Solve for q. |q + 8| ≤ 2 Write a compound inequality like 1 3. Use integers, proper fractions, or improper fractions in simplest form. ______

Absolute Value Definition: We have the inequality |q + 8| ≤ 2. To solve for q, we need to consider the definition of absolute value, which states that |x| = x if x ≥ 0 and |x| = -x if x < 0. Therefore, the inequality |q + 8| ≤ 2 means that q + 8 is at most 2 units away from 0 on the number line, in either direction.Non-Negative Case: First, we consider the case where q + 8 is non-negative. In this case, we can drop the absolute value without changing the sign: q + 8 ≤ 2 Now, we solve for q by subtracting 8 from both sides: q + 8 - 8 ≤ 2 - 8 q ≤ -6Negative Case: Next, we consider the case where q + 8 is negative. In this case, we must change the sign when we remove the absolute value: -(q + 8) ≤ 2 Now, we distribute the negative sign and solve for q: -q - 8 ≤ 2 -q ≤ 2 + 8 q ≥ -10 (after multiplying both sides by -1, we must reverse the inequality sign)Compound Inequality: Combining both cases, we have a compound inequality that represents all possible values of q that satisfy the original inequality: -10 ≤ q ≤ -6 This compound inequality includes both the case where q + 8 is non-negative and the case where q + 8 is negative.

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Solve for $q$ . $|q + 8| \leq 2$ Write a compound inequality like 1 < x < 3 or like x < 1 or x > 3. Use integers, proper fractions, or improper fractions in simplest form. $\newline$ ______

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Algebra 2

Solve absolute value inequalities

Full solution

Q. Solve for

q

|q + 8| \leq 2

Write a compound inequality like

1 < x < 3

or like

x < 1

x > 3

. Use integers, proper fractions, or improper fractions in simplest form.

\newline

______

Absolute Value Definition: We have the inequality $|q + 8| \leq 2$ . To solve for $q$ , we need to consider the definition of absolute value, which states that $|x| = x$ if $x \geq 0$ and $|x| = -x$ if x < 0. Therefore, the inequality $|q + 8| \leq 2$ means that $q + 8$ is at most $2$ units away from $0$ on the number line, in either direction.
Non-Negative Case: First, we consider the case where $q + 8$ is non-negative. In this case, we can drop the absolute value without changing the sign: $\newline$ $q + 8 \leq 2$ $\newline$ Now, we solve for $q$ by subtracting $8$ from both sides: $\newline$ $q + 8 - 8 \leq 2 - 8$ $\newline$ $q \leq -6$
Negative Case: Next, we consider the case where $q + 8$ is negative. In this case, we must change the sign when we remove the absolute value: $\newline$ $-(q + 8) \leq 2$ $\newline$ Now, we distribute the negative sign and solve for $q$ : $\newline$ $-q - 8 \leq 2$ $\newline$ $-q \leq 2 + 8$ $\newline$ $q \geq -10$ (after multiplying both sides by $-1$ , we must reverse the inequality sign)
Compound Inequality: Combining both cases, we have a compound inequality that represents all possible values of $q$ that satisfy the original inequality: $\newline$ $-10 \leq q \leq -6$ $\newline$ This compound inequality includes both the case where $q + 8$ is non-negative and the case where $q + 8$ is negative.