mathmath333
  • mathmath333
Solve
Mathematics
  • Stacey Warren - Expert brainly.com
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katieb
  • katieb
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mathmath333
  • mathmath333
\(\large \color{black}{\begin{align} |2x-4|+|3x+9|=16\hspace{.33em}\\~\\ \end{align}}\)
anonymous
  • anonymous
If x < -3 then we have 4-2x-3x-9 = 16 so -5x = 21 i.e. x = -21/5. If -3 <= x < 2 then we have 4-2x+3x+9 = 16 so x = 3 but this is not in this range so is not a solution. If x >= 2 then 2x-4+3x+9 = 16 so 5x=11 so x = 11/5. So the only solutions are x = -21/5 and x = 11/5.
Michele_Laino
  • Michele_Laino
we have to distinguish these four cases: \[\large \begin{gathered} \left\{ \begin{gathered} 2x - 4 \geqslant 0 \hfill \\ 3x + 9 \geqslant 0 \hfill \\ \end{gathered} \right.,\quad \left\{ \begin{gathered} 2x - 4 \geqslant 0 \hfill \\ 3x + 9 < 0 \hfill \\ \end{gathered} \right., \hfill \\ \hfill \\ \left\{ \begin{gathered} 2x - 4 < 0 \hfill \\ 3x + 9 \geqslant 0 \hfill \\ \end{gathered} \right.,\quad \left\{ \begin{gathered} 2x - 4 < 0 \hfill \\ 3x + 9 < 0 \hfill \\ \end{gathered} \right. \hfill \\ \end{gathered} \]

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anonymous
  • anonymous
2 Attachments
anonymous
  • anonymous
This is the right way to solve it
Michele_Laino
  • Michele_Laino
for example, if we develop the first case, we get: \[\large \left\{ \begin{gathered} 2x - 4 \geqslant 0 \hfill \\ 3x + 9 \geqslant 0 \hfill \\ 2x - 4 + 3x + 9 = 16 \hfill \\ \end{gathered} \right. \Rightarrow \left\{ \begin{gathered} x \geqslant 2 \hfill \\ x \geqslant - 3 \hfill \\ x = \frac{{11}}{5} \hfill \\ \end{gathered} \right.\] so one solution is x=11/5
mathmath333
  • mathmath333
ok what about the other
anonymous
  • anonymous
Yes!
anonymous
  • anonymous
What other?
Michele_Laino
  • Michele_Laino
second case: \[\large \left\{ \begin{gathered} 2x - 4 \geqslant 0 \hfill \\ 3x + 9 < 0 \hfill \\ 2x - 4 - 3x - 9 = 16 \hfill \\ \end{gathered} \right. \Rightarrow \left\{ \begin{gathered} x \geqslant 2 \hfill \\ x < - 3 \hfill \\ 2x - 4 - 3x - 9 = 16 \hfill \\ \end{gathered} \right.\] which is the empty set, so we have no solutions
mathmath333
  • mathmath333
ok
Michele_Laino
  • Michele_Laino
third case: \[\large \left\{ \begin{gathered} 2x - 4 < 0 \hfill \\ 3x + 9 \geqslant 0 \hfill \\ - 2x + 4 + 3x + 9 = 16 \hfill \\ \end{gathered} \right. \Rightarrow \left\{ \begin{gathered} x < 2 \hfill \\ x \geqslant - 3 \hfill \\ x = 3 \hfill \\ \end{gathered} \right.\] which is again the empty set, so again we have no solutions, since x=3 is not acceptable
mathmath333
  • mathmath333
ok
Michele_Laino
  • Michele_Laino
fourth case: \[\large \left\{ \begin{gathered} 2x - 4 < 0 \hfill \\ 3x + 9 < 0 \hfill \\ - 2x + 4 - 3x - 9 = 16 \hfill \\ \end{gathered} \right. \Rightarrow \left\{ \begin{gathered} x < 2 \hfill \\ x < - 3 \hfill \\ x = - \frac{{21}}{5} \hfill \\ \end{gathered} \right.\] this value is acceptable, so the solution, is x=-21/5
Michele_Laino
  • Michele_Laino
reassuming, the solutions of our equation, are: \[\Large x = - \frac{{21}}{5},\quad x = \frac{{11}}{5}\]
mathmath333
  • mathmath333
thnks
Michele_Laino
  • Michele_Laino
:)

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