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anonymous

  • one year ago

HELP! WILL MEDAL AND FAN!!!! A system of equations is shown below: -3x + 7y = -16 -9x + 5y = 16 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. (6 points) Part B: Show that the equivalent system has the same solution as the original system of equations.

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  1. mathstudent55
    • one year ago
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    Part A: You can start this way: Multiply the first equation by 2 on both sides. Now add it to the second equation. What do you get?

  2. anonymous
    • one year ago
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    -3x + 7y = -16 X2 X2

  3. anonymous
    • one year ago
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    -6x + 7y = -32

  4. mathstudent55
    • one year ago
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    Remember to distribute the 2 on the left side. \(2 \times (-3x + 7y) = 2 \times (-16) \)

  5. anonymous
    • one year ago
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    do i solve this or something?

  6. mathstudent55
    • one year ago
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    No, not yet. We are working on multiplying the first equation by 2 on both sides. On the left side, you need to distribute the 2. See below. \(\color{red}{2 \times} (\color{green}{-3x} + \color{blue}{7y}) = 2 \times (-16)\) \(\color{red}{2 \times } \color{green}{-3x} + \color{red}{2 \times } \color{blue}{7y} = -32\) \(-6x + 14y = -32\)

  7. mathstudent55
    • one year ago
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    So far, we have multiplied the first equation by 2 on both sides. Now we add that new equation to the second original equation: \(~~~~~~~~~~-6x + 14y = -32\) \(~~~+~~~-9x + ~~5y = 16\) ---------------------------- \(~~~~~~~~~~-15x + 19y = -16\)

  8. anonymous
    • one year ago
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    are we still on Part A?

  9. anonymous
    • one year ago
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    i dont have a lot of time and i still have 2 more questions to do after this one

  10. mathstudent55
    • one year ago
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    The new equation just above is the sum of one original equation and a multiple of the other original equation. Now we take the new equation and one of the original equations, and that completes Part A. -3x + 7y = -16 -15x + 19y = -16

  11. mathstudent55
    • one year ago
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    That is the end of part A. We have a new system that is equivalent to the original one.

  12. anonymous
    • one year ago
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    okay so what should i type in for part a?

  13. mathstudent55
    • one year ago
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    Everything we did above.

  14. mathstudent55
    • one year ago
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    Skipping the explanations.

  15. mathstudent55
    • one year ago
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    I'll show you.

  16. anonymous
    • one year ago
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    can it be put into not that because i cant write like that in my assignment

  17. mathstudent55
    • one year ago
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    Part A. -3x + 7y = -16 -9x + 5y = 16 2(-3x + 7y) = 2(-16) -6x + 14y = -32 + -9x + 5y = 16 --------------------- -15x + 19y = -16 -3x + 7y = -16 -15x + 19y = -16

  18. mathstudent55
    • one year ago
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    Ok. The response above is Part A.

  19. anonymous
    • one year ago
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    Okay thank you can you help me with part b too?

  20. mathstudent55
    • one year ago
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    Yes. Now in Part B, we need to show both systems are equivalent to each other and both have the same solution. We could solve both systems, but there is a faster way. We solve only the original system. Then we try the solution of the original system in the second system. If the solution works, that shows the systems are equivalent.

  21. mathstudent55
    • one year ago
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    Let's solve the original system. We can use the addition method. We need to add the two equations together and make sure one of the variables adds to zero.

  22. mathstudent55
    • one year ago
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    Since the first term of the first equation is -3x, if we multiply it by -3, we get 9x. 9x added to the first term of the second equation, -9x, will add to zero. That is what we want, so we start by multiplying the first equation by -3 on both sides. -3x + 7y = -16 ------ *3 on both sides ----- > -9x -21y = 48 -9x + 5y = 16 ------ copied just as it is ---> 9x + 5y = 16 ---------------- -- sum of the two equations --> -16y = 64 -16y = 64 y = -4 Now that we know y = -4, we substitute -4 in the first equation for y. Then we find x. -3x + 7y = -16 -3x + 7(-4) = -16 -3x -28 = -16 -3x = 12 x = -4 x is also -4. The solution of the original system of equations is: x = -4, y = -4

  23. anonymous
    • one year ago
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    so that is our part b correct?

  24. mathstudent55
    • one year ago
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    One more step. We need to check the solution we got for the first system in the second system.

  25. anonymous
    • one year ago
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    ok

  26. mathstudent55
    • one year ago
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    We can solve this system, but it's faster to just check the solution we already know. The question now is: Is the solution x = -4 and y = -4 also a solution to the new system of equations? Let's test (-4, -4) in both equations of the second system. -3x + 7y = -16 -3(-4) + 7(-4) =? -16 12 + (-28) =? -16 -16 = -16 The solution works on the first equation. -15x + 19y = -16 -15(-4) + 19(-4) =? -16 60 + (-76) =? -16 60 - 76 =? -16 -16 = -16 The solution works on the second equation. This just showed that the two systems are equivalent. Part B is done.

  27. anonymous
    • one year ago
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    okay i get it thanks so much!

  28. anonymous
    • one year ago
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    can u help with another? @mathstudent55

  29. anonymous
    • one year ago
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    i just posted it

  30. mathstudent55
    • one year ago
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    You're welcome. I'll look.

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