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Math Help - Solving by elimination or subsitution.

  1. #1
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    Solving by elimination or subsitution.

    Hey guys it has been so long since I have done a lot of algebra and I am working on this take home assignment and am trying hard to figure out how to tackle some of these.

    The instructions state: "Solve the following systems of equations by elimination, substitution, or graphically.

    A 3x-y=6 & x+4y=3 B y+2=-x^2+1 & x2+1=2y+6 C x-2y+3z=10 & 2x-3y-z=8 & 5x-9y+8z=20


    For A I got X being 2 1/3 and Y being 3. I am struggling on the other two. Any help is appreciated.
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  2. #2
    MHF Contributor red_dog's Avatar
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    B)
    \left\{\begin{array}{ll}y+2=-x^2+1\\2y+6=x^2+1\end{array}\right.
    By elimination:
    Adding the equations we'll eliminate x^2 and we get
    3y+8=2\Rightarrow 3y=-6\Rightarrow y=-2
    Now, plug y in the first (or second) equation and find x.

    By substitution:
    From the first equation y=-x^2-1.
    Substitute y in the second equation and solve for x.

    Graphically:
    Attached Thumbnails Attached Thumbnails Solving by elimination or subsitution.-grafic2.jpg  
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  3. #3
    MHF Contributor red_dog's Avatar
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    C)
    \left\{\begin{array}{lll}x-2y+3z=10\\2x-3y-z=8\\5z-9y+8z=20\end{array}\right.
    By elimination:
    We'll eliminate z:
    Multiply the second equation by 3 and add to the first:
    7x-11y=21 (1)
    Multiply the second by 8 and add to rhe third:
    21x-33y=84 (2)
    Now, multiply (1) by -3 and add to (2):
    0=21, which is obviously false. So the system has no solution.

    By substitution:
    From the first equation x=2y-3z+10
    Substitute x in the others two equations and you'll obtain a system in y and z. Apply the same method to this system.
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  4. #4
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    Quote Originally Posted by Ty Durdan View Post
    ...

    The instructions state: "Solve the following systems of equations by elimination, substitution, or graphically.

    A 3x-y=6 & x+4y=3 ...

    For A I got X being 2 1/3 and Y being 3.
    Hello,

    I don't know how you did A) - but:

    \left\{\begin{array}{ll}3x-y=6 \\ x+4y=3 \end{array}\right. . Multiply the 1rst equation by 4. You'll get:

    \left\{\begin{array}{ll}12x-4y=24 \\ x+4y=3 \end{array}\right. and then add columnwise(?):

    13x = 27~\Longleftrightarrow~ \boxed{x=\frac{27}{13}}. Plug in this value into one of the equations to calculate y:

    3x-y=6~\Longrightarrow~3 \cdot \frac{27}{13}-y=6~\Longrightarrow~ y=\frac{81}{13}-\frac{78}{13}=\boxed{\frac{3}{13}}
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