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Math Help - Substitution Method

  1. #1
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    Substitution Method

    Solve and check each of the following problems using the substitution method.

    1. 3x + y = 5
    -2x + 3y = 4

    2. y = 5x
    2x + y = 28

    3. x + 3y = 23
    3x - 2y = 3

    4. x = 1/5y
    3x + 2y = 26

    I need help in figuring out the answers to these problems. If you could please explain to me how to solve it I'd appreciate it very much.
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  2. #2
    Bar0n janvdl's Avatar
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    1. y = 5 - 3x
    Set this into (-2x + 3y = 4)

    -2x + 3(5 - 3x) = 4
    -2x + 15 - 9x = 4
    -11x = -11
    x = 1

    Set x = 1 into (y = 5 - 3x)
    y = 5 - 3(1)
    y = 2

    ---------------------------------
    2. y = 5x
    Set y = 5x into the 2nd equation (2x + y = 28)
    2x + 5x = 28
    x = 4
    Set x = 4 into the 1st equation
    y = 5(4) = 20

    ----------------------------------

    3. x + 3y = 23

    x = 23 - 3y
    Set into 2nd equation(3x - 2y = 3)
    3(23 - 3y) - 2y = 3
    69 - 9y - 2y = 3
    -11y = -66
    Thus y = 6
    Set y = 6 into 1st equation
    x + 3(6) = 23
    x = 5

    ------------------------------------

    4. x = 1/5y
    3x + 2y = 26

    Set (x = 1/5y) into 2nd equation

    3(1/5y) + 2y = 26
    3/5y + 2y = 26
    3y + 10y + 130
    13y = 130
    y = 10

    Set y = 10 into 1st equation

    x = 1/5 x 10
    x = 2

    ------------------------------------------------

    Hope this helps....
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  3. #3
    Bar0n janvdl's Avatar
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    Never insert a calculated value into the same equation you got its value from. You'll only get an answer stating that the LHS = RHS. Which is true, but its not what you want.
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  4. #4
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    Talking

    Thanks so much Janvdl!
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