A community for students.
Here's the question you clicked on:
 0 viewing
mathmath333
 one year ago
Solve
mathmath333
 one year ago
Solve

This Question is Closed

mathmath333
 one year ago
Best ResponseYou've already chosen the best response.0\(\large \color{black}{\begin{align} 2x4+3x+9=16\hspace{.33em}\\~\\ \end{align}}\)

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0If x < 3 then we have 42x3x9 = 16 so 5x = 21 i.e. x = 21/5. If 3 <= x < 2 then we have 42x+3x+9 = 16 so x = 3 but this is not in this range so is not a solution. If x >= 2 then 2x4+3x+9 = 16 so 5x=11 so x = 11/5. So the only solutions are x = 21/5 and x = 11/5.

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.2we have to distinguish these four cases: \[\large \begin{gathered} \left\{ \begin{gathered} 2x  4 \geqslant 0 \hfill \\ 3x + 9 \geqslant 0 \hfill \\ \end{gathered} \right.,\quad \left\{ \begin{gathered} 2x  4 \geqslant 0 \hfill \\ 3x + 9 < 0 \hfill \\ \end{gathered} \right., \hfill \\ \hfill \\ \left\{ \begin{gathered} 2x  4 < 0 \hfill \\ 3x + 9 \geqslant 0 \hfill \\ \end{gathered} \right.,\quad \left\{ \begin{gathered} 2x  4 < 0 \hfill \\ 3x + 9 < 0 \hfill \\ \end{gathered} \right. \hfill \\ \end{gathered} \]

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0This is the right way to solve it

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.2for 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
 one year ago
Best ResponseYou've already chosen the best response.0ok what about the other

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.2second 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

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.2third 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

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.2fourth 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
 one year ago
Best ResponseYou've already chosen the best response.2reassuming, the solutions of our equation, are: \[\Large x =  \frac{{21}}{5},\quad x = \frac{{11}}{5}\]
Ask your own question
Sign UpFind more explanations on OpenStudy
Your question is ready. Sign up for free to start getting answers.
spraguer
(Moderator)
5
→ View Detailed Profile
is replying to Can someone tell me what button the professor is hitting...
23
 Teamwork 19 Teammate
 Problem Solving 19 Hero
 Engagement 19 Mad Hatter
 You have blocked this person.
 ✔ You're a fan Checking fan status...
Thanks for being so helpful in mathematics. If you are getting quality help, make sure you spread the word about OpenStudy.