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Destinyyyy

  • one year ago

Can someone tell me what I did wrong? If possible, solve the system of equations. Use any method.

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  1. Destinyyyy
    • one year ago
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    3x+5y=6 (1) 5x+7y=6 (2) -5(3x+5y=6) -15x-25y=-30 3(5x+7y=6) 15x+21y=30 -4y=0 y=0

  2. pooja195
    • one year ago
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    3(5x+7y=6) 15x+21y=30

  3. pooja195
    • one year ago
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    Theres something wrong with that.Keep in mind OPPOSITES CANCEL.

  4. Destinyyyy
    • one year ago
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    What? O.o

  5. pooja195
    • one year ago
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    15+15= 30 we want to cancel the x value

  6. pooja195
    • one year ago
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    -5(3x+5y=6) -15x-25y=-30 same here -15+-15=-30

  7. Destinyyyy
    • one year ago
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    -15x-25y=-30 15x+21y=30 4y=0

  8. pooja195
    • one year ago
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    what did u do?

  9. Destinyyyy
    • one year ago
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    I have no clue what you doing. I am doing substitution method. The x is not suppose to just disappear like you have.

  10. pooja195
    • one year ago
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    -5(3x+5y=6)<---wrong -15x-25y=-30 5(3x+5y=6) 15+25y=30 15+25y=30 -15x-25y=-30

  11. pooja195
    • one year ago
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    Get it?

  12. anonymous
    • one year ago
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    the mistake is in the second eq: \[3(5x+7y=6)\] you get: \[15x+21y=18\] because 3*6=18, not 30 !!

  13. pooja195
    • one year ago
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    They made a mistake on both.

  14. pooja195
    • one year ago
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    There shouldnt even be 2 sets of equations

  15. pooja195
    • one year ago
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    @Destinyyyy 15+25y=30 -15x-25y=-30 solve this

  16. Destinyyyy
    • one year ago
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    @Greg_D I see that now thank you.. Cannot believe I did that -.-

  17. pooja195
    • one year ago
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    Im gonna let them finish up.

  18. pooja195
    • one year ago
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    But there should not be 2 sets.

  19. anonymous
    • one year ago
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    it was a system of 2 eqs from the begining of the problem... the only mistake i can see is the one i pointed out earlier 3*6 is not 30

  20. Destinyyyy
    • one year ago
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    I have no clue what your talking about @pooja195

  21. Destinyyyy
    • one year ago
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    I know the final answer is (-3,3) ... But im not sure how they get -3

  22. anonymous
    • one year ago
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    @pooja195 there are a lot of ways to solve the system, the one @Destinyyyy is using is fine, there was just that little mistake

  23. anonymous
    • one year ago
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    once you get y=3, just replace that in any of the original equations...

  24. Destinyyyy
    • one year ago
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    Im only allowed to use elimination, graph, substitution.

  25. Destinyyyy
    • one year ago
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    Thank you @Greg_D

  26. anonymous
    • one year ago
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    so you arent allow to use "any" method

  27. pooja195
    • one year ago
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    -5 (3x+5y=6) 3(5x+7y=6 ) -15x-25y=-30 15x+21y=18 -4y=-12 divde both sides by -4 y=3 15x+21(3)=18 15x+63=18 subtract 63 both sides 15x=-45 divde both sides by 15 x=-3

  28. Destinyyyy
    • one year ago
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    Its any method of the three. I should of said that

  29. pooja195
    • one year ago
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    notice how there is not 2 sets of equations in no way shape or for should u have 2 sets of equations even when doing elimination

  30. Destinyyyy
    • one year ago
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    Um the 3 is suppose to go into one of the original equation.

  31. Destinyyyy
    • one year ago
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    I know how to solve it. I just didnt see where I messed up. @Greg_D showed me my mistake.

  32. anonymous
    • one year ago
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    look if: \[y=3\] and: \[3x+5y=6\] then: \[3x+15=6\] you can get x from there...

  33. anonymous
    • one year ago
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    the same is done in the last part of @pooja195 last response :)

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