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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

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  1. Kash_TheSmartGuy
    • one year ago
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    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
    • one year ago
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    You get it, right?

  3. anonymous
    • one year ago
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    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
    • one year ago
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    @Michele_Laino

  5. anonymous
    • one year ago
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    ?

  6. anonymous
    • one year ago
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    is it right?

  7. Michele_Laino
    • one year ago
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    We can rewrite your equation as follows: \[\left| {2x + 6} \right| = 4\] I have divided both sides by -3

  8. Michele_Laino
    • one year ago
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    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
    • one year ago
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    the solutions of your equations are given by the solutions of those system of inequality above

  10. Michele_Laino
    • one year ago
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    equation*

  11. Michele_Laino
    • one year ago
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    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
    • one year ago
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    |dw:1435429357830:dw|

  13. Michele_Laino
    • one year ago
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    |dw:1435429414248:dw|

  14. Michele_Laino
    • one year ago
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    since x=-1, belongs to the interval (-3, +infinity), then x=-1 is a solution of your original equation

  15. Michele_Laino
    • one year ago
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    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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