Solve the inequality.

[tex]\[
\begin{array}{c}
-2x + 8 \ \textless \ 14 \quad \text{and} \quad -3x - 9 \geq -12 \\
[?] \ \textless \ x \leq \square
\end{array}
\][/tex]



Answer :

To solve the compound inequality [tex]\( -2x + 8 < 14 \)[/tex] and [tex]\( -3x - 9 \geq -12 \)[/tex], let’s break it down step-by-step for each inequality separately.

1. Solve the first inequality: [tex]\( -2x + 8 < 14 \)[/tex]

- First, subtract 8 from both sides to isolate the term involving [tex]\( x \)[/tex]:
[tex]\[ -2x < 14 - 8 \][/tex]
[tex]\[ -2x < 6 \][/tex]

- Next, divide both sides by [tex]\(-2\)[/tex] to solve for [tex]\( x \)[/tex]. Remember that dividing by a negative number reverses the inequality direction:
[tex]\[ x > \frac{6}{-2} \][/tex]
[tex]\[ x > -3 \][/tex]

2. Solve the second inequality: [tex]\( -3x - 9 \geq -12 \)[/tex]

- First, add 9 to both sides to isolate the term involving [tex]\( x \)[/tex]:
[tex]\[ -3x \geq -12 + 9 \][/tex]
[tex]\[ -3x \geq -3 \][/tex]

- Next, divide both sides by [tex]\(-3\)[/tex] to solve for [tex]\( x \)[/tex]. Again, remember that dividing by a negative number reverses the inequality direction:
[tex]\[ x \leq \frac{-3}{-3} \][/tex]
[tex]\[ x \leq 1 \][/tex]

3. Combine the results:

- From the first inequality, we have [tex]\( x > -3 \)[/tex].
- From the second inequality, we have [tex]\( x \leq 1 \)[/tex].

Combining these two results, we get the following compound inequality:
[tex]\[ -3 < x \leq 1 \][/tex]

Therefore, the solution to the compound inequality [tex]\( -2x + 8 < 14 \)[/tex] and [tex]\( -3x - 9 \geq -12 \)[/tex] is:
[tex]\[ -3 < x \leq 1 \][/tex]