## anonymous one year ago How do I do this? What is the answer? Solve for x: -3|2x + 6| = -12 1)x = 1 and x = 5 2)x = -1 and x = -5 3)x = -9 and x = 3 4)No solution

• This Question is Open
1. Kash_TheSmartGuy

First, distribute:$-3\left| 2x+6 \right|=-12$$6x+18=-12$$6x+(18-18)=(-12)-18$$6x=-30$$\frac{ 6x }{ 6 }=\frac{ -30 }{ 6 }$$x=-5$I would go for B.

2. Kash_TheSmartGuy

You get it, right?

3. anonymous

are you sure? I need to get this question right. And yes I get it but I just want to make sure that the answer is b.

4. Kash_TheSmartGuy

@Michele_Laino

5. anonymous

?

6. anonymous

is it right?

7. Michele_Laino

We can rewrite your equation as follows: $\left| {2x + 6} \right| = 4$ I have divided both sides by -3

8. Michele_Laino

now we have to consider these 2 cases: $\Large \left\{ \begin{gathered} 2x + 6 \geqslant 0 \hfill \\ 2x + 6 = 4 \hfill \\ \end{gathered} \right.\; \cup \,\left\{ {\begin{array}{*{20}{c}} {2x + 6 < 0} \\ { - \left( {2x + 6} \right) = 4} \end{array}} \right.$

9. Michele_Laino

the solutions of your equations are given by the solutions of those system of inequality above

10. Michele_Laino

equation*

11. Michele_Laino

for example I solve the first system: I get this: $\Large \left\{ \begin{gathered} 2x + 6 \geqslant 0 \hfill \\ 2x + 6 = 4 \hfill \\ \end{gathered} \right. \Rightarrow \left\{ {\begin{array}{*{20}{c}} {x \geqslant - 3} \\ {x = - 1} \end{array}} \right.$ am I right?

12. Michele_Laino

|dw:1435429357830:dw|

13. Michele_Laino

|dw:1435429414248:dw|

14. Michele_Laino

since x=-1, belongs to the interval (-3, +infinity), then x=-1 is a solution of your original equation

15. Michele_Laino

now, please do the same with the second system: $\Large \left\{ {\begin{array}{*{20}{c}} {2x + 6 < 0} \\ { - \left( {2x + 6} \right) = 4} \end{array}} \right.$