Question

solve the inequality. graph the solution, if possible. 3. $6m \leq -42$ 4. $\frac{x}{2} - 5 \geq -9$ 5. $4p + 3 \geq 2(2p + 1)$ 6. $-2 \leq 5 - 3a \leq 13$ 7. $-7.5 < 2 - h$ or $6h + 5 > 71$ 8. $4|-3b + 5| - 9 < 7$

Explanation:

Problem 3: $6m \leq -42$

Step1: Isolate $m$ by dividing by 6

$\frac{6m}{6} \leq \frac{-42}{6}$

Step2: Simplify both sides

$m \leq -7$

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Problem 4: $\frac{x}{2} - 5 \geq -9$

Step1: Add 5 to both sides

$\frac{x}{2} - 5 + 5 \geq -9 + 5$
$\frac{x}{2} \geq -4$

Step2: Multiply by 2 to isolate $x$

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

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Problem 5: $4p + 3 \geq 2(2p + 1)$

Step1: Expand the right-hand side

$4p + 3 \geq 4p + 2$

Step2: Subtract $4p$ from both sides

$4p + 3 - 4p \geq 4p + 2 - 4p$
$3 \geq 2$
This is always true, so all real numbers are solutions.

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Problem 6: $-2 \leq 5 - 3a \leq 13$

Step1: Subtract 5 from all parts

$-2 - 5 \leq 5 - 3a - 5 \leq 13 - 5$
$-7 \leq -3a \leq 8$

Step2: Divide by -3 (reverse inequalities)

$\frac{-7}{-3} \geq a \geq \frac{8}{-3}$
$-\frac{8}{3} \leq a \leq \frac{7}{3}$

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Problem 7: $-7.5 < 2 - h$ or $6h + 5 > 71$

First inequality: $-7.5 < 2 - h$

Step1: Subtract 2 from both sides

$-7.5 - 2 < 2 - h - 2$
$-9.5 < -h$

Step2: Multiply by -1 (reverse inequality)

$9.5 > h$ or $h < 9.5$

Second inequality: $6h + 5 > 71$

Step1: Subtract 5 from both sides

$6h + 5 - 5 > 71 - 5$
$6h > 66$

Step2: Divide by 6

$\frac{6h}{6} > \frac{66}{6}$
$h > 11$
Combined solution: $h < 9.5$ or $h > 11$

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Problem 8: $4|-3b + 5| - 9 < 7$

Step1: Add 9 to both sides

$4|-3b + 5| - 9 + 9 < 7 + 9$
$4|-3b + 5| < 16$

Step2: Divide by 4

$\frac{4|-3b + 5|}{4} < \frac{16}{4}$
$|-3b + 5| < 4$

Step3: Rewrite absolute value inequality

$-4 < -3b + 5 < 4$

Step4: Subtract 5 from all parts

$-4 - 5 < -3b + 5 - 5 < 4 - 5$
$-9 < -3b < -1$

Step5: Divide by -3 (reverse inequalities)

$\frac{-9}{-3} > b > \frac{-1}{-3}$
$\frac{1}{3} < b < 3$

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Answer:

1. $m \leq -7$
2. $x \geq -8$
3. All real numbers are solutions
4. $-\frac{8}{3} \leq a \leq \frac{7}{3}$
5. $h < 9.5$ or $h > 11$
6. $\frac{1}{3} < b < 3$

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Graph Notes (for reference):

1. Closed dot at $-7$, shade left.
2. Closed dot at $-8$, shade right.
3. Shade entire number line.
4. Closed dots at $-\frac{8}{3}$ and $\frac{7}{3}$, shade between them.
5. Open dot at $9.5$ (shade left) and open dot at $11$ (shade right).
6. Open dots at $\frac{1}{3}$ and $3$, shade between them.