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

Understand Absolute Value: We are given the inequality |s + 2| ≤ 1. To solve for s, 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 |s + 2| ≤ 1 means that s + 2 is at most 1 unit away from 0 on the number line, either in the positive or negative direction.Split into Two Inequalities: We can split the inequality into two separate inequalities to account for both the positive and negative scenarios. The first scenario is when s + 2 is non-negative, and the second scenario is when s + 2 is negative. For the first scenario, we have: s + 2 ≤ 1 Subtracting 2 from both sides gives us: s ≤ 1 - 2 s ≤ -1Positive Scenario Solution: For the second scenario, we consider the negative version of s + 2, which gives us: -(s + 2) ≤ 1 Multiplying both sides by -1 and remembering to reverse the inequality sign gives us: s + 2 ≥ -1 Subtracting 2 from both sides gives us: s ≥ -1 - 2 s ≥ -3Negative Scenario Solution: Combining both scenarios into a compound inequality, we get: -3 ≤ s ≤ -1 This compound inequality represents all the values of s that satisfy the original inequality |s + 2| ≤ 1.

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Solve for $s$ . $\newline$ $|s + 2| \leq 1$ $\newline$ 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

s

\newline

|s + 2| \leq 1

\newline

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

______

Understand Absolute Value: We are given the inequality $|s + 2| \leq 1$ . To solve for $s$ , 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 $|s + 2| \leq 1$ means that $s + 2$ is at most $1$ unit away from $0$ on the number line, either in the positive or negative direction.
Split into Two Inequalities: We can split the inequality into two separate inequalities to account for both the positive and negative scenarios. The first scenario is when $s + 2$ is non-negative, and the second scenario is when $s + 2$ is negative. $\newline$ For the first scenario, we have: $\newline$ $s + 2 \leq 1$ $\newline$ Subtracting $2$ from both sides gives us: $\newline$ $s \leq 1 - 2$ $\newline$ $s \leq -1$
Positive Scenario Solution: For the second scenario, we consider the negative version of $s + 2$ , which gives us: $\newline$ $-(s + 2) \leq 1$ $\newline$ Multiplying both sides by $-1$ and remembering to reverse the inequality sign gives us: $\newline$ $s + 2 \geq -1$ $\newline$ Subtracting $2$ from both sides gives us: $\newline$ $s \geq -1 - 2$ $\newline$ $s \geq -3$
Negative Scenario Solution: Combining both scenarios into a compound inequality, we get: $\newline$ $-3 \leq s \leq -1$ $\newline$ This compound inequality represents all the values of $s$ that satisfy the original inequality $|s + 2| \leq 1$ .