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anonymous

  • one year ago

I need help. A system of equations is shown below: 5x - 5y = 10 3x - 2y = 2 Part A: Create an equivalent system of equations by replacing one equation with the sum of that equation and a multiple of the other. Show the steps to do this. Part B: Show that the equivalent system has the same solution as the original system of equations.

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  1. Michele_Laino
    • one year ago
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    we can multiply the first equation by 2 and the second equation by -5, so we get the subsequent equivalent system: \[\Large \left\{ \begin{gathered} 10x - 10y = 20 \hfill \\ - 15x + 10y = - 10 \hfill \\ \end{gathered} \right.\]

  2. anonymous
    • one year ago
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    Right..

  3. Michele_Laino
    • one year ago
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    now we add those two equations together, so we get: \[\Large - 5x = 10\]

  4. Michele_Laino
    • one year ago
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    am I right?

  5. anonymous
    • one year ago
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    Yes

  6. Michele_Laino
    • one year ago
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    so the requested equivalent system, in part A, can be this: \[\Large \left\{ \begin{gathered} 5x - 5y = 10 \hfill \\ - 5x = 10 \hfill \\ \end{gathered} \right.\]

  7. anonymous
    • one year ago
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    So it would be - 5x = 10 and that would be an equivalent system of the equation?

  8. Michele_Laino
    • one year ago
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    yes!

  9. anonymous
    • one year ago
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    Yay! For Part B, I just check my work an write that as the proof?

  10. Michele_Laino
    • one year ago
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    another way to get an equivalent system is multipling the firs equation by -2/5, so Iget this: \[\left\{ \begin{gathered} - 2x + 2y = - 4 \hfill \\ 3x - 2y = 2 \hfill \\ \end{gathered} \right.\]

  11. Michele_Laino
    • one year ago
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    then I add the two equation together, so i can write: x=-2 so another equivalent system is: \[\Large \left\{ \begin{gathered} x = - 2 \hfill \\ 3x - 2y = 2 \hfill \\ \end{gathered} \right.\]

  12. Michele_Laino
    • one year ago
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    which equivalent system do you prefer?

  13. anonymous
    • one year ago
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    I prefer the first way you did, the 2nd one was kind of confusing but I could probably understand it if I put the time. But either way is great, thank you so much! Thats for Part A right? Then for Part B I use the way you used but I like do the opposite of everything and divide to get the equation again? Would that work or no?

  14. Michele_Laino
    • one year ago
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    ok, we can use the first system, namely: \[\Large \left\{ \begin{gathered} 5x - 5y = 10 \hfill \\ - 5x = 10 \hfill \\ \end{gathered} \right.\]

  15. Michele_Laino
    • one year ago
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    now, please solve the second equation for, x, what do you get?

  16. anonymous
    • one year ago
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    You do 15x - 10y and it would equal to 20?

  17. Michele_Laino
    • one year ago
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    no, it is simple, you have to compute this: \[\Large x = \frac{{10}}{{ - 5}} = ...?\]

  18. anonymous
    • one year ago
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    -2?

  19. Michele_Laino
    • one year ago
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    perfect!

  20. Michele_Laino
    • one year ago
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    now, we have to substitute x=-2 into the first equation, namely: \[\Large 5 \times \left( { - 2} \right) - 5y = 10\]

  21. Michele_Laino
    • one year ago
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    or: \[ \Large - 10 - 5y = 10\]

  22. anonymous
    • one year ago
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    Why is there no x?

  23. Michele_Laino
    • one year ago
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    since I have replaced x with its value, which is -2

  24. anonymous
    • one year ago
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    OH My bad!

  25. Michele_Laino
    • one year ago
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    so, what is y?

  26. anonymous
    • one year ago
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    -2.5

  27. Michele_Laino
    • one year ago
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    we have these steps: \[\Large \begin{gathered} - 10 - 5y = 10 \hfill \\ - 5y = - 20 \hfill \\ y = \frac{{ - 20}}{{ - 5}} \hfill \\ \end{gathered} \]

  28. anonymous
    • one year ago
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    \[y = 4?\]

  29. Michele_Laino
    • one year ago
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    oops.. I have made an error, here are the right steps: \[\Large \begin{gathered} - 10 - 5y = 10 \hfill \\ - 5y = 20 \hfill \\ y = \frac{{20}}{{ - 5}} = ...? \hfill \\ \end{gathered} \]

  30. anonymous
    • one year ago
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    oh -4

  31. Michele_Laino
    • one year ago
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    ok! so the solution of the equivalent system, is: \[\Large x = - 2,\quad y = - 4\]

  32. anonymous
    • one year ago
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    Yes

  33. Michele_Laino
    • one year ago
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    now, we have to substitute those value of x and y into the first equation of the original system, namely: \[\Large 5x - 5y = 10\]

  34. Michele_Laino
    • one year ago
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    values*

  35. Michele_Laino
    • one year ago
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    so we can write this: \[\Large 5 \times \left( { - 2} \right) - 5 \times \left( { - 4} \right) = 10\]

  36. anonymous
    • one year ago
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    Nice!

  37. anonymous
    • one year ago
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    Thats all I need to do?

  38. Michele_Laino
    • one year ago
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    we got an identity, since we can write: \[\Large \begin{gathered} 5 \times \left( { - 2} \right) - 5 \times \left( { - 4} \right) = 10 \hfill \\ - 10 + 20 = 10 \hfill \\ 10 = 10 \hfill \\ \end{gathered} \]

  39. Michele_Laino
    • one year ago
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    no, we have to substitute those values of x=-2 and y=-4, into the second equation of your original system, namely: \[\Large 3x - 2y = 2\]

  40. Michele_Laino
    • one year ago
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    so we get: \[\Large 3 \times \left( { - 2} \right) - 2 \times \left( { - 4} \right) = 2\]

  41. Michele_Laino
    • one year ago
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    again we got an identity, since we can write these steps: \[\Large \begin{gathered} 3 \times \left( { - 2} \right) - 2 \times \left( { - 4} \right) = 2 \hfill \\ - 6 + 8 = 2 \hfill \\ 2 = 2 \hfill \\ \end{gathered} \]

  42. anonymous
    • one year ago
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    So we can say the 2 equations are identity then

  43. Michele_Laino
    • one year ago
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    yes! after our substitution they became 2 identities. That means the solution of the equivalent system is also the solution of your original system

  44. anonymous
    • one year ago
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    Perfect

  45. Michele_Laino
    • one year ago
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    :)

  46. anonymous
    • one year ago
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    Thank you so much you have a new fan now haha :)

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