Solve 5|x+3|-12 <= 13. (A) There are no solutions. (B) -2 <= x <= 2 (C) -8 <= x <= 8 (D) -8 <= x <= 2

Isolate absolute value expression: Isolate the absolute value expression. To solve the inequality 5|x+3| - 12 <= 13, we first need to isolate the absolute value expression on one side of the inequality. We do this by adding 12 to both sides of the inequality. 5|x+3| - 12 + 12 <= 13 + 12 5|x+3| <= 25Remove coefficient: Remove the coefficient of the absolute value expression. Next, we divide both sides of the inequality by 5 to solve for the absolute value expression |x+3|. 5|x+3| / 5 = -5Solve first inequality: Solve the first inequality. We solve the first inequality x+3 = -5 by subtracting 3 from both sides. x+3 - 3 >= -5 - 3 x >= -8Combine solutions: Combine the solutions to find the solution set. The solution set for the inequality is the intersection of the solutions to the two inequalities we found in steps 4 and 5. This means x must be greater than or equal to -8 and less than or equal to 2. -8 <= x <= 2

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Solve 5|x+3|-12 <= 13.
(A) There are no solutions.
(B) -2 <= x <= 2
(C) -8 <= x <= 8
(D) -8 <= x <= 2

Solve $5|x+3|-12 \leq 13$ . $\newline$ (A) There are no solutions. $\newline$ (B) $-2 \leq x \leq 2$ $\newline$ (C) $-8 \leq x \leq 8$ $\newline$ (D) $-8 \leq x \leq 2$

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Math Problems

Grade 6

Integer inequalities with absolute values

Full solution

Q. Solve

5|x+3|-12 \leq 13

\newline

(A) There are no solutions.

\newline

(B)

-2 \leq x \leq 2

\newline

(C)

-8 \leq x \leq 8

\newline

(D)

-8 \leq x \leq 2

Isolate absolute value expression: Isolate the absolute value expression. $\newline$ To solve the inequality $5|x+3| - 12 \leq 13$ , we first need to isolate the absolute value expression on one side of the inequality. We do this by adding $12$ to both sides of the inequality. $\newline$ $5|x+3| - 12 + 12 \leq 13 + 12$ $\newline$ $5|x+3| \leq 25$
Remove coefficient: Remove the coefficient of the absolute value expression. $\newline$ Next, we divide both sides of the inequality by $5$ to solve for the absolute value expression $|x+3|$ . $\newline$ $\frac{5|x+3|}{5} \leq \frac{25}{5}$ $\newline$ $|x+3| \leq 5$
Set up inequalities: Set up two separate inequalities. $\newline$ The absolute value inequality $|x+3| \leq 5$ means that the quantity inside the absolute value, $x+3$ , is less than or equal to $5$ and greater than or equal to $-5$ . We can write this as two separate inequalities: $\newline$ $x+3 \leq 5$ and $x+3 \geq -5$
Solve first inequality: Solve the first inequality. $\newline$ We solve the first inequality $x+3 \leq 5$ by subtracting $3$ from both sides. $\newline$ $x+3 - 3 \leq 5 - 3$ $\newline$ $x \leq 2$
Solve second inequality: Solve the second inequality. $\newline$ We solve the second inequality $x+3 \geq -5$ by subtracting $3$ from both sides. $\newline$ $x+3 - 3 \geq -5 - 3$ $\newline$ $x \geq -8$
Combine solutions: Combine the solutions to find the solution set. $\newline$ The solution set for the inequality is the intersection of the solutions to the two inequalities we found in steps $4$ and $5$ . This means $x$ must be greater than or equal to $-8$ and less than or equal to $2$ . $\newline$ $-8 \leq x \leq 2$