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Math Help - Elimination and substitution

  1. #1
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    Elimination and substitution

    I have this problem: y=1x and y=.5x+4
    I got (2,4) and was wondering if you guys can show me the steps for both elimination and substitution. Also how would I graph this? just put one dot on the graph?
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    Quote Originally Posted by Jubbly View Post
    I have this problem: y=1x and y=.5x+4
    I got (2,4) and was wondering if you guys can show me the steps for both elimination and substitution. Also how would I graph this? just put one dot on the graph?
    How did you get (2,4)? If you try checking your work by substituting, you get

    x=2\Rightarrow y=2

    for the first, and

    x=2\Rightarrow y=5

    for the second. Your point is not on either of the curves, so it certainly could not be in the solution set.
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  3. #3
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    Well, I did 1x=.5x+4
    I subtracted .5x from both sides making 1x and .5x
    now its .5x=4
    divided .5x into 4 and got 2
    so x=2?

    Can you show me a step by step of how your doing it and how your substituting?
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    Quote Originally Posted by Jubbly View Post
    divided .5x into 4 and got 2
    4 divided by 0.5 is 8, not 2:

    \frac4{1/2}=4\cdot\frac21=4\cdot2=8\text.

    The correct solution is (8,\,8)\text.
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  5. #5
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    Oh, sorry about that.

    If you don't mind can you show me the elimination and substitution equation you used for this. Thanks!
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    Certainly. We have

    \left\{\begin{array}{rcl}<br />
y&=&x\\<br />
y&=&\frac12x+4<br />
\end{array}\right.

    Substitution:

    Substituting x=y into the second equation produces

    y=\frac12y+4,

    and solving for y gives

    \frac12y=4\Rightarrow y=8.

    Back-substituting, we get x=8.


    Elimination:

    Let's first rearrange the equations a little,

    \left\{\begin{array}{rcl}<br />
y-x&=&0\\<br />
2y-x&=&8<br />
\end{array}\right..

    Subtract the first equation from the second,

    \left\{\begin{array}{rcl}<br />
y-x&=&0\\<br />
y&=&8<br />
\end{array}\right.

    and subtract the second equation from the first:

    \left\{\begin{array}{rcl}<br />
 -x&=&-8\\<br />
 y&=&8<br />
 \end{array}\right.\Rightarrow\left\{\begin{array}{  rcl}<br />
  x&=&8\\<br />
  y&=&8<br />
  \end{array}\right.
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