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Mathematics LibreTexts

9.2: Rational Inequalities

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  • Darlene Diaz
  • Santiago Canyon College via ASCCC Open Educational Resources Initiative

In earlier chapters, we discussed many types of inequalities, and one of the most critical parts of inequalities is we know the solutions are always intervals of numbers. Hence, we now discuss rational inequalities where the solutions are intervals, but we take one critical detail into account. Since rational inequalities and equations contain excluded values, we must take this into consideration when obtaining the solution. In fact, we use the excluded values and \(x\)-intercepts, if any, to determine the intervals in which make the inequality true. Again, with inequalities, we are trying to find intervals of numbers in which the statements is true.

Steps for solving rational inequalities

Step 1. Rewrite the inequality so that only zero is on the right side.

Step 2. Determine where the rational expression is undefined or equals zero .

Step 3. Graph the values found in Step 2. on a number line split into intervals.

Step 4. Take a test point within each interval and determine the sign of the result.

Step 5. Determine the solution, where the solution is the interval in which makes the inequality true.

Be careful when determining the intervals. Recall, where the expression is undefined is not included in the interval, i.e., always use parenthesis for the values in which the expression is undefined.

Example 9.2.1

Solve \(\dfrac{x-3}{x+1}>0\).

Step 1. Rewrite the inequality so that only zero is on the right side. Since \(\dfrac{x-3}{x+1}>0\) already has zero on the right side, this step is done.

Step 2. Determine where the rational expression is undefined or equals zero . To obtain where the rational expression equals zero , we set the numerator equal to zero: \[\begin{aligned} x − 3 &= 0\\ x &= 3\end{aligned}\] To obtain where the expression is undefined , we find its excluded value(s) by setting the denominator equal to zero: \[\begin{aligned}x + 1 &= 0\\ x &= −1\end{aligned}\]

Step 3. Graph the values found in Step 2. on a number line into split into intervals. Label \(-1\) and \(3\) on a blank real number line:

clipboard_ec35505fdab89a647162909a8a0d5c757.png

Step 4. Take a test point within each interval and determine the sign of the result. We take test values on each side of \(−1\) and \(3\). Let’s choose fairly easy numbers such as \(−2,\: 0,\) and \(4\). We plug-n-chug these numbers into \(\dfrac{x − 3}{x + 1}\) and determine whether the value is positive or negative:

clipboard_e2e93ebff9b9650706a48ab917a75d2cc.png

\[\begin{array}{rl} \text{letting }x=-2\Longrightarrow &\dfrac{-2-3}{-2+1}=5>0 \\ \text{letting }x=0\Longrightarrow &\dfrac{0-3}{0+1}=-3<0 \\ \text{letting }x=4\Longrightarrow &\dfrac{4-3}{4+1}=0.2>0\end{array}\nonumber\]

Step 5. Determine the solution, where the solution is the interval in which makes the inequality true. Since \(\dfrac{x − 3}{x + 1} > 0\) (from Step 1. ), then we are looking for where the test values are positive. Looking at the number line above, we see these are the values to the left of \(−1\) and to the right of \(3\). Thus, the solution is \((−∞, −1) ∪ (3, ∞)\).

We always use a parenthesis for the value that is excluded since this value makes the rational expression undefined. Using a bracket or parenthesis for where the expression equals zero is determined by the original inequality sign, e.g., \(<, >\) use parenthesis, and \(≥, ≤\) use brackets.

Example 9.2.2

Solve \(\dfrac{2x+3}{x-2}\leq 1\).

Step 1. Rewrite the inequality so that only zero is on the right side. We rewrite \(\dfrac{2x+3}{x-2}\leq 1\) so that there is zero on the right side, and as one fraction: \[\begin{array}{rl}\dfrac{2x+3}{x-2}\leq 1&\text{Subtract }1\text{ from each side} \\ \dfrac{2x+3}{x-2}-1\leq 0&\text{Rewrite as one fraction where LCD: }(x-2) \\ \dfrac{2x+3}{x-2}-\dfrac{x-2}{x-2}\leq 0&\text{Subtract across numerators} \\ \dfrac{2x+3-x+2}{x-2}\leq 0&\text{Simplify} \\ \dfrac{x+5}{x-2}\leq 0&\text{We use this inequality to obtain the solution} \end{array}\nonumber\]

Step 2. Determine where the rational expression is undefined or equals zero . To obtain where the rational expression equals zero , we set the numerator equal to zero: \[\begin{aligned} x + 5 &= 0\\ x &= −5\end{aligned}\] To obtain the excluded values, we set the denominator equal to zero: \[\begin{aligned}x − 2 &= 0 \\x &= 2\end{aligned}\]

Step 3. Graph the values found in Step 2. on a number line into split into intervals. Label \(−5\) and \(2\) on a blank number line:

clipboard_ebf94cd54672d729194844a616f565515.png

Step 4. Take a test point within each interval and determine the sign of the result. We take test values on each side of \(−5\) and \(2\). Let’s choose fairly easy numbers such as \(−6,\: 0,\) and \(3\). We plug these numbers into \(\dfrac{x + 5}{x − 2}\) and determine whether the value is positive or negative:

clipboard_e189fef5cb9eebad0ec7571393a055829.png

\[\begin{array}{rl} \text{letting }x=-6\Longrightarrow &\dfrac{-6+5}{-6-2}=\dfrac{1}{8}>0 \\ \text{letting }x=0\Longrightarrow &\dfrac{0+5}{0-2}=-\dfrac{5}{2}<0 \\ \text{letting }x=3\Longrightarrow &\dfrac{3+5}{3-2}=8>0\end{array}\nonumber\]

Step 5. Determine the solution, where the solution is the interval in which makes the inequality true. Since \(\dfrac{x + 5}{x − 2} ≤ 0\) (from Step 1. ), then we are looking for where the test values are negative or equal to zero. Looking at the number line above, we see these are the values between \(−5\) and \(2\). Thus, the solution is \([−5, 2)\).

We used a bracket on \(−5\) since the original inequality was \(≤\) and a parenthesis on \(2\) since \(2\) was an excluded value.

Rational Inequalities Homework

Solve the rational inequalities. Write the solution in interval notation.

Exercise 9.2.1

\(\dfrac{x-3}{x+1}<0\)

Exercise 9.2.2

\(\dfrac{x-7}{x+6}>0\)

Exercise 9.2.3

\(\dfrac{x-3}{x+1}<1\)

Exercise 9.2.4

\(\dfrac{x+27}{x+4}<9\)

Exercise 9.2.5

\(x+\dfrac{45}{x}<14\)

Exercise 9.2.6

\(\dfrac{(x-9)(x+9)}{x}\geq 0\)

Exercise 9.2.7

\(\dfrac{(x+12)(x-4)}{x-1}\geq 0\)

Exercise 9.2.8

\(\dfrac{3x}{7-x}<x\)

Exercise 9.2.9

\(\dfrac{4x}{3-x}\geq 4x\)

Exercise 9.2.10

\(\dfrac{8}{x-3}>\dfrac{6}{x-1}\)

Find the values of \(x\) that satisfy the given condition for the function.

Exercise 9.2.11

Solve \(R(x)\geq 0\) if \(R(x)=\dfrac{x+6}{x-4}\).

Exercise 9.2.12

Solve \(R(x)\leq 0\) if \(R(x)=\dfrac{x-4}{x+8}\).

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of 1 pt 1013 (3 complete) ▼ Score: 66 67%, 2 of 3 pt 24)9.1.3 Question Help *

The distribution of the scores on a certain exam is N(60,5), which means that the exam scores are Normally distributed with a mean of 60 and standard

MyStatLab questions

a. Sketch the curve and label, on the x-axis, the position of the mean, the mean plus or minus one standard deviation, the mean plus or minus two standard deviations and the mean plus or minus three standard deviations

b. Find the probability that a randomly selected score will be less than 55.

Shade the region under the Normal curve whose area corresponds to this probability of 5 C.

The correct option is: C

The population mean is μ = 60%

The population standard deviation is σ = 5

Construct the graph which indicates the mean and standard deviation.

The normal distribution is a continuous probability distribution that is symmetrical on both sides of the mean and any particular normal distribution is completely specified by the numbers mean and standard deviation.

The empirical rule states that for a normal distribution, nearly all of the data will fall within three standard deviation of the mean that is (68%, 95% and 99.7%)

Fundamentals

The Empirical rule for data sets having a distribution that is approximately bell shaped and the following properties can apply,

About 68% of all values fall within 1 standard deviation of the mean

About 95% of all values fall within 2 standard deviation of the mean

About 99.7% of all values fall within 3 standard deviation of the mean

MyStatLab answers

The correct graph is,

The empirical rule assumption the normal distribution mean indicates the center line of the graph.

b. What is the probability that a randomly selected score will be less than 557 Using the Empirical Rule, the probability that a randomly selected score will be less than 55 is about Type an integer or a decimal.)

Enter your answer in the answer box and then click Check Answer Clear Al part

Calculate the probability that a randomly selected score will be less than 55.

According to the Empirical rule the 68% of the area is fall in the between one standard deviation.

Un shaded area = 1- Shaded area

Now, calculate the probability that less than 55.

P =(Lessthan55)

= 0.3/2​=0.16​

The probability that a randomly selected score will be less than 55 is 0.16.

The correct graph is

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Home > CC2 > Chapter 9 > Lesson 9.2.1

Lesson 9.1.1, lesson 9.1.2, lesson 9.1.3, lesson 9.2.1, lesson 9.2.2, lesson 9.2.3, lesson 9.2.4.

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  • NCERT Solutions
  • NCERT Class 9
  • NCERT 9 Maths
  • Chapter 9: Areas Of Parallelograms And Triangle
  • Exercise 9.2

NCERT Solutions for Class 9 Maths Chapter 9 - Area of Parallelograms and Triangles Exercise 9.2

* According to the latest update on NCERT Syllabus 2023-24, this chapter has been removed.

Section 9.3 of the textbook, under which Exercise 9.2 appears, deals with the concept of “Parallelograms on the same Base and Between the same Parallels.” It also explains the “Theorem 9.1: Parallelograms on the same base and between the same parallels are equal in area” with proof. The questions and exercises given in this section of NCERT Solutions try to explain this concept to you with the help of simple examples.

These questions from the textbook are designed as per the NCERT syllabus and guidelines. Subject experts have designed the NCERT Solutions for Class 9 Maths Chapter 9 – Area of Parallelograms and Triangles with the aim to help students understand the concepts well and score high marks.

NCERT Solutions for Class 9 Maths Chapter 9 – Area of Parallelograms and Triangles Exercise 9.2

Access answers to ncert class 9 maths chapter 9 – area of parallelograms and triangles exercise 9.2.

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Access Other Exercise Solutions of Class 9 Maths Chapter 9 Area of Parallelograms and Triangles

Exercise 9.1 Solutions 1 Short Type Answer with Reasoning

Exercise 9.3 Solutions 12 Short Type Answers and 4 Long Type Answers

Exercise 9.4 Solutions 4 Short Type Answers, 1 Long Type Answer and 3 Very Long Type Answers

NCERT Solutions for Class 9 Maths Chapter 9 Area of Parallelograms and Triangles Exercise 9.2

1. In Fig. 9.15, ABCD is a parallelogram, AE ⊥ DC and CF ⊥ AD. If AB = 16 cm, AE = 8 cm and CF = 10 cm, find AD.

Ncert solutions class 9 chapter 9-2

AB = CD = 16 cm (Opposite sides of a parallelogram)

CF = 10 cm and AE = 8 cm

Area of parallelogram = Base × Altitude

= CD×AE = AD×CF

⇒ 16×8 = AD×10

⇒ AD = 128/10 cm

⇒ AD = 12.8 cm

2. If E, F, G and H are respectively the mid-points of the sides of a parallelogram ABCD, show that ar (EFGH) = 1/2 ar(ABCD).

Ncert solutions class 9 chapter 9-3

E, F, G and H are the mid-points of the sides of a parallelogram ABCD, respectively.

ar (EFGH) = ½ ar(ABCD)

Construction,

H and F are joined.

AD || BC and AD = BC (Opposite sides of a parallelogram)

⇒ ½ AD = ½ BC

AH || BF and and DH || CF

⇒ AH = BF and DH = CF (H and F are mid points)

∴, ABFH and HFCD are parallelograms.

We know that, ΔEFH and parallelogram ABFH, both lie on the same FH the common base and in-between the same parallel lines AB and HF.

∴ area of EFH = ½ area of ABFH — (i)

And, area of GHF = ½ area of HFCD — (ii)

Adding (i) and (ii),

area of ΔEFH + area of ΔGHF = ½ area of ABFH + ½ area of HFCD

⇒ area of EFGH = area of ABFH

∴ ar (EFGH) = ½ ar(ABCD)

3. P and Q are any two points lying on the sides DC and AD, respectively of a parallelogram ABCD.

Show that ar(APB) = ar(BQC).

Ncert solutions class 9 chapter 9-4

ΔAPB and parallelogram ABCD lie on the same base AB and in-between same parallel AB and DC.

ar(ΔAPB) = ½ ar(parallelogram ABCD) — (i)

ar(ΔBQC) = ½ ar(parallelogram ABCD) — (ii)

From (i) and (ii), we have

ar(ΔAPB) = ar(ΔBQC)

4. In Fig. 9.16, P is a point in the interior of a parallelogram ABCD. Show that

(i) ar(APB) + ar(PCD) = ½ ar(ABCD)

(ii) ar(APD) + ar(PBC) = ar(APB) + ar(PCD)

[Hint : Through P, draw a line parallel to AB.]

Ncert solutions class 9 chapter 9-5

(i) A line GH is drawn parallel to AB passing through P.

In a parallelogram,

AB || GH (by construction) — (i)

AD || BC ⇒ AG || BH — (ii)

From equations (i) and (ii),

ABHG is a parallelogram.

ΔAPB and parallelogram ABHG are lying on the same base AB and in-between the same parallel lines AB and GH.

∴ ar(ΔAPB) = ½ ar(ABHG) — (iii)

ΔPCD and parallelogram CDGH are lying on the same base CD and in-between the same parallel lines CD and GH.

∴ ar(ΔPCD) = ½ ar(CDGH) — (iv)

Adding equations (iii) and (iv),

ar(ΔAPB) + ar(ΔPCD) = ½ [ar(ABHG)+ar(CDGH)]

⇒ ar(APB)+ ar(PCD) = ½ ar(ABCD)

(ii) A line EF is drawn parallel to AD passing through P.

In the parallelogram,

AD || EF (by construction) — (i)

AB || CD ⇒ AE || DF — (ii)

AEDF is a parallelogram.

ΔAPD and parallelogram AEFD are lying on the same base AD and in-between the same parallel lines AD and EF.

∴ar(ΔAPD) = ½ ar(AEFD) — (iii)

ΔPBC and parallelogram BCFE are lying on the same base BC and in-between the same parallel lines BC and EF.

∴ar(ΔPBC) = ½ ar(BCFE) — (iv)

ar(ΔAPD)+ ar(ΔPBC) = ½ {ar(AEFD)+ar(BCFE)}

⇒ar(APD)+ar(PBC) = ar(APB)+ar(PCD)

5. In Fig. 9.17, PQRS and ABRS are parallelograms and X is any point on side BR. Show that (i) ar (PQRS) = ar (ABRS) (ii) ar (AXS) = ½ ar (PQRS) 

Ncert solutions class 9 chapter 9-7

(i) Parallelogram PQRS and ABRS lie on the same base SR and in-between the same parallel lines SR and PB.

∴ ar(PQRS) = ar(ABRS) — (i)

(ii) ΔAXS and parallelogram ABRS are lying on the same base AS and in-between the same parallel lines AS and BR.

∴ ar(ΔAXS) = ½ ar(ABRS) — (ii)

From (i) and (ii), we find that,

ar(ΔAXS) = ½ ar(PQRS)

6. A farmer had a field in the form of a parallelogram PQRS. She took any point A on RS and joined it to points P and Q. In how many parts is the fields divided? What are the shapes of these parts? The farmer wants to sow wheat and pulses in equal portions of the field separately. How should she do it?

Ncert solutions class 9 chapter 9-8

The field is divided into three parts each in triangular shape.

Let, ΔPSA, ΔPAQ and ΔQAR be the triangles.

Area of (ΔPSA + ΔPAQ + ΔQAR) = Area of PQRS — (i)

Area of ΔPAQ = ½ area of PQRS — (ii)

Here, the triangle and parallelogram are on the same base and in-between the same parallel lines.

From (i) and (ii),

Area of ΔPSA +Area of ΔQAR = ½ area of PQRS — (iii)

From (ii) and (iii), we can conclude that,

The farmer must sow wheat or pulses in ΔPAQ or either in both ΔPSA and ΔQAR.

NCERT Solutions for Class 9 Maths covers all the questions from Exercise 9.2 of the chapter “Area of Parallelograms and Triangles” from the textbook. Exercise 9.2 contains a total of 6 questions. Of this, 5 are short answer questions, and 1 is a long answer question. In this exercise, the questions may require you to prove or provide the answer to the questions as specified.

Meanwhile, here we also give some benefits to solving these exercises:

  • Find it easier to face the exams
  • Gain more practice solving questions and be more confident
  • Learn the concept more easily
  • Understand the topic thoroughly

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