Solved Examples on Compound Linear Inequalities



Samuel Dominic Chukwuemeka (SamDom For Peace) For ACT Students
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For JAMB and CMAT Students
Calculators are not allowed. So, the questions are solved in a way that does not require a calculator.

Solve, Check, and Graph these inequalities.
Write each solution in set notation.
Write each solution in interval notation.

(1.) $-7 \lt 2x + 3 \lt 9$


$-7 \lt 2x + 3 \lt 9$

This is a case of "AND".
It is an inequality in "compact form".
We can solve this one in two ways: Separately or Together.

$ \boldsymbol{1st\:\: way:\:\: Separately} \\[3ex] -7 \lt 2x + 3 \\[3ex] 2x + 3 \gt -7 \\[3ex] 2x \gt -7 - 3 \\[3ex] 2x \gt -10 \\[3ex] x \gt -\dfrac{10}{2} \\[5ex] x \gt -5 \\[3ex] (-5, \infty) \\[3ex] AND \\[3ex] 2x + 3 \lt 9 \\[3ex] 2x \lt 9 - 3 \\[3ex] 2x \lt 6 \\[3ex] x \lt 3 \\[3ex] (-\infty, 3) \\[3ex] $ Because of the "AND", the solution is the intersection of the two solutions.

$ Solution\:\: is\:\: -5 \lt x \lt 3 \\[3ex] Solution\:\: is\:\: (-5, 3) \\[5ex] \boldsymbol{2nd\:\: way:\:\: Together} \\[3ex] -7 \lt 2x + 3 \lt 9 \\[3ex] Subtract\:\: 3\:\: from\:\: each\:\: side \\[3ex] -7 - 3 \lt 2x + 3 - 3 \lt 9 - 3 \\[3ex] -10 \lt 2x \lt 6 \\[3ex] Divide\:\: each\:\: side\:\: by\:\: 2 \\[3ex] -\dfrac{10}{2} \lt \dfrac{2x}{2} \lt \dfrac{6}{2} \\[3ex] -5 \lt x \lt 3 \\[3ex] Set\:\:Notation:\:\: \{x | -5 \lt x \lt 3\} \\[3ex] Interval\:\:Notation:\:\: (-5, 3) \\[3ex] $ To check the solution, we have to check it with only one number.
This is because of the "AND".
That number should satisfy both inequalities.

Check
$ \underline{1st\:\:Part} \\[3ex] -7 \lt 2x + 3 \\[3ex] Solution\:\: is\:\: -5 \lt x \lt 3 \\[3ex] Let\:\: x = 0 \\[3ex] 2x + 3 \\[3ex] = 2(0) + 3 \\[3ex] = 0 + 3 \\[3ex] = 3 \\[3ex] -7 \lt 3 $ $ \underline{2nd\:\:Part} \\[3ex] 2x + 3 \lt 9 \\[3ex] Solution\:\: is\:\: -5 \lt x \lt 3 \\[3ex] Must\:\:use\:\: x = 0 \\[3ex] 2x + 3 \\[3ex] = 2(0) + 3 \\[3ex] = 0 + 3 \\[3ex] = 3 \\[3ex] 3 \lt 9 $
$-7 \lt 3 \lt 9$
(2.) $3x - 1 \lt -5 \:\:\:OR\:\:\: 3x - 1 \ge 5$


$3x - 1 \lt -5 \:\:\:OR\:\:\: 3x - 1 \gt 5$

Let us solve each inequality one at a time.

$ 3x - 1 \lt -5 \\[3ex] 3x \lt -5 + 1 \\[3ex] 3x \lt -4 \\[3ex] x \lt -\dfrac{4}{3} \\[5ex] \left(-\infty, -\dfrac{4}{3}\right) \\[5ex] OR \\[3ex] 3x - 1 \ge 5 \\[3ex] 3x \ge 5 + 1 \\[3ex] 3x \ge 6 \\[3ex] x \ge \dfrac{6}{3} \\[5ex] x \ge 2 \\[3ex] [2, \infty) \\[3ex] Set\:\:Notation:\:\: \left\{x | x \lt -\dfrac{4}{3} \:\:OR\:\: x \ge 2\right\} \\[5ex] Interval\:\:Notation:\:\: \left(-\infty, -\dfrac{4}{3}\right) \cup [2, \infty) \\[5ex] $ To check the solution, we have to check each one to it's own.
This is because of the "OR"

Check
$ \underline{LHS} \\[3ex] 3x - 1 \\[3ex] x \lt -\dfrac{4}{3} \\[5ex] Let\:\: x = -2 \\[3ex] 3(-2) - 1 \\[3ex] -6 - 1 \\[3ex] -7 \\[3ex] OR \\[3ex] 3x - 1 \\[3ex] x \gt 2 \\[3ex] Let\:\: x = 3 \\[3ex] 3(3) - 1 \\[3ex] 9 - 1 \\[3ex] 8 $ $ \underline{RHS} \\[3ex] -5 \\[3ex] OR \\[3ex] 5 $
$-7 \lt -5$
OR
$8 \ge 5$
(3.) $-1 \lt \dfrac{1}{2}(2p + 3) \le 14$


$-1 \lt \dfrac{1}{2}(2p + 4) \le 14$

This is a case of "AND".
It is an inequality in "compact form".
We can solve this one in two ways: Separately or Together.

$ \boldsymbol{1st\:\: way:\:\: Separately} \\[3ex] -1 \lt \dfrac{1}{2}(2p + 4) \\[5ex] \dfrac{1}{2}(2p + 4) \gt -1 \\[5ex] LCD = 2 \\[3ex] Multiply\:\: both\:\: sides\:\: by\:\: 2 \\[3ex] 2 * \dfrac{1}{2}(2p + 4) \gt 2 * -1 \\[5ex] 2p + 4 \gt -2 \\[3ex] 2p \gt -2 - 4 \\[3ex] 2p \gt -6 \\[3ex] p \gt -\dfrac{6}{2} \\[5ex] p \gt -3 \\[3ex] (-3, \infty) \\[3ex] AND \\[3ex] \dfrac{1}{2}(2p + 4) \le 14 \\[5ex] LCD = 2 \\[3ex] Multiply\:\: both\:\: sides\:\: by\:\: 2 \\[3ex] 2 * \dfrac{1}{2}(2p + 4) \le 2 * 14 \\[5ex] 2p + 4 \le 28 \\[3ex] 2p \le 28 - 4 \\[3ex] 2p \le 24 \\[3ex] p \le \dfrac{24}{2} \\[5ex] p \le 12 \\[3ex] (-\infty, 12] \\[3ex] $ Because of the "AND", the solution is the intersection of the two solutions.

$ Set\:\:Notation:\:\: \{p | -3 \lt p \le 12\} \\[3ex] Interval\:\:Notation:\:\: (-3, 12] \\[5ex] \boldsymbol{2nd\:\: way:\:\: Together} \\[3ex] -1 \lt \dfrac{1}{2}(2p + 4) \le 14 \\[5ex] LCD = 2 \\[3ex] Multiply\:\: each\:\: side\:\: by\:\: 2 \\[3ex] 2 * -1 \lt 2 * \dfrac{1}{2}(2p + 4) \le 2 * 14 \\[5ex] -2 \lt 1(2p + 4) \le 28 \\[3ex] -2 \lt 2p + 4 \le 28 \\[3ex] Subtract\:\: 4\:\: from\:\: each\:\: side \\[3ex] -2 - 4 \lt 2p + 4 - 4 \le 28 - 4 \\[3ex] -6 \lt 2p \le 24 \\[3ex] Divide\:\: each\:\: side\:\: by\:\: 2 \\[3ex] -\dfrac{6}{2} \lt \dfrac{2p}{2} \le \dfrac{24}{2} \\[3ex] -3 \lt p \le 12 \\[3ex] Set\:\:Notation:\:\: \{p | -3 \lt p \le 12\} \\[3ex] Interval\:\:Notation:\:\: (-3, 12] \\[3ex] $ To check the solution, we have to check it with only one number.
This is because of the "AND".
That number should satisfy both inequalities.

Check
$ \underline{1st\:\:Part} \\[3ex] -1 \lt \dfrac{1}{2}(2p + 4) \\[5ex] Solution\:\: is\:\: -3 \lt p \le 12 \\[3ex] Let\:\: p = 1 \\[3ex] 2p + 4 \\[3ex] = 2(1) + 4 \\[3ex] = 1 + 4 \\[3ex] = 5 \\[3ex] -1 \lt 5 $ $ \underline{2nd\:\:Part} \\[3ex] \dfrac{1}{2}(2p + 3) \le 14 \\[5ex] Solution\:\: is\:\: -3 \lt p \le 12 \\[3ex] Must\:\:use\:\: p = 1 \\[3ex] 2p + 4 \\[3ex] = 2(1) + 4 \\[3ex] = 1 + 4 \\[3ex] = 5 \\[3ex] 5 \le 14 $
$-1 \lt 5 \le 14$
(4.) WASSCE Solve: $-\dfrac{1}{4} \lt \dfrac{3}{4}(3x - 2) \lt \dfrac{1}{2}$

$ A.\:\: \dfrac{5}{9} \lt x \lt \dfrac{8}{9} \\[5ex] B.\:\: -\dfrac{8}{9} \lt x \lt \dfrac{7}{9} \\[5ex] C.\:\: -\dfrac{8}{9} \lt x \lt \dfrac{5}{9} \\[5ex] D.\:\: -\dfrac{7}{9} \lt x \lt \dfrac{8}{9} \\[5ex] $

$-\dfrac{1}{4} \lt \dfrac{3}{4}(3x - 2) \lt \dfrac{1}{2}$

This is a case of "AND".
It is an inequality in "compact form".
We can solve this one in two ways: Separately or Together.

$ \boldsymbol{1st\:\: way:\:\: Separately} \\[3ex] -\dfrac{1}{4} \lt \dfrac{3}{4}(3x - 2) \\[5ex] \dfrac{3}{4}(3x - 2) \gt -\dfrac{1}{4} \\[5ex] LCD = 4 \\[3ex] Multiply\:\: both\:\: sides\:\: by\:\: 4 \\[3ex] 4 * \dfrac{3}{4}(3x - 2) \gt 4 * -\dfrac{1}{4} \\[5ex] 3(3x - 2) \gt -1(1) \\[3ex] 9x - 6 \gt -1 \\[3ex] 9x \gt -1 + 6 \\[3ex] 9x \gt 5 \\[3ex] x \gt \dfrac{5}{9} \\[5ex] \left(\dfrac{5}{9}, \infty\right) \\[5ex] AND \\[3ex] \dfrac{3}{4}(3x - 2) \lt \dfrac{1}{2} \\[5ex] LCD = 4 \\[3ex] Multiply\:\: both\:\: sides\:\: by\:\: 4 \\[3ex] 4 * \dfrac{3}{4}(3x - 2) \lt 4 * \dfrac{1}{2} \\[5ex] 3(3x - 2) \lt 2(1) \\[3ex] 9x - 6 \lt 2 \\[3ex] 9x \lt 2 + 6 \\[3ex] 9x \lt 8 \\[3ex] x \lt \dfrac{8}{9} \\[5ex] \left(-\infty, \dfrac{8}{9}\right) \\[5ex] $ Because of the "AND", the solution is the intersection of the two solutions.

$ Set\:\:Notation:\:\: \left\{x | \dfrac{5}{9} \lt x \lt \dfrac{8}{9}\right\} \\[5ex] Interval\:\:Notation:\:\: \left(\dfrac{5}{9}, \dfrac{8}{9}\right) \\[5ex] \boldsymbol{2nd\:\: way:\:\: Together} \\[3ex] -\dfrac{1}{4} \lt \dfrac{3}{4}(3x - 2) \lt \dfrac{1}{2} \\[5ex] LCD = 4 \\[3ex] Multiply\:\: each\:\: side\:\: by\:\: 4 \\[3ex] 4 * -\dfrac{1}{4} \lt 4 * \dfrac{3}{4}(3x - 2) \lt 4 * \dfrac{1}{2} \\[5ex] -1 \lt 3(3x - 2) \lt 2(1) \\[3ex] -1 \lt 9x - 6 \lt 2 \\[3ex] Add\:\: 6\:\:to\:\:both\:\:sides \\[3ex] -1 + 6 \lt 9x - 6 + 6 \lt 2 + 6 \\[3ex] 5 \lt 9x \lt 8 \\[3ex] Divide\:\: each\:\: side\:\: by\:\: 9 \\[3ex] \dfrac{5}{9} \lt \dfrac{9x}{9} \lt \dfrac{8}{9} \\[5ex] \dfrac{5}{9} \lt x \lt \dfrac{8}{9} \\[5ex] Set\:\:Notation:\:\: \left\{x | \dfrac{5}{9} \lt x \lt \dfrac{8}{9}\right\} \\[5ex] Interval\:\:Notation:\:\: \left(\dfrac{5}{9}, \dfrac{8}{9}\right) \\[5ex] $ To check the solution, we have to check it with only one number.
This is because of the "AND".
That number should satisfy both inequalities.

Check
$ \underline{1st\:\:Part} \\[3ex] -\dfrac{1}{4} \lt \dfrac{3}{4}(3x - 2) \\[5ex] Solution\:\: is\:\: \dfrac{5}{9} \lt x \lt \dfrac{8}{9} \\[5ex] Let\:\: x = \dfrac{6}{9} \\[5ex] \dfrac{3}{4}(3x - 2) \\[5ex] = \dfrac{9x}{4} - \dfrac{3}{2} \\[5ex] = \dfrac{9}{4} * \dfrac{6}{9} - \dfrac{3}{2} \\[5ex] = \dfrac{6}{4} - \dfrac{3}{2} \\[5ex] = \dfrac{3}{2} - \dfrac{3}{2} \\[5ex] = 0 \\[3ex] -\dfrac{1}{4} \lt 0 $ $ \underline{2nd\:\:Part} \\[3ex] \dfrac{3}{4}(3x - 2) \lt \dfrac{1}{2} \\[5ex] Solution\:\: is\:\: \dfrac{5}{9} \lt x \lt \dfrac{8}{9} \\[5ex] Must\:\:use\:\: x = \dfrac{6}{9} \\[5ex] \dfrac{3}{4}(3x - 2) \\[5ex] = \dfrac{9}{4}x - \dfrac{3}{2} \\[5ex] = \dfrac{9}{4} * \dfrac{6}{9} - \dfrac{3}{2} \\[5ex] = \dfrac{6}{4} - \dfrac{3}{2} \\[5ex] = \dfrac{3}{2} - \dfrac{3}{2} \\[5ex] = 0 \\[3ex] 0 \lt \dfrac{1}{2} $
$-\dfrac{1}{4} \lt 0 \lt \dfrac{1}{2}$