4x-2y=14

y=2x+7

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- Aug 24th 2010, 12:15 PMmafai44solve by substitution
4x-2y=14

y=2x+7 - Aug 24th 2010, 12:37 PMyeKciM
put y second in first equation for example ... (can be another way around)

$\displaystyle 4x -2y= 14 $

$\displaystyle 2x -y= -7 $

$\displaystyle D=\begin{vmatrix}

4 &-2 \\

2 &-1

\end{vmatrix} = -4+4 = 0 $

$\displaystyle D_x= \begin{vmatrix}

14 &-2 \\

-7 & 1

\end{vmatrix} = 14-14 = 0 $

$\displaystyle D_Y= \begin{vmatrix}

4 &14 \\

2& -7

\end{vmatrix} =-28-28 =-56 $

so there is no solution for this system of equations :D - Aug 24th 2010, 12:38 PMArchie Meade
$\displaystyle x+3=7$ means x is 4

$\displaystyle x+y=7$ 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...

$\displaystyle 4x-2y=14$

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

$\displaystyle 4x-2(2x+7)=14$

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. - Aug 24th 2010, 10:25 PMIsomorphism
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 - Aug 24th 2010, 11:43 PMdevdel
{4x - 2y = 14 ⇔ 4x = 2y + 14 ⇔ 4x - 14 = 2y ⇔ y = (4x - 14)/2 = 2x - 7 (1)

{y = 2x + 7 (2)

(1) ≠ (2)