GMAT Math : Solving inequalities

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Example Question #31 : Solving Inequalities

Give the solution set of the inequality

$x^{2} + 4x > 5x + 6$

Possible Answers:

$(-\infty, -2) \cup (3, \infty)$

$(-\infty, -3) \cup (2, \infty)$

$(-\infty, -1) \cup (6, \infty)$

$(-\infty, \infty)$

$(-\infty, -6) \cup (1, \infty)$

Correct answer:

$(-\infty, -2) \cup (3, \infty)$

Explanation:

To solve a quadratic inequality, move all expressions to the left first:

$x^{2} + 4x > 5x + 6$

$x^{2} + 4x- 5x - 6 > 5x + 6 - 5x - 6$

$x^{2} -x - 6 > 0$

(x-3)(x+2)> 0

The boundary points of the solution set will be the points at which:

x - 3 =0 : that is, x = 3 , or

x+ 2= 0 : that is, x = -2 .

None of these values will be included in the solution set, since equality is not allowed by the inequality symbol.

Test the intervals

$(-\infty, -2) , (-2, 3), (3, \infty)$

by choosing a value in each interval and testing the truth of the inequality.

$(-\infty, -2)$ : Test x = -3

$x^{2} + 4x > 5x + 6$

$(-3)^{2} + 4 (-3) > 5(-3) + 6$

9-12> -15+ 6

-3 > -9

True; include the interval $(-\infty, -2)$ .

(-2, 3) : Test x = 0

$x^{2} + 4x > 5x + 6$

$(0)^{2} + 4 (0) > 5(0) + 6$

0> 6

False; exclude the interval (-2, 3) .

$(3, \infty)$ : Test x = 4

$x^{2} + 4x > 5x + 6$

$(4)^{2} + 4 (4) > 5(4) + 6$

16+16> 20+6

32> 26

True; include the interval $(3, \infty)$ .

The solution set is $(-\infty, -2) \cup (3, \infty)$ .

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Example Question #32 : Solving Inequalities

Give the solution set of the inequality

$\frac{2}{x}< \frac{3}{x-3}$

Possible Answers:

$(-6, 0) \cup (3, \infty)$

$(- \infty, -6 ) \cup (0, 3)$

$(- \infty, 0) \cup ( 3, \infty)$

$(- \infty, \infty)$

(0, 3)

Correct answer:

$(-6, 0) \cup (3, \infty)$

Explanation:

To solve a rational inequality, move all expressions to the left first:

$\frac{2}{x}< \frac{3}{x-3}$

$\frac{2}{x}- \frac{3}{x-3} < \frac{3}{x-3}- \frac{3}{x-3}$

$\frac{2}{x}- \frac{3}{x-3} <0$

$\frac{2 (x-3)}{x(x-3)}- \frac{3 x}{x(x-3)} <0$

$\frac{2 x-6 }{x(x-3)}- \frac{3 x}{x(x-3)} <0$

$\frac{2 x-6 -3x }{x(x-3)} <0$

$\frac{-x-6 }{x(x-3)} <0$

$\frac{x+6 }{x(x-3)} >0$

The boundary points of the solution set will be the points at which:

x+6 = 0 - that is, x = -6 ;

x= 0 ; and

x-3 = 0 - that is, x = 3 .

None of these values will be included in the solution set, since equality is not allowed by the inequality symbol.

Test the intervals

$(- \infty, -6), (-6, 0), (0, 3),(3, \infty)$

by choosing a value in each interval and testing the truth of the inequality.

$(- \infty, -6)$ - test x= -7

$\frac{2}{x}< \frac{3}{x-3}$

$\frac{2}{-7}< \frac{3}{-7-3}$

$\frac{2}{-7}< \frac{3}{-10}$

$-\frac{2}{7}< -\frac{3}{10}$

False - exclude $(- \infty, -6)$

(-6, 0) - test x = -1

$\frac{2}{x}< \frac{3}{x-3}$

$\frac{2}{-1}< \frac{3}{-1-3}$

$\frac{2}{-1}< \frac{3}{-4}$

$-2<- \frac{3}{4}$

True - include (-6, 0)

(0, 3) - test x = 1

$\frac{2}{x}< \frac{3}{x-3}$

$\frac{2}{1}< \frac{3}{1-3}$

$\frac{2}{1}< \frac{3}{-2}$

$2<- \frac{3}{2}$

False - exclude (0, 3)

$(3, \infty)$ - test x = 4

$\frac{2}{x}< \frac{3}{x-3}$

$\frac{2}{4}< \frac{3}{4-3}$

$\frac{2}{4}< \frac{3}{1}$

$\frac{1}{2} < 3$

True - include $(3, \infty)$

The solution set is $(-6, 0) \cup (3, \infty)$ .

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Example Question #33 : Solving Inequalities

Give the solution set of the inequality

$x^{2} +10x + 7 \le 10x + 32$

Possible Answers:

$\varnothing$

[-5, 5]

$\left \{ -5\right \}$

$(-\infty, -5] \cup [5 , \infty)$

$\left \{ 5\right \}$

Correct answer:

[-5, 5]

Explanation:

The square of a real number must be nonnegative, so this is a true statement regardless of the value of . The solution set is the set of all real numbers $(-\infty, \infty)$

To solve a quadratic inequality, move all expressions to the left first

$x^{2} +10x + 7 \le 10x + 32$

$x^{2} +10x + 7 - 10x - 32 \le 10x + 32 - 10x - 32$

$x^{2} -25 \le 0$

$(x+5)(x-5) \le 0$

The boundary points of the solution set will be the points at which:

x+5 = 0 ; that is, x = -5 ; or,

x-5 = 0 ; that is, x = 5.

These values will be included in the solution set, since equality is allowed by the inequality symbol.

Test the intervals

$(-\infty, -5), (-5, 5), (5, \infty)$ by choosing a value in each interval and testing the truth of the inequality.

$(-\infty, -5)$ : test x = -6

$x^{2} +10x + 7 \le 10x + 32$

$(-6) ^{2} +10 \cdot (-6) + 7 \le 10 \cdot(-6) + 32$

$36 -60 + 7 \le -60 + 32$

$-17 \le -28$

False - exclude this interval

(-5, 5) : test x= 0

$x^{2} +10x + 7 \le 10x + 32$

$0^{2} +10 \cdot 0 + 7 \le 10 \cdot 0 + 32$

$7 \le 32$

True - include this interval

$(5, \infty)$ : test x = 6

$x^{2} +10x + 7 \le 10x + 32$

$6^{2} +10 \cdot 6 + 7 \le 10 \cdot 6 + 32$

$36 +60 + 7 \le 60 + 32$

$103 \le 92$

False - exclude this interval

[-5, 5] is the solution set.

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Example Question #31 : Solving Inequalities

Solve for :

|12 - 5x | < 160

Possible Answers:

$\left ( - \infty , 12 \frac{11}{12}\right ) \cup \left ( 13\frac{3}{4} , \infty \right )$

$\varnothing$

$\left (12 \frac{11}{12}, 13\frac{3}{4} \right )$

$\left ( - \infty , -29 \frac{3}{5} \right ) \cup \left ( 34 \frac{ 2}{5}, \infty \right )$

$\left ( -29 \frac{3}{5}, 34 \frac{ 2}{5} \right )$

Correct answer:

$\left ( -29 \frac{3}{5}, 34 \frac{ 2}{5} \right )$

Explanation:

|12 - 5x | < 160

can be rewritten as the inequality

-160<12 - 5x < 160

-160- 12 <12 - 5x - 12 < 160 - 12

-172 < - 5x < 148

$\frac{-172}{-5} > \frac{- 5x }{-5} > \frac{148}{-5}$ (note the change in direction of the inequality symbols)

$34 \frac{ 2}{5} > x > -29 \frac{3}{5}$

This is the set $\left ( -29 \frac{3}{5}, 34 \frac{ 2}{5} \right )$ .

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Example Question #31 : Solving Inequalities

Solve the following inequality:

$-3+9x+14\leq 6x+1-2x$

Possible Answers:

$x\leq-2$

$x\geq2$

$x\geq \frac{4}{3}$

$x\leq-\frac{4}{3}$

$x\leq1$

Correct answer:

$x\leq-2$

Explanation:

Like any other equation, we solve the inequality by first grouping like terms. Grouping the terms on the left side of the equation and the constants on the right side of the equation, we have:

$-3+9x+14\leq 6x+1-2x$

$9x-6x+2x\leq 1+3-14$

$5x\leq -10$

$x=\frac{-10}{5}$

$x\leq-2$

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GMAT Math : Solving inequalities

Study concepts, example questions & explanations for GMAT Math

All GMAT Math Resources

Example Questions

Example Question #31 : Solving Inequalities

Example Question #32 : Solving Inequalities

Example Question #33 : Solving Inequalities

Example Question #31 : Solving Inequalities

Example Question #31 : Solving Inequalities

All GMAT Math Resources

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