Question

One last question :) I will medal and fan :) Thank you for helping if you do!
Solve for x: −3|x + 7| = −12

x = 5 over 3, x = −19 over 3
x = −3, x = −11
x = −3, x = 11
No solution

Answer 1

I say no solution?

Answer 2

???

Answer 3

For any number \(z\), \(|z|\) is always positive. Your equation has a negative number multiplied by a positive number (the left side), which is always going to give a negative number. This guarantees that there's at least one solution.

Answer 4

To find the solution(s), you can write
\[\begin{align*}
-3|x+7|&=-12\\[1ex]
|x+7|&=4
\end{align*}\]Then, invoking the definition of the absolute value function, you can analyze the left side a bit more:
\[|x|=\begin{cases}x&\text{if }x\ge0\\[1ex]-x&\text{if }x<0\end{cases}\implies|x+7|=\begin{cases}x+7&\text{if }x+7\ge0\\[1ex]-(x+7)&\text{if }x+7<0\end{cases}=\begin{cases}x+7&\text{if }x\ge-7\\[1ex]-x-7&\text{if }x<-7\end{cases}\]Consider one case at a time.

(1) If \(x\ge-7\), then the equation becomes
\[|x+7|=4\implies x+7=4\]
(2) If \(x<-7\), then you get
\[|x+7|=4\implies -x-7=4\]and you can see that there are indeed exactly two (not zero) solutions.

Answer 5

Oops, copying the cases that are cut off:
\[\cdots\implies|x+7|=\begin{cases}x+7&\text{if }x+7\ge0\\[1ex]-(x+7)&\text{if }x+7<0\end{cases}=\begin{cases}x+7&\text{if }x\ge-7\\[1ex]-x-7&\text{if }x<-7\end{cases}\]