Question

Solve

Answer 1

\(\large \color{black}{\begin{align} |2x-4|+|3x+9|=16\hspace{.33em}\\~\\
\end{align}}\)

Answer 2

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.

Answer 3

we have to distinguish these four cases:
\[\large \begin{gathered}
\left\{ \begin{gathered}
2x - 4 \geqslant 0 \hfill \\
3x + 9 \geqslant 0 \hfill \\
\end{gathered} \right.,\quad \left\{ \begin{gathered}
2x - 4 \geqslant 0 \hfill \\
3x + 9 < 0 \hfill \\
\end{gathered} \right., \hfill \\
\hfill \\
\left\{ \begin{gathered}
2x - 4 < 0 \hfill \\
3x + 9 \geqslant 0 \hfill \\
\end{gathered} \right.,\quad \left\{ \begin{gathered}
2x - 4 < 0 \hfill \\
3x + 9 < 0 \hfill \\
\end{gathered} \right. \hfill \\
\end{gathered} \]

Answer 4

This is the right way to solve it

Answer 5

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

Answer 6

ok what about the other

Answer 7

Yes!

Answer 8

What other?

Answer 9

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

Answer 10

ok

Answer 11

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

Answer 12

ok

Answer 13

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

Answer 14

reassuming, the solutions of our equation, are:
\[\Large x = - \frac{{21}}{5},\quad x = \frac{{11}}{5}\]

Answer 15

thnks

Answer 16

:)