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Solve for ss.\newlines+21|s + 2| \leq 1\newlineWrite a compound inequality like 1<x<31 < x < 3 or like x<1x < 1 or x>3x > 3. Use integers, proper fractions, or improper fractions in simplest form.\newline______

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Q. Solve for ss.\newlines+21|s + 2| \leq 1\newlineWrite a compound inequality like 1<x<31 < x < 3 or like x<1x < 1 or x>3x > 3. Use integers, proper fractions, or improper fractions in simplest form.\newline______
  1. Understand Absolute Value: We are given the inequality s+21|s + 2| \leq 1. To solve for ss, we need to consider the definition of absolute value, which states that x=x|x| = x if x0x \geq 0 and x=x|x| = -x if x<0x < 0. Therefore, the inequality s+21|s + 2| \leq 1 means that s+2s + 2 is at most 11 unit away from 00 on the number line, either in the positive or negative direction.
  2. 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+2s + 2 is non-negative, and the second scenario is when s+2s + 2 is negative.\newlineFor the first scenario, we have:\newlines+21s + 2 \leq 1\newlineSubtracting 22 from both sides gives us:\newlines12s \leq 1 - 2\newlines1s \leq -1
  3. Positive Scenario Solution: For the second scenario, we consider the negative version of s+2s + 2, which gives us:\newline(s+2)1-(s + 2) \leq 1\newlineMultiplying both sides by 1-1 and remembering to reverse the inequality sign gives us:\newlines+21s + 2 \geq -1\newlineSubtracting 22 from both sides gives us:\newlines12s \geq -1 - 2\newlines3s \geq -3
  4. Negative Scenario Solution: Combining both scenarios into a compound inequality, we get:\newline3s1-3 \leq s \leq -1\newlineThis compound inequality represents all the values of ss that satisfy the original inequality s+21|s + 2| \leq 1.

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