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Math Help - Substitution

  1. #1
    Junior Member
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    Substitution

    I'm really lost with these substitution problems everytime I try to solve it it comes out with a weird number and doesn't work when double checking. Thanks for your help! Also is there anyway to do this on a graphing calculator?

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


    y=-3x+12
    y=-7x-36


    -6y+8x=36
    x=4y+11

    y=6x-5
    y=-x+9
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  2. #2
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    Quote Originally Posted by Jubbly View Post

    3x+5y=2
    x+4y=-4
    The first step is finding a variable in one of the equations to solve for. I like how the x variable in your second equation has no coefficient, so I think I will solve for it:

     x+4y=-4
     x=-4y-4 ***

    Now that I have x in terms of y, I'm going to plug that x value into the OTHER equation:

     3x+5y=2
     3( -4y-4 )+5y=2

    So now we have an equation with just one variable. This should be easy to solve if we're careful with distribution and minus signs:

     3(-4y-4 )+5y=2
     -12y -12 +5y=2
     -7y -12 =2
     -7y =14
     y =-2

    Alrighty, we have a y value. Now we want the x value. If only we had x written down in terms of y someplace .. oh hey, peek up at the *** above:

     x=-4y-4
     x=-4(-2)-4
     x=8-4
     x=4

    So our solution may be (4, -2).

    Let's check just to make certain we're correct:

    3x+5y=2
    3(4)+5(-2)=2
    12-10=2
    2=2

     x+4y=-4
     (4)+4(-2)=-4
     4-8=-4
     -4=-4

    We've got it. And more importantly, we know we are right because of our check.
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  3. #3
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    Thanks so much for your help!

    For step one how did you get -4y-4? Nevermind this question, I figured it out!
    Thanks so much for your help still!

    I also have another question if you don't mind how do I solve for this equation y=-3x+12 and y=-7x-36 they both have Ys and I don't know how to do with them
    Last edited by Jubbly; February 7th 2009 at 12:04 PM.
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  4. #4
    Newbie stmsnyder1's Avatar
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    RE: For Step 1 How Did You Get -4y-4

    Hi,


    x+4y=-4

    It actually should be x=-4-4y. Take 4 y and subtract it.
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  5. #5
    Member TheMasterMind's Avatar
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    Quote Originally Posted by Jubbly View Post
    I'm really lost with these substitution problems everytime I try to solve it it comes out with a weird number and doesn't work when double checking. Thanks for your help! Also is there anyway to do this on a graphing calculator?

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


    y=-3x+12
    y=-7x-36


    -6y+8x=36
    x=4y+11

    y=6x-5
    y=-x+9
    Linear systems with variables isolated(no co-efficient infront) should be solved with the substitution method rather then graphing or the elimination method.

    y=-3x+12

    y=-7x-36

    you have isolated y in both equations, this you can substitute it into one equation to solve for a variable(in this case x)

    -7x-36=-3x+12

    <br />
-7x+3x=12+36

    -4x=48

    x=48/-4

    \boxed{<br />
x=-12}

    Now that you have the value of x you can substitute it into one of the original equations and solve for the other variable (in this case y)

    y=-3(-12)+12

    y=36+12

    \boxed{y=48}

    To check substitute the numbers back into the equation and see if the LHS and RHS are equal to each other. You can see now that this is the best way to solve all of the systems of equations you have there!
    Last edited by TheMasterMind; February 7th 2009 at 02:56 PM.
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