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-   -   Solving Linear Systems by substitution! HELP! (https://www.askmehelpdesk.com/showthread.php?t=489597)

  • Jul 18, 2010, 08:09 PM
    MathMomma
    Solving Linear Systems by substitution! HELP!
    I CAN NOT FIGURE OUT THIS QUESTION the solution is given as (-1,-5) and I am not getting those two answers!

    4(1-2x) = 3(3-y) - 12
    2(3x-1)-(y+4) = - 7
  • Jul 18, 2010, 09:00 PM
    KISS

    There is a trick that cuts out a LOT of the algebra.

    2(3x-1)-(y+4) = - 7

    See the -(y+4) --> Make it 3-y

    First:
    2(3x-1) -4-y = -7

    Add 7 to both sides

    2(3x-1) + (3-y) = 0

    Check original solution since you have it.

    2(3(-1)-1) + (3+5) =0; Good, didn't change answer

    Solve 2(3x-1) + (3-y) = 0 for 3-y

    3-y = -2(3x-1)

    Now substitute the -2(3x-1) for 3-y into your first equation.

    Who says you have to substitute y?

    I know, it's a trick.
  • Jul 18, 2010, 09:50 PM
    KISS

    Your # 1 eqn is: 4(1-2x) = 3(3-y) - 12

    substituting:

    4(1-2x) = 3(-2(3x-1)) - 12

    check with -1 for x since we have it.

    4(1+2) = 3(-2(-4)) - 12
    12 = 24 -12

    So, you just have to do this algebraic mess:

    4(1-2x) = 3(-2(3x-1)) - 12

    to find x.

    4-8x = -6(3x-1) - 12

    My way:

    0 = -6(3x-1) - 12 -4 + 8x

    Count x's -18 +8 or -10x
    Count #'s: 6 - 16 or -10

    0 = -10x-10

    10x = -10 ---> x = -1

    Substitute into 2(3x-1)-(y+4) = - 7

    make it equal to zero; e.g. add 7 to both sides

    2(-4) -y - 4 +7 = 0
    Count because it's easier.

    y's = -1
    #'s -8 -4 +7 = -12-7 = -5

    -1y - 5 = 0
    -y = 5
    y = -5
  • Jul 18, 2010, 09:55 PM
    KISS

    You can do it the hard way. I can find the mistakes. See a full solution above.
  • Jul 18, 2010, 10:09 PM
    Unknown008

    Maybe cleaning the equations from the very start will help?

    4(1-2x) = 3(3-y) - 12
    2(3x-1)-(y+4) = - 7

    These, when expanding become:
    4 - 8x = 9 - 3y - 12
    6x - 2 - y - 4 = -7

    Now, simplify both;
    - 8x = -3y -7
    6x - y = -1

    Hm... too many negatives. I multiply both by -1;
    8x = 3y + 7
    y - 6x = 1

    Now, you can make y the subject of the formula to sub into the first equation, like this:
    y = 1 + 6x

    8x = 3(1 + 6x) + 7
    8x = 3 + 18x + 7

    Now, just simplify to get x = -1.

    Then, replace that value into " y - 6x = 1 " or " 8x = 3y + 7 " or " y = 1 + 6x " to get the value of y.

    There isn't much brackets to worry about, so long as you keep them simple :)
  • Jul 18, 2010, 10:18 PM
    KISS

    And, of course there are a few other ways to do it:

    The second eqn has y nearly all by itself, so you can solve for y in that eqn. But you can also solve for -y and substitute into the first.

    You don't have to find x and then y. You can find y then x.

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