Question

How do you solve the system `y-4/5x=1` and `y-x=1` using substitution?

Answer 1

`x=0`
`y=1`

Explanation:

`y-x=1|+x`
`y=color(red)(1+x)`

`color(red)(y)-4/5x=1`
`color(red)(1+x)-4/5x=1|-1`
`1/5x=0|*5`
`x=color(blue)(0)`

`y-color(blue)(x)=1|+x`
`y-color(blue)(0)=1`
`y=1`