Math Help - solve by substitution

1. solve by substitution

4x-2y=14
y=2x+7

2. Originally Posted by mafai44
4x-2y=14
y=2x+7
put y second in first equation for example ... (can be another way around)

$4x -2y= 14$

$2x -y= -7$

$D=\begin{vmatrix}
4 &-2 \\
2 &-1
\end{vmatrix} = -4+4 = 0$

$D_x= \begin{vmatrix}
14 &-2 \\
-7 & 1
\end{vmatrix} = 14-14 = 0$

$D_Y= \begin{vmatrix}
4 &14 \\
2& -7
\end{vmatrix} =-28-28 =-56$

so there is no solution for this system of equations

3. Originally Posted by mafai44
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.

4. 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

5. {4x - 2y = 14 ⇔ 4x = 2y + 14 ⇔ 4x - 14 = 2y ⇔ y = (4x - 14)/2 = 2x - 7 (1)
{y = 2x + 7 (2)

(1) ≠ (2)