General Mathematics Quarter 1 Module 6: Solving Rational Equations and Inequalities

General Mathematics

Quarter 1 – Module 6: Solving Rational Equations and Inequalities

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CO_Q1_General Mathematics SHS Module 6

General Mathematics Alternative Delivery Mode Quarter 1 – Module 6: Solving Rational Equations and Inequalities First Edition, 2021

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Published by the Department of Education Secretary: Leonor Magtolis Briones Undersecretary: Diosdado M. San Antonio

Development Team of the Module

Writer: January B. Regio

Editors: Elizabeth B. Dizon, Anicia J. Villaruel, and Roy O. Natividad

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General Mathematics Quarter 1 – Module 6: Solving Rational Equations and Inequalities

Senior High School

Introductory Message

This Self-Learning Module (SLM) is prepared so that you, our dear learners, can continue your studies and learn while at home. Activities, questions, directions, exercises, and discussions are carefully stated for you to understand each lesson.

Each SLM is composed of different parts. Each part shall guide you step-by-step as you discover and understand the lesson prepared for you.

Pre-tests are provided to measure your prior knowledge on lessons in each SLM. This will tell you if you need to proceed on completing this module or if you need to ask your facilitator or your teacher’s assistance for better understanding of the lesson. At the end of each module, you need to answer the post-test to self-check your learning. Answer keys are provided for each activity and test. We trust that you will be honest in using these.

In addition to the material in the main text, Notes to the Teacher are also provided to our facilitators and parents for strategies and reminders on how they can best help you on your home-based learning.

Please use this module with care. Do not put unnecessary marks on any part of this SLM. Use a separate sheet of paper in answering the exercises and tests. And read the instructions carefully before performing each task.

If you have any questions in using this SLM or any difficulty in answering the tasks in this module, do not hesitate to consult your teacher or facilitator.

Thank you.

What I Need to Know

This module was designed and written for learners like you to determine a method and set of steps for solving rational equations and inequalities. Learners like you can also explore and develop new methods that you have synthesized and apply these techniques for performing operations with rational expressions.

In this module, you will able to explain the appropriate methods in solving rational equations and inequalities you used. You will also be able to check and explain extraneous solutions.

After going through this module, you are expected to:

1. Apply appropriate methods in solving rational equations and inequalities.
2. Solve rational equations and inequalities using algebraic techniques for simplifying and manipulating of expressions.
3. Determine whether the solutions found are acceptable for the problem by checking the solutions.

CO_Q1_General mathematics SHS Module 6

What I Know

Choose the letter of the best answer. Write the chosen letter on a separate sheet of paper.

1. Which of the following expressions is an equality between two expressions containing one or more variables?
   a. rational function
   b. rational equation
   c. rational inequality
   d. irrational expression

   
   

2. What do you call a root obtained in the process of solving an equation which looks correct but after analyzing it turns out as incorrect?
   a. extraneous solution
   b. rational expression
   c. least common denominator
   d. quotient

   
   

3. What do you call an inequality which involves one or more rational expressions?
   a. rational function
   b. rational equation
   c. rational inequality
   d. irrational expression

   
   

4. What is the usual technique to a solve rational equation?
   a. multiply both sides of the equation by its greatest common factor
   b. multiply both sides of the equation by its least common denominator
   c. multiply both sides of the equation by its inverse factor
   d. multiply both sides of the equation by its greatest common denominator

   
   

For items 5-9: Refer to the rational equation below.

$\frac{x}{3} + \frac{1}{4} = \frac{x}{2}$

5. What is the LCD of the denominator 3, 4 and 2?
   a. 3
   b. 6
   c. 8
   d. 12

CO_Q1_General mathematics SHS Module 6

6. What property will be used if you multiply the LCD on both sides of the equation?
   a. Distributive Property
   b. Associative Property
   c. Commutative Property
   d. Additive Property

   
   

7. What will be the new form of the equation after applying the property and simplifying?
   a. $4x + 3 = 6x$
   b. $3x + 4 = 2x$
   c. $6x + 4 = 3x$
   d. $12x + 3 = 12x$

   
   

8. What will be the solution on the given rational equation?
   a. $\frac{2}{3}$
   b. $\frac{3}{2}$
   c. 2
   d. 3

   
   

9. How will you check if your solution is correct?
   a. by eliminating the rational expressions.
   b. by dividing both sides of the equation by LCD.
   c. by applying Commutative Property.
   d. by substituting the answer or solution in the original equation.

   
   

10. What will you if you obtained a solution that made the expression in the equation undefined?
    a. Accept even if it is untrue value.
    b. Do not reject since it will satisfy the equation in a long run.
    c. Discard the solution since it is unreal.
    d. Continue the solution even if it will give undefined answer.

    
    

11. Which of the following is NOT an inequality sign?
    a. ≤
    b. √
    c. ≥
    d. <

    
    

12. Express the graph of solution set into interval notation.

    
    

[chart: number line; item 12 graph of solution set. Labeled marks from -4 to 4. A label -3 sits above a closed dot at -3, and a label 1 sits above an open dot at 1. The segment between them is shaded.]

a. $\{x \mid -3 \leq x < 1\}$
b. $\{x \mid -3 \leq x \leq 1\}$
c. $\{x \mid 3 < x \leq 1\}$
d. $\{x \mid 3 \leq x < 1\}$

CO_Q1_General mathematics SHS Module 6

13. Below are the steps in solving rational inequality EXCEPT
    a. Put the inequality in general form.
    b. Set the numerator and denominator equal to one and solve.
    c. Plot the critical values on a number line, breaking the number line into intervals and take a test number from each interval by substituting into the original inequality.
    d. Determine if the endpoints of the intervals in the solution should be included in the intervals.

    
    

14. Solve for the solutions of the rational inequality $\frac{(x+3)}{(x-2)} \leq 1$.
    a. [∞, 2)
    b. (∞, 2]
    c. (-∞, 2)
    d. [-∞, 2)

    
    

15. How will you know that the critical points for item no. 14 will satisfy the inequality?
    a. If it makes a true statement, then the interval from which it came from is not in the solution.
    b. If it makes a false statement, then the interval from which it came from is in the solution.
    c. If it makes a true statement, then the interval from which it came from is in the solution.
    d. If it makes a false statement, then the interval from which it came from is either in the solution or not.

    
    

CO_Q1_General mathematics SHS Module 6

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Rational inequality is an inequality which contains one or more rational expressions. It can be used in engineering and production quality assurance as well as in businesses to control inventory, plan production lines, produce pricing models, and for shipping/warehousing goods and materials.

Solving an inequality is much like solving a rational equation except that there are additional steps that focus on illustrating the solution set of an inequality on a number line.

Let us use problem number 2 in the previous activity to discuss and deepen your knowledge and skills in solving rational inequality.

$$\frac{3x + 1}{x – 1} \geq 2$$

Step 1. Put the rational inequality in the general form where > can be replaced by <, ≤ and ≥.

$$\frac{R(x)}{Q(x)} > 0$$

$$\frac{3x + 1}{x – 1} – 2 \geq 0$$

Step 2. Write the inequality into a single rational expression on the left-hand side.

$$\frac{3x + 1 – 2(x – 1)}{x – 1} \geq 0$$

$$\frac{x + 3}{x – 1} \geq 0$$

Note: Remember that one side must always be zero and the other side is always a single fraction, so simplify the fractions if there is more than one fraction.

Step 3. Set the numerator and denominator equal to zero and solve. The values you get are called critical values.

Numerator: $x + 3 = 0$ $x = -3$

Denominator: $x – 1 = 0$ $x = 1$

Step 4. Plot the critical values on a number line, breaking the number line into intervals.

[chart: number line; critical values -3 and 1 marked above the line, with -3 and 1 labeled; integers -4 to 4 labeled below the line; x-axis arrow at right]

Step 5. Substitute critical values to the inequality to determine if the endpoints of the intervals in the solution should be included or not.

$$\frac{3x + 1}{x – 1} \geq 2$$

when $x = -3$

$$\frac{3(-3) + 1}{(-3) – 1} \geq 2$$

$$\frac{-8}{-4} \geq 2$$

$$2 \geq 2 \ ✓ \ (x = -3 \text{ is included in the solution})$$

when $x = 1$

$$\frac{3(1) + 1}{(1) – 1} \geq 2$$

$$\frac{4}{0} \geq 2$$

$$undefined \geq 2 \ ☓ \ (x = 1 \text{ is not included in the solution})$$

See the illustration below.

[chart: number line; critical values -3 (closed circle) and 1 (open circle) labeled above the line; integers -4 to 4 labeled below the line; x-axis arrow at right]

Step 6. Select test values in each interval and substitute those values into the inequality.

$$\frac{3x + 1}{x – 1} \geq 2$$

[chart: number line; integers -8 to 8 labeled below the line; critical values -3 and 1 labeled above the line; three downward arrows mark test points in the intervals]

when $x = -5$

$$\frac{3(-5) + 1}{(-5) – 1} \geq 2$$

$$\frac{-14}{-6} \geq 2$$

$$\frac{7}{3} \text{ or } 2.33 \geq 2 \ (x = -5 \text{ TRUE})$$

when $x = -1$

$$\frac{3(-1) + 1}{(-1) – 1} \geq 2$$

$$\frac{-2}{-2} \geq 2$$

$$1 \geq 2 \ (x = -1 \text{ FALSE})$$

when $x = 3$

$$\frac{3(3) + 1}{(3) – 1} \geq 2$$

$$\frac{10}{2} \geq 2$$

$$5 \geq 2 \ (x = 5 \text{ TRUE})$$

Note: a. If the test value makes the inequality TRUE, then the entire interval is a solution to the inequality. b. If the test value makes the inequality FALSE, then the entire interval is not a solution to the inequality.

Step 7. Use interval notation to write the final answer.

$$(-\infty, -3] \cup (1, \infty)$$

Let’s learn more!

Solve each rational equation and inequality.

1. $\frac{4x + 1}{x + 1} – 3 = \frac{12}{x^2 – 1}$

   
   

2. $\frac{2x – 8}{x – 2} \geq 0$

   
   

Solution:

Rational Equation	$\frac{4x + 1}{x + 1} – 3 = \frac{12}{x^2 – 1}$
1. Find the Least Common Denominator (LCD).	LCD: $(x + 1)(x – 1)$
2. Multiply both sides of the equation by its the LCD.	$(x + 1)(x – 1)\left[\frac{4x + 1}{x + 1} – 3 = \frac{12}{x^2 – 1}\right]$
3. Apply the Distributive Property and then simplify.	$(x – 1)(4x + 1) – 3(x + 1)(x – 1) = 12$ simplify: $4x^2 – 3x – 1 – 3x^2 + 3 = 12$

Rational Equation (table continued from the previous page)	$\frac{4x + 1}{x + 1} – 3 = \frac{12}{x^2 – 1}$
	$x^2 – 3x + 2 = 12$ $x^2 – 3x – 10 = 0$ Factor $(x – 5)(x + 2) = 0$
4. Find all the possible values of x.	$x – 5 = 0$ $x = 5$ $x + 2 = 0$ $x = -2$
5. Check each value by substituting into original equation and reject any extraneous root/s	Checking: $\frac{4x + 1}{x + 1} – 3 = \frac{12}{x^2 – 1}$ when $x = 5$ $\frac{4(5) + 1}{5 + 1} – 3 = \frac{12}{5^2 – 1}$ $\frac{21}{6} – 3 = \frac{12}{24}$ $\frac{3}{6} = \frac{12}{24}$ $\frac{1}{2} = \frac{1}{2}$ ✓ when $x = -2$ $\frac{4(-2) + 1}{(-2) + 1} – 3 = \frac{12}{(-2)^2 – 1}$ $\frac{-7}{-1} – 3 = \frac{12}{3}$ $4 = 4$ ✓ Note: No extraneous root

Rational Inequality	$\frac{2x – 8}{x – 2} \geq 0$
1. Put the rational inequality in general form. $\frac{R(x)}{Q(x)} > 0$ where > can be replaced by <, ≤ and ≥	This inequality is already in general form. We are all set to go.
2. Write the inequality into a single rational expression on the left side. (You can refer to the review section for solving unlike denominators)	This inequality is already in a single rational expression wherein 0 is on one side. $\frac{2x – 8}{x – 2} \geq 0$
3. Set the numerator and denominator equal to zero and solve. The values you get are called critical values.	Numerator: $2x – 8 = 0$ $2x = 8$ $x = 4$

Rational Inequality (table continued from the previous page)	$\frac{2x – 8}{x – 2} \geq 0$
	Denominator: $x – 2 = 0$ $x = 2$
4. Plot the critical values on a number line, breaking the number line into intervals.	[chart: number line; critical values 2 and 4 labeled above the line; integers 2, 3, and 4 labeled below the line]
5. Substitute critical values to the inequality to determine if the endpoints of the intervals in the solution should be included or not.	$\frac{2x – 8}{x – 2} \geq 0$ when $x = 2$ $\frac{2(2) – 8}{2 – 2} \geq 0$ $\frac{-4}{0} \geq 0$ $undefined \geq 0$ ☓ ($x = 2$ is not included in the solution) when $x = 4$ $\frac{2(4) – 8}{4 – 2} \geq 0$ $\frac{0}{2} \geq 0$ $0 \geq 0$ ✓ ($x = 4$ is included in the solution)
6. Select test values in each interval and substitute those values into the inequality. Note: If the test value makes the inequality true, then the entire interval is a solution to the inequality. If the test value makes the inequality false, then the entire interval is not a solution to the inequality.	[chart: number line; integers 2, 3, and 4 labeled below the line; critical values 2 (open circle) and 4 (closed circle) labeled above; arrows mark test points in each interval] $\frac{2x – 8}{x – 2} \geq 0$ when $x = 1$ $\frac{2(1) – 8}{1 – 2} \geq 0$ $\frac{-6}{-1} \geq 0$ $6 \geq 0$ ✓ $true$ when $x = 3$ $\frac{2(3) – 8}{3 – 2} \geq 0$ $\frac{-2}{1} \geq 0$ $-2 \geq 0$ ☓ $false$ when $x = 5$

Rational Inequality (table continued from the previous page)	$\frac{2x – 8}{x – 2} \geq 0$
	$\frac{2(5) – 8}{5 – 2} \geq 0$ $\frac{2}{3} \geq 0$ ✓ $true$
7. Use interval notation or set notation to write the final answer.	$(-\infty, 2) \cup [4, \infty)$

What’s More

Activity 1.

Solve the following rational equations and inequalities using the guided procedure on the table below.

Rational Equation	$\frac{x – 2}{x^2 – 4} + \frac{1}{x + 2} = \frac{1}{x – 2}$
1. Find the Least Common Denominator (LCD).
2. Multiply both sides of the equation by its the LCD.
3. Apply the Distributive Property and then simplify.
4. Find all the possible values of x.	$x = 6$
5. Check each value by substituting into the original equation and reject any extraneous root/s

Rational Equation	$\frac{x – 6}{x^2 – 4x – 12} + \frac{2}{x + 2} = \frac{1}{x – 6}$
1. Find the Least Common Denominator (LCD).
2. Multiply both sides of the equation by its the LCD.
3. Apply the Distributive Property and then simplify.
4. Find all the possible values of x.	$x = 10$
5. Check each value by substituting into original equation and reject any extraneous root/s

Rational Inequality	$\frac{2(x – 4)}{x} < -4$
1. Put the rational inequality in general form. $\frac{R(x)}{Q(x)} > 0$ where > can be replaced by <, ≤ and ≥
2. Write the inequality into a single rational expression on the left side. (You can refer to the review section for solving unlike denominators)
3. Set the numerator and denominator equal to zero and solve. The values you get are called critical values.
4. Plot the critical values on a number line, breaking the number line into intervals.
5. Substitute critical values to the inequality to determine if the endpoints of the intervals in the solution should be included or not.

Rational Inequality (table continued from the previous page)	$\frac{2(x – 4)}{x} < -4$
6. Select test values in each interval and substitute those values into the inequality. Note: If the test value makes the inequality true, then the entire interval is a solution to the inequality. If the test value makes the inequality false, then the entire interval is not a solution to the inequality.
7. Use interval notation or set notation to write the final answer.	$(0, \frac{3}{4})$

Rational Inequality	$\frac{x^2 + x – 6}{x^2 – 3x – 4} \leq 0$
1. Put the rational inequality in general form. $\frac{R(x)}{Q(x)} > 0$ where > can be replaced by <, ≤ and ≥
2. Write the inequality into a single rational expression on the left side. (You can refer to the review section for solving unlike denominators)
3. Set the numerator and denominator equal to zero and solve. The values you get are called critical values.
4. Plot the critical values on a number line, breaking the number line into intervals.
5. Substitute critical values to the inequality to determine if the endpoints of the intervals in the solution should be included or not.
6. Select test values in each interval and substitute those values into the inequality. Note: If the test value makes the inequality true, then the entire interval is a solution to the inequality. If the test value makes the inequality false, then the entire interval is not a solution to the inequality.
7. Use interval notation or set notation to write the final answer.	$[-3, -1) \cup [2, 4)$

Activity 2

Solve each problem below and choose the letter that corresponds to the solution to each problem. Place the correct answer in the corresponding lines.

What did the bible verses 1John 4:7-21 is all about?

---

1 2 3 4 5 6 7 8 9

1. (x+2)/3 = (2x−4)/2
2. 7/(4x) − 3/x² = 1/(2x²)
3. 2x/(x+1) + 5/(2x) = 2
4. (x²−1)/(x−3) = 8/(x−3)
5. 1/(x−6) + x/(x−2) = 4/(x²−8x+12)
6. 5x/(x−1) < 4
7. x/(x−2) − 7 = 2/(x−2)
8. (x²+x−12)/(x−1) ≤ 0
9. (3x+1)/(x−2) ≥ 5

A. -3 C. -1 and 6 D. -5 E. (2, 11/2] G. 4 I. 3 L. (-4, 1) N. -3 and 3 O. 2 S. -1 V. (-∞, -4] ∪ (1, 3] Y. (-∞, -4) ∪ [1, 3)

What I Have Learned

Complete the following statements by writing the correct word or words and formulas.

1. A _____________________ is an equation containing at least one rational expression with a polynomial in the numerator and denominator.

   
   

2. To determine if the endpoints of the intervals in the solution should be included or not you need to _____________________ the critical values to the inequality.

   
   

3. In order to get the critical values you need to set _______________________ and _______________________ equal to zero.

   
   

4. The first step in solving rational inequality is to put the inequality in general form where in one side must always be _________________________ and the other side is in a _______________________ fraction.

   
   

5. If the test value makes the inequality _______________________, then the entire interval is a solution to the inequality.

   
   

Topics for Discussion

1. Explain the similarities and differences in solving between rational equations and inequalities?
2. Why do extraneous solutions sometimes occur and don’t work in the original form of the equation?

What I Can Do

Read the situation carefully and solve the problem.

The new COVID-19 testing facility in Lucena City is operating with two laboratory technicians. Technician A takes 2 hours to finish 50 samples of specimens from CoVID-19 patients. Technician B takes 3 hours to finish 45 samples of specimens from COVID-19 patients. Working together, how long should it take them to finish 150 samples of specimens from COVID-19 patients?

Hint: Think about how many samples of specimens each technician can finish in one hour. This is their testing rate.

Assessment

Choose the letter of the best answer. Write the chosen letter on a separate sheet of paper.

1. It is an equation containing at least one fraction whose numerator and denominator are polynomials.
   a. rational function
   b. rational equation
   c. rational inequality
   d. irrational equation

   
   

2. The usual technique to eliminate denominator in solving a rational equation is to multiply both sides of the equation by its _________________.
   a. inverse factor
   b. greatest common factor
   c. least common denominator
   d. greatest common denominator

   
   

3. An inequality which involves one or more rational expressions is called _________________.
   a. rational function
   b. rational equation
   c. rational inequality
   d. irrational equation

   
   

4. You can only use cross multiplication in solving rational equation if and only if you have one fraction equal to one fraction, that is, if the fractions are _________________________.
   a. negative
   b. positive
   c. inequal
   d. proportional

   
   

5. If the test value makes the inequality _________________________, then the entire interval is not a solution to the inequality.
   a. true
   b. false
   c. proportional
   d. reciprocal

   
   

For items 6-7: Refer to the rational equation below.

5/(2x − 4) + 2/(x + 3) = 3/(x − 2)

6. To solve the equation, we multiply both sides by
   a. x − 2
   b. x + 3
   c. (x + 2)(x − 3)
   d. (x − 2)(x + 3)

   
   

7. Which of the following will be the solution to the given rational equation?
   a. 11/3
   b. 3/11
   c. −11/3
   d. −3/11

   
   

For items 8-10: Refer to the rational inequality below.

(x+12)/(x+2) ≤ 2

8. What are the critical values in the given rational inequality?
   a. x = −2 and 8
   b. x = −2 and −8
   c. x = −2 and 12
   d. x = −2 and −12

   
   

9. Which of the critical value or values is/are included as endpoints of the intervals?
   a. −2
   b. 2
   c. −8
   d. 8

   
   

10. Which of the following is the solution in the given inequality?
    a. (−∞, −2) ∪ (8, ∞)
    b. (−∞, −2] ∪ [8, ∞)
    c. (−∞, −2) ∪ [8, ∞)
    d. (−∞, −2) ∪ [−8, ∞)

    
    

For items 11-13, solve for the solutions of the following rational equations.

11. 2/(x+2) + 1/(x−2) = 3/x
    a. 6
    b. −6
    c. 8
    d. −8

    
    

12. 8/x² + 1 = 9/x
    a. −1 and −8
    b. 1 and 8
    c. −1 and 8
    d. 1 and −8

    
    

13. 1/x² − 16 = 0
    a. ±1
    b. ±2
    c. ±1/2
    d. ±1/4

    
    

For items 14-15, solve for the solutions of the following rational equations.

14. 5/(x−3) > 3/(x+1)
    a. (−∞, −7) ∪ (−1, ∞)
    b. (−∞, −1) ∪ (3, ∞)
    c. (−7, −1) ∪ (3, ∞)
    d. (−7, −1] ∪ [3, ∞)

    
    

15. (x−3)(x+2)/(x−1) ≤ 0
    a. (−∞, −2) ∪ (1,3]
    b. (−∞, −2] ∪ (1,3]
    c. (−∞, −2] ∪ [1,3)
    d. (−∞, −2) ∪ [1,3)

    
    

Additional Activities

Practice Worksheet: Solving Rational Equations and Inequalities

Solve each equation. Check extraneous solutions for rational equations. Write your answer in interval notation for rational inequalities.

LEVEL 1
1. 8/(x+1) = 4/3	2. 2x + 3 = x/4
3. (x−4)/(x+5) ≤ 0	4. (x+3)/(3x−6) > 0
5. 4/x + 1/(3x) = 9	6. 1/(x²−4) ≤ 0

LEVEL 2
7. 20/x − 20/(x−2) = 4/x	8. 2/(x²−x) = 1/(x−1)
9. (x−9)/(3x+2) ≥ 3	10. (x+32)/(x+6) ≤ 6
11. 4/x + 1/x² = 1/(5x²)	12. 1 + 2/(x+1) < 2/x

LEVEL 3
13. 3x/(x+1) = 12/(x²−1) + 2	14. 2/(4x²−9) + 1/(2x−3) = 3/(2x+3)
15. (x²+1)(x−2)/((x−1)(x+1)) ≥ 0	16. (x+7)(x−3)/(x−5)² > 0
17. (x−3)/(2x+10) + 2x − 12 = (x²+3x−18)/(2x+10)	18. (12x³+16x²−3x−4)/(8x³+12x²+10x+15) < 0

Answer Key

[transcriber’s note: this page is printed upside down; the boxes are transcribed in upright reading order, top row first, then bottom row]

What I Know

1. B
2. A
3. C
4. B
5. D
6. A
7. A
8. B
9. D
10. C
11. B
12. A
13. B
14. D
15. C

What’s More

1. G
2. O
3. D
4. I
5. S
6. L
7. O
8. V
9. E

What I Have Learned

1. Extraneous solutions
2. Substitute
3. Numerator, denominator
4. zero, single
5. true

What I Can Do

It will take 3 hours and 45 minutes for the two laboratory technicians to finish 150 samples of specimens.

Assessment

1. B
2. C
3. C
4. D
5. B
6. D
7. A
8. A
9. D
10. C
11. A
12. B
13. D
14. C
15. A

Additional Activities

1. x = 5
2. x = −12/7
3. (−5, 4]
4. (−∞, −3) ∪ (3, ∞)
5. x = 13/27 [?]
6. (−2, 2)
7. x = −8
8. x = 2
9. [−13/5, −3/2) [?]
10. (−∞, −6) ∪ [−4/5, ∞)
11. x = −1/5
12. (−2, −1) ∪ (0, 1)
13. x = −2 and 5
14. x = 7/2
15. (−1, 1) ∪ [2, ∞)
16. (−∞, −7) ∪ (3, 5) ∪ (5, ∞)
17. x = 7
18. (−3/2, −4/3) ∪ (−1/2, 1/2) [?]

References

Aunzo, Rodulfo, Flores Maricar, Gagani Ray Ferdinand M, and Quennie Ypanto. 2016. General Mathematics Activity-based, Scaffolding of Student . Quezon City: C&E Publishing, Inc.

2016. General Mathematics Learner’s Material . Meralco Avenue, Pasig City, Philippines 1600: Lexicon Press Inc.

      
      

2017. General Mathematics Teacher’s Guide. Meralco Avenue, Pasig City, Philippines 1600: Lexicon Press Inc.

      
      

Oronce, Orlando A. 2016. General Mathematics. Sampaloc, Manila: Rex Bookstore, Inc.

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