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Math Help - solve by substitution

  1. #1
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    solve by substitution

    4x-2y=14
    y=2x+7
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  2. #2
    Senior Member yeKciM's Avatar
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    Quote Originally Posted by mafai44 View Post
    4x-2y=14
    y=2x+7
    put y second in first equation for example ... (can be another way around)

    4x -2y= 14

    2x -y= -7

    D=\begin{vmatrix}<br />
4 &-2 \\ <br />
2 &-1 <br />
\end{vmatrix} = -4+4 = 0

     D_x= \begin{vmatrix}<br />
14 &-2 \\ <br />
-7 & 1<br />
\end{vmatrix} = 14-14 = 0

    D_Y= \begin{vmatrix}<br />
4 &14 \\ <br />
 2& -7<br />
\end{vmatrix} =-28-28  =-56

    so there is no solution for this system of equations
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  3. #3
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    Quote Originally Posted by mafai44 View Post
    4x-2y=14
    y=2x+7
    x+3=7 means x is 4

    x+y=7 does not tell us what x or y are

    If x=0, y=7
    If x=1, y=6 etc

    That's the scenario you have on the first line... 4x-2y=14.

    If we knew what y is "in terms of x", we'd then have.... a certain number of x=value

    The 2nd line achieves this situation if you use the given y back into the first line...

    4x-2y=14

    y=2x+7... use this y in the first line

    4x-2(2x+7)=14

    Unfortunately, both of your equations are the equations of seperate parallel lines,
    so substituting does not lead to a solution as there isn't any.
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  4. #4
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    you can also see that there is no solution by basic manipulation.

    Suppose, on the contrary, there was a solution x = a and y = b. Then 4a - 2b = 14 and b = 2a + 7. But 4a - 2b = 14 means 2(2a - b) = 14 which gives us 2a - b = 7.

    So we have 2a - b = 7 and b = 2a + 7.

    From the latter equation above, 7 = b - 2a = -(2a - b) = -7. But this is not possible! So our assumption that "a solution to our system of equations exists" must have been wrong. So we have no solutions for 4x-2y=14 and y=2x+7
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  5. #5
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    {4x - 2y = 14 ⇔ 4x = 2y + 14 ⇔ 4x - 14 = 2y ⇔ y = (4x - 14)/2 = 2x - 7 (1)
    {y = 2x + 7 (2)

    (1) ≠ (2)
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