Substitution Method to solve linear simultaneous equation

[Yazan975]

View attachment 8419

What I have done:

I changed all fractions to common denom and that gave me

5y-5x=1 (1) *I numbered the fractions
5y+2x=5 (2)

Then: 5y=5-2x

Substitute into equation 1
(5-2x)-5x=1
5-7x=1
x=4/7

Thing is my answer says I should be getting x=0

Any hints?

 

[Olinguito]

When you multiply the first equation by $6$ and simplify, you should get $y-5x=1$.