Thread: solve by substitution

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

Originally Posted by mafai44

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

put y second in first equation for example ... (can be another way around)











so there is no solution for this system of equations

Originally Posted by mafai44

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

means x is 4

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...



... use this y in the first line



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.

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

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

(1) ≠ (2)