To solve each equation for [tex]\( x \)[/tex] and match the solutions:
1. Equation: [tex]\(-a x - 20 = -14\)[/tex]
- First, isolate [tex]\( -a x \)[/tex]:
[tex]\[
-a x - 20 + 20 = -14 + 20 \implies -a x = 6
\][/tex]
- Then, solve for [tex]\( x \)[/tex]:
[tex]\[
x = -\frac{6}{a}
\][/tex]
Therefore, [tex]\(-a x - 20 = -14 \implies x = -\frac{6}{a}\)[/tex].
2. Equation: [tex]\(4 = \frac{6}{a} x + 5\)[/tex]
- First, isolate [tex]\(\frac{6}{a} x\)[/tex]:
[tex]\[
4 - 5 = \frac{6}{a} x + 5 - 5 \implies -1 = \frac{6}{a} x
\][/tex]
- Then, solve for [tex]\( x \)[/tex]:
[tex]\[
x = -\frac{a}{6}
\][/tex]
Therefore, [tex]\(4 = \frac{6}{a} x + 5 \implies x = -\frac{a}{6}\)[/tex].
3. Equation: [tex]\(7 + 2 a x = 13\)[/tex]
- First, isolate [tex]\(2 a x\)[/tex]:
[tex]\[
7 + 2 a x - 7 = 13 - 7 \implies 2 a x = 6
\][/tex]
- Then, solve for [tex]\( x \)[/tex]:
[tex]\[
x = \frac{3}{a}
\][/tex]
Therefore, [tex]\(7 + 2 a x = 13 \implies x = \frac{3}{a}\)[/tex].
Matching the solutions to each equation:
[tex]\[
\begin{array}{ll}
-a x - 20 = -14 & \quad x = -\frac{6}{a} \\
4 = \frac{6}{a} x + 5 & \quad x = -\frac{a}{6} \\
7 + 2 a x = 13 & \quad x = \frac{3}{a}
\end{array}
\][/tex]
