A community for students.

Here's the question you clicked on:

55 members online
  • 0 replying
  • 0 viewing

mathmath333

  • one year ago

Solve

  • This Question is Closed
  1. mathmath333
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

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

  2. anonymous
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    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
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 2

    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
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    2 Attachments
  5. anonymous
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    This is the right way to solve it

  6. Michele_Laino
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 2

    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
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    ok what about the other

  8. anonymous
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    Yes!

  9. anonymous
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    What other?

  10. Michele_Laino
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 2

    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
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    ok

  12. Michele_Laino
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 2

    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
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    ok

  14. Michele_Laino
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 2

    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
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 2

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

  16. mathmath333
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    thnks

  17. Michele_Laino
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 2

    :)

  18. Not the answer you are looking for?
    Search for more explanations.

    • Attachments:

Ask your own question

Sign Up
Find more explanations on OpenStudy
Privacy Policy

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
  • 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.

This is the testimonial you wrote.
You haven't written a testimonial for Owlfred.