## mathmath333 one year ago Solve

1. mathmath333

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

2. 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.

3. Michele_Laino

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

4. anonymous

5. anonymous

This is the right way to solve it

6. 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

7. mathmath333

8. anonymous

Yes!

9. anonymous

What other?

10. 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

11. mathmath333

ok

12. 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

13. mathmath333

ok

14. 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

15. Michele_Laino

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

16. mathmath333

thnks

17. Michele_Laino

:)