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Math Help - subs method / elimination

  1. #1
    Junior Member
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    subs method / elimination

    is this where you pick numbers and plug in to see what works?


    substitution problem
    6x+8y=-16
    -7x+y=60


    elimination method
    5x+3y=-13
    7x-2y=17
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  2. #2
    Senior Member ecMathGeek's Avatar
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    Quote Originally Posted by rj2001 View Post
    is this where you pick numbers and plug in to see what works?


    substitution problem
    6x+8y=-16
    -7x+y=60


    elimination method
    5x+3y=-13
    7x-2y=17
    Substitution refers to "substituting" one variable in one equation for the same variable in another.
    For example:
    In -7x + y = 60, let's solve for y: y = 7x + 60 (I added 7x to both sides)

    With this, we have
    y = 7x + 60
    6x + 8y = -16

    Since y = 7x + 60, we can plug this in for the "y" in 6x + 8y = -16, and we get:
    6x + 8(7x + 60)= -16

    Now we have one equation with only "x"s. Solving for x, we get:
    6x + 56x + 480 = -16
    62x = -496
    x = -8

    Now we can solve for the value of y in y = 7x + 60 since x = -8
    y = 7(-8) + 60 = -56 + 60 = 4

    So we have x = -8, y = 4, which is the point (-8,4)
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  3. #3
    Senior Member ecMathGeek's Avatar
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    Quote Originally Posted by rj2001 View Post
    is this where you pick numbers and plug in to see what works?


    substitution problem
    6x+8y=-16
    -7x+y=60


    elimination method
    5x+3y=-13
    7x-2y=17
    Elimination refers to "eliminating" one variable from both equations by "adding" both equations together. Usually this also involves multiplying one or both equations by some constant such that if we "add" the two equations one of the variables is removed.

    For example, in the system of equations
    5x + 3y = -13
    7x - 2y = 17

    If we want to eliminate the "y"s, we can multiply the top equation by 2 and the bottom equation by 3 and then add the equations together.
    2(5x + 3y) = 2(-13)
    3(7x - 2y) = 3(17)

    These become
    10x + 6y = -26
    21x - 6y = 51

    Adding both equations together ("x"s with "x"s, "y"s with "y"s, and constants with constants) we get:
    (10x + 21x) + (6y - 6y) = (-26 + 51)
    31x + 0y = 25
    31x = 25
    x = 25/31

    Now we solve for y in 10x + 6y = -26
    6y = -10x - 26
    y = -5/3*x - 13/3

    And plugging in x = 25/31, we get
    y = -5/3(25/31) - 13/3 = -125/93 - 403/93 = -528/93 = -176/31

    (25/31,-176/31) ... Check to make sure the numbers you put are correct. I might have made a mistake, but I doubt your answer should involve such complex fractions.
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