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ineedyouubiebs

  • 2 years ago

Where do the graphs of the following equations intersect: x-y=3 x=-2y-6 please someone who is actually willing to help me i struggle on this

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  1. omoshiroi
    • 2 years ago
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    ok. first you need to have one of the equations in terms of x or y. Lets take x-y=3. solve for x ineedyouubiebs

  2. ineedyouubiebs
    • 2 years ago
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    so it doesnt matter which one i start with?

  3. omoshiroi
    • 2 years ago
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    no, it does not matter. but x-y=3 is easier to work with.

  4. Eminy
    • 2 years ago
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    Solve for y, x-y=3 -x -y=-x+3 times -1 y=x-3 x=-2y-6 -6 x-6=-2y divde buy -2 -1/2x+3=y or y=-1/2x+3

  5. ineedyouubiebs
    • 2 years ago
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    would it be (-2y-6)-y=3 ?? please correct me if im wrong

  6. dpaInc
    • 2 years ago
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    @ineedyouubiebs , that would be correct... ^^^

  7. ineedyouubiebs
    • 2 years ago
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    omgggg yesss!!

  8. omoshiroi
    • 2 years ago
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    yes that is correct

  9. ineedyouubiebs
    • 2 years ago
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    @dpaInc and how would i add like terms to that...i mean would i have to distribute?

  10. dpaInc
    • 2 years ago
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    technically you are distributing a "1" in 1(-2y-6)-y=3 but since it is only a positive 1, just get rid of the parenthesis to get: -2y - 6 - y = 3 now combine like terms and solve....

  11. ineedyouubiebs
    • 2 years ago
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    y=-3

  12. dpaInc
    • 2 years ago
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    correct... now solve for x...

  13. ineedyouubiebs
    • 2 years ago
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    and how would i do that?

  14. dpaInc
    • 2 years ago
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    you have x = -2y - 6 from the beginning....

  15. dpaInc
    • 2 years ago
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    just plug in the value that you found for y: x = -2(-3) - 6

  16. ineedyouubiebs
    • 2 years ago
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    x=6-6

  17. dpaInc
    • 2 years ago
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    simplify that....

  18. ineedyouubiebs
    • 2 years ago
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    x=0

  19. dpaInc
    • 2 years ago
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    you're basically done... and whenever you need to see if your answer is correct or not, plug in the values of both x and y into BOTH equations. if they both turn out to be true statements, your answer is correct.

  20. ineedyouubiebs
    • 2 years ago
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    okay thank you soo much=D can you help me in another problem?

  21. dpaInc
    • 2 years ago
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    sure.... close this question and post a new one.... :)

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