Question

How do you solve `x^2 - 2y = 1` and `x^2 + 5y = 29` using substitution?

Answer 1

`x=3`
`y=4`

Explanation:

`x^2-2y=1` → `x^2=1+2y`
`x^2+5y=29`

Then we substitute the `x^2` in the second equation with `1+2y`.

`1+2y+5y=29`
`7y=28`
`y=4`

If `y` is 4 then
`x^2=1+2xx4`
`x^2=9`
`x=sqrt9=3`