Question

Solve the absolute value inequality. Express the answer using interval notation.

9 − |2x + 7| ≤ 6

Answer 1

\(\bf 9 - |2x + 7| \le 6 \implies -9+9 - |2x + 7| \le 6-9\implies |2x + 7| \le -3\\ \quad \\
|2x + 7| \le -3\implies
\begin{cases}
+(2x + 7) \le -3\\ \quad \\
\bf -(2x + 7) \le -3
\end{cases}\)

that'd give you 2 values for "x", one for each scenario

Answer 2

But an absolute value cannot be less than a negative number, since it is constantly positive.

Answer 3

hmmm I made a quick mistake... kinda ate the minus

so \(\bf 9 - |2x + 7| \le 6 \implies -9+9 - |2x + 7| \le 6-9\implies -|2x + 7| \le -3\\ \quad \\
\textit{recall that whenever multiplying by a negative value you have to }\\
{\color{red}{ flip}}\textit{ the inequality sign}\\ \quad \\
|2x+7|\ge 3 \implies
|2x + 7| \ge 3\implies
\begin{cases}
+(2x + 7) \ge -3\\ \quad \\
\bf -(2x + 7) \ge -3
\end{cases}\)

Answer 4

hmm
shoot.. .anyhow.. even more typos... one sec \(\bf 9 - |2x + 7| \le 6 \implies -9+9 - |2x + 7| \le 6-9\implies -|2x + 7| \le -3\\ \quad \\
\textit{recall that whenever multiplying by a negative value you have to }\\
{\color{red}{ flip}}\textit{ the inequality sign}\\ \quad \\
|2x+7|\ge 3 \implies
|2x + 7| \ge 3\implies
\begin{cases}
+(2x + 7) \ge 3\\ \quad \\
\bf -(2x + 7) \ge 3
\end{cases}\)

so.... let's try to do the 1st one

Answer 5

\(\bf +(2x + 7) \ge -3\implies 2x+7\ge 3\) so... what should we do to isolate the "x"?

Answer 6

ack... typos are killing me anyhow \(\bf +(2x + 7) \ge 3\implies 2x+7\ge 3\)

Answer 7

you'd do it more or less in the same way you'd solve a linear equation, as if you were using

Answer 8

as if you were using \(\bf 2x+7= 3\)