and possible alternative approaches.

It can also analyze textbook material with practice math questions to aid students’ studies. 

While HomeworkAI excels in helping students complete math assignments with high precision, this AI homework tool can also help students in other school subjects, such as biology, physics, chemistry, literature, and history.

Try HomeworkAI for free, or choose from a low-cost subscription plan for unlimited uses. 

  • Allows students to work at their own pace at home. 
  • User-friendly platform is easy to navigate. 
  • It can help students excel in many subjects, including math.
  • This may cause students to rely too much on online math-solving platforms. 
  • Rarely, solutions may be outdated or incorrect. 

Try HomeworkAI and get instant help for your math homework >>>

5. Genius Tutor – Best AI Math Solver for High-Level Learning

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Genius Tutor is a versatile AI tutor and homework helper that can help students build their math skills and gain confidence in their academic abilities.

While the AI math solver is geared toward all types of learners, it is best suited for high school and college-level students.

The AI math problem solver provides a step-by-step breakdown for math questions of all types, showing the exact process of figuring out math problems and concepts.

Genius Tutor also highlights and explains important theorems, formulas, and rules so that students know when and how to use them.

Genius Tutor not only helps students complete math homework assignments in record time but can also help them prepare for exams.

No credit card is needed to try Genius Tutor, and budget-friendly paid subscriptions are available after the free trial.

  • Can help students with all mathematical disciplines.
  • Provides in-depth guides that foster lifelong learning. 
  • Gives instant feedback on a variety of homework questions. 
  • May not provide accurate solutions to highly complex math problems. 

Genius Tutor’s AI math solver can instantly elevate your math learning experience >>>

Mathway – Best AI math solver for solving algebra problems

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Algebra is a complex branch of mathematics that many students struggle with in high school and college.

Mathway offers a sophisticated AI math solver designed to solve algebra homework questions, from word problems to complex mathematical operations that form meaningful expressions.

The math solver AI tool combines an algebra calculator with a conversational chatbot. Simply type in a math problem or upload a photo and get instant step-by-step solutions. 

Mathway also offers AI-driven math problem solvers for other branches of math, such as calculus, statistics, chemistry, and physics.

  • The clean interface is easy to use. 
  • You can upload documents on a computer or mobile device. 
  • It makes it easy to master algebraic concepts. 
  • Additional features require a paid upgrade. 
  • Does not always provide detailed explanations. 

Air Math – Best AI math solver for mobile uses

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Nowadays, many students rely on their smartphones or other mobile devices for homework help. Air Math is a smart AI math solver app available on Apple and Android devices.

Once installed, the Air Math app allows students to snap and solve math homework questions in under three seconds.

The innovative math AI solver can solve everything from geometry questions to word problems.

The 24/7 instant solutions include step-by-step solutions to teach students how to solve the problem independently. 

If you still have problems understanding the explanations, Air Math can connect you with professional math experts worldwide at any time. 

  • Free to use.
  • Offers support on mobile devices.
  • You can ask expert math tutors for additional assistance. 
  • Students can download the Chrome Extension on the web. 
  • The app may not accurately read handwritten math questions. 

StudyMonkey – Best AI math solver for in-depth explanations 

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StudyMonkey is a free AI homework helper that provides academic assistance in many areas, including mathematics.

The powerful AI math solver saves students time and headaches by instantly generating solutions to complex math problems, preventing long homework sessions. 

Type in the math problem, and StudyMonkey provides an accurate answer, detailed explanation, and steps to solve the problem to make it easier to understand.

This platform also retains a history of past questions asked, allowing students to review and revisit solutions anytime, aiding in effective long-term learning. 

  • Can handle math problems from first grade to expert. 
  • Offers a free plan. 
  • Math features are limited.
  • Can not upload images or documents. 
  • You must pay for a subscription to ask more than three questions daily.

Interactive Mathematics – Best AI math solver for comprehensive chat support

Interactive mathematics

Many students are familiar with chatbots, making Interactive Mathematics a popular option for homework help.

The state-of-the-art AI math problem solver claims to be more accurate than ChatGPT and more powerful than a math calculator. Its speed also surpasses human math tutors. 

Using Interactive Mathematics for homework help is also very simple.

You can type in your math question or upload an image, and the tool immediately sets to work, with the added benefit of offering solutions through a chatbot-style conversation that simulates a real-time, interactive math problem-solving session.

  • You can help students improve their grades.
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  • Math Article

Algebraic Expressions Questions

Algebraic Expressions Questions and solutions are provided here to help students of Class 7 to Class 10. As we know, algebraic expressions is one of the most important concepts of mathematics as it deals with the representation of real-life situations mathematically. Also, we can perform various arithmetic operations on algebraic expressions. Let’s understand how to solve various problems related to algebraic expressions and get the practice questions to improve your problem-solving skills.

What is an algebraic expression?

An algebraic expression is formed from variables and constants using different operations. Also, algebraic expressions are made up of terms, and each term is the product of factors. These factors may be numerical or algebraic. Let us consider an example of an algebraic expression.

4x 2 + 3x – 5

Terms: 4x 2 , 3x, 5

Arithmetic operations: +, –

Coefficients: 4, 3

Constant: 5

Variable: x

Learn more about algebraic expressions .

1. Simplify the algebraic expression and write the coefficients:

2x 2 (x + 2) – 3x (x 2 – 3) – 5x(x + 5)

= 2x 3 + 4x 2 – 3x 3 + 9x – 5x 2 – 25x

= 2x 3 – 3x 3 – 5x 2 + 4x 2 + 9x – 25x

= -x 3 – x 2 – 16x

2. Evaluate algebraic expression ax 2 + by 2 – cz for x = 1, y = -1, z = 2, a = -2, b = 1, c = -2:

Given algebraic expression is:

ax 2 + by 2 – cz

Substituting x = 1, y = -1, z = 2, a = -2, b = 1 and c = -2 in the given expression, we get;

ax 2 + by 2 – cz = (-2)(1) 2 + (1)(-1) 2 – (-2)(2)

= -2 + 1 + 4

3. Add 3xy + 5yz – 7xz + 1 and -4xy + 2yz – 2xz + 5xyz + 1.

(3xy + 5yz – 7xz + 1) + (-4xy + 2yz – 2xz + 5xyz + 1)

Let us group the like terms and then add them.

= (3xy – 4xy) + (5yz + 2yz) + (-7xz – 2xz) + 5xyz + (1 + 1)

= -xy + 7yz – 9xz + 5xyz + 2

4. Write the following statements in terms of algebraic expressions.

(i) Add 4 to the product of a number and 7.

(ii) Subtract 4 from the product of a number and 7.

(iii) Add a number to the product of that number and 6.

(iv) Subtract a number from the product of that number and 8.

(v) Add a number to the product of that number and negative 5.

(i) Add 4 to the product of a number and 7 = 7x + 4

(ii) Subtract 4 from the product of a number and 7 = 7x – 4

(iii) Add a number to the product of that number and 6 = 6x + 6

(iv) Subtract a number from the product of that number and 8 = 8x – x

(v) Add a number to the product of that number and negative 5 = -5x + x

5. Simplify the expression: −2a(a + b) − 2a − (a + b)(−2a) − a − 2

−2a(a + b) − 2a − (a + b)(−2a) − a − 2

Now, expand the terms.

-2a 2 – 2ab – 2a – (-2a 2 – 2ab) – a – 2

= -2a 2 – 2ab – 2a + 2a 2 + 2ab – a – 2

Group the like terms and simplify.

= (-2a 2 + 2a 2 ) + (2ab – 2ab) + (-2a – a) – 2

= 0 + 0 – 3a – 2

= -3a – 2

Therefore, −2a(a + b) − 2a − (a + b)(−2a) − a − 2= -3a – 2.

6. Add the following algebraic expressions.

5x 3 + 7 + 6x – 5x 2 , 2x 2 – 8 – 9x, 4x – 2x 2 + 3x 3 , 3x 3 – 9x – x 2 and x – x 2 – x 3 – 4

(5x 3 + 7 + 6x – 5x 2 ) + (2x 2 – 8 – 9x) + (4x – 2x 2 + 3x 3 ) + (3x 3 – 9x – x 2 ) + (x – x 2 – x 3 – 4)

Combining the like terms, we get;

(5x 3 + 3x 3 + 3x 3 – x 3 ) + (– 5x 2 + 2x 2 – 2x 2 – x 2 – x 2 ) + (6x – 9x + 4x – 9x + x) + (7 – 8 – 4)

= 10x 3 – 7x 2 – 7x – 5

7. Subtract the sum of – 3x 3 y 2 + 2x 2 y 3 and – 3x 2 y 3 – 5y 4 from x 4 + x 3 y 2 + x 2 y 3 + y 4 .

Let us find the sum of – 3x 3 y 2 + 2x 2 y 3 and – 3x 2 y 3 – 5y 4 .

i.e., (– 3x 3 y 2 + 2x 2 y 3 ) + (– 3x 2 y 3 – 5y 4 )

= -3x 3 y 2 + (2x 2 y 3 – 3x 2 y 3 ) – 5y 4

= -3x 3 y 2 – x 2 y 3 – 5y 4

Now, subtract the above sum from x 4 + x 3 y 2 + x 2 y 3 + y 4 .

i.e., (x 4 + x 3 y 2 + x 2 y 3 + y 4 ) – (-3x 3 y 2 – x 2 y 3 – 5y 4 )

= x 4 + (x 3 y 2 + 3x 3 y 2 ) + (x 2 y 3 + x 2 y 3 )+ (y 4 + 5y 4 )

= x 4 + 4x 3 y 3 + 2x 2 y 3 + 6y 4

8. What should be added to 3pq + 5p 2 q 2 + p 3 to get p 3 + 2p 2 q 2 + 4pq?

In order to find the required expression, we should subtract 3pq + 5p 2 q 2 + p 3 from p 3 + 2p 2 q 2 + 4pq.

Thus, the required expression will be:

p 3 + 2p 2 q 2 + 4pq − (3pq + 5p 2 q 2 + p 3 )

= p 3 + 2p 2 q 2 + 4pq − 3pq − 5p 2 q 2 − p 3

= p 3 − p 3 + 2p 2 q 2 − 5p 2 q 2 + 4pq − 3pq

= −3p 2 q 2 + pq

Therefore, −3p 2 q 2 + pq to be added to 3pq + 5p 2 q 2 + p 3 , to get p 3 + 2p 2 q 2 + 4pq.

9. Subtract 4.5x 5 – 3.4x 2 + 5.7 from 5x 4 – 3.2x 2 – 7.3x.

Let us subtract 4.5x 5 – 3.4x 2 + 5.7 from 5x 4 – 3.2x 2 – 7.3x.

(5x 4 – 3.2x 2 – 7.3x) – (4.5x 5 – 3.4x 2 + 5.7)

= 5x 4 – 3.2x 2 – 7.3x – 4.5x 5 + 3.4x 2 – 5.7

= 5x 4 + (3.4x 2 – 3.2x 2 ) – 7.3x – 4.5x 5 – 5.7

= 5x 4 + 0.2x 2 – 7.3x – 4.5x 5 – 5.7

= -4.5x 5 + 5x 4 + 0.2x 2 – 7.3x – 5.7

10. Factorise the expression 10x 2 + 5x + 2xy + y.

10x 2 + 5x + 2xy + y

Take the common factors out.

= 5x(2x + 1) + y(2x + 1)

Again, take the common terms out.

= (2x + 1)(5x + y)

Therefore, 10x 2 + 5x + 2xy + y = (2x + 1)(5x + y).

Practice Questions on Algebraic Expressions

  • Find the value of the expression a 2 + 3b 2 + 6ab for a = 1 and b = – 2.
  • Find the number of terms of the expression 3x 2 y – 2y 2 z – z 2 x + + 4xy – 5.
  • Simplify the expression 50x 3 – 21x + 107 + 41x 3 – x + 1 – 93 + 71x – 31x 3 .
  • How much is y 4 – 12y 2 + y + 14 greater than 17y 3 + 34y 2 – 51y + 68?

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Demetrios p. kanoussis ph.d.

Dr. Demetrios P. Kanoussis is a professional Electrical Engineer and Mathematician. He earned his bachelor's degree in Electrical Engineering from National Technical University of Athens (N.T.U.A), Greece, in 1980.His bachelor's thesis was: "Propagation of electromagnetic waves in wave guides with cross section a sector of a circle".

He earned his Master's degree in Mathematics, in 1983, from Tennessee Technological University, (T.T.U), USA. His master's thesis was: " Boundary-Initial value problems in elliptic coordinates".

From the same university, (T.T.U), he earned his Ph.D degree in Engineering, in 1986. His Ph.D dissertation was: "Plasma density distribution of systems with hyperboloids of revolution boundary confinements".

Demetrios P. Kanoussis has a long teaching experience and has taught various courses in the areas of Applied Mathematics and Electrical Engineering. During his stay at T.T.U, he taught several courses in the EE Department, mainly in the areas of Electromagnetic Fields, Physical Electronics and Plasma Engineering.

As a professional electrical engineer, Demetrios P. Kanoussis has been working for several years in the design and the implementation of various projects, mainly in the field of integrated control systems.

Dr. Kanoussis is the author or coauthor of numerous research articles. His original scientific research and contribution has been published, among other places, in various international journals. His original research work can be found, concentrated, in the Research Gate network.

In addition to his professional activities, teaching and research, Dr. Kanoussis is the author of numerous textbooks in Electrical Engineering and Mathematics.

Besides reading math and science books, in his spare time, Demetrios Kanoussis enjoys swimming, forest walking, taking care and pruning his olive trees and producing his own olive oil.

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COMMENTS

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  2. How to Solve an Algebraic Expression: 10 Steps (with Pictures)

    First, move everything that isn't under the radical sign to the other side of the equation: √ (2x+9) = 5. Then, square both sides to remove the radical: (√ (2x+9)) 2 = 5 2 =. 2x + 9 = 25. Now, solve the equation as you normally would by combining the constants and isolating the variable: 2x = 25 - 9 =. 2x = 16.

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    r + 15 = 2s, where r = the number of points Ron scored and s = the number of points Shawn scored. Any two different variables are acceptable. Problem : Translate the following word statement into an algebraic expression or equation: "John's age divided by 3 is equal to John's age minus 14." = j - 14, where j = John's age.

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  21. Algebraic Expressions Questions

    Practice Questions on Algebraic Expressions. Find the value of the expression a 2 + 3b 2 + 6ab for a = 1 and b = - 2. Find the number of terms of the expression 3x 2 y - 2y 2 z - z 2 x + + 4xy - 5. Simplify the expression 50x 3 - 21x + 107 + 41x 3 - x + 1 - 93 + 71x - 31x 3. Add the following expressions: t - t 2 - t 3 ...

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    Section 1.6 : Rational Expressions. For problems 1 - 3 reduce each of the following to lowest terms. x2−6x −7 x2 −10x +21 x 2 − 6 x − 7 x 2 − 10 x + 21 Solution. x2 +6x +9 x2 −9 x 2 + 6 x + 9 x 2 − 9 Solution. 2x2−x −28 20−x −x2 2 x 2 − x − 28 20 − x − x 2 Solution. For problems 4 - 7 perform the indicated ...