anonymous
  • anonymous
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
Algebra
chestercat
  • chestercat
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Kash_TheSmartGuy
  • 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.
Kash_TheSmartGuy
  • Kash_TheSmartGuy
You get it, right?
anonymous
  • 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.

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Kash_TheSmartGuy
  • Kash_TheSmartGuy
anonymous
  • anonymous
?
anonymous
  • anonymous
is it right?
Michele_Laino
  • Michele_Laino
We can rewrite your equation as follows: \[\left| {2x + 6} \right| = 4\] I have divided both sides by -3
Michele_Laino
  • Michele_Laino
now we have to consider these 2 cases: \[\Large \left\{ \begin{gathered} 2x + 6 \geqslant 0 \hfill \\ 2x + 6 = 4 \hfill \\ \end{gathered} \right.\; \cup \,\left\{ {\begin{array}{*{20}{c}} {2x + 6 < 0} \\ { - \left( {2x + 6} \right) = 4} \end{array}} \right.\]
Michele_Laino
  • Michele_Laino
the solutions of your equations are given by the solutions of those system of inequality above
Michele_Laino
  • Michele_Laino
equation*
Michele_Laino
  • 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?
Michele_Laino
  • Michele_Laino
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Michele_Laino
  • Michele_Laino
|dw:1435429414248:dw|
Michele_Laino
  • Michele_Laino
since x=-1, belongs to the interval (-3, +infinity), then x=-1 is a solution of your original equation
Michele_Laino
  • Michele_Laino
now, please do the same with the second system: \[\Large \left\{ {\begin{array}{*{20}{c}} {2x + 6 < 0} \\ { - \left( {2x + 6} \right) = 4} \end{array}} \right.\]

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