Question

How do you solve `y^2=x^2-64` and `3y=x+8` using substitution?

Answer 1

The solutions are the pairs `(x,y)=(-8,0)` and `(x,y)=(10,6)`

Explanation:

Solve `3y=x+8` with respect to `x` and substitute in `y^2=x^2-64`
hence you get

`y^2=(3y-8)^2-64`

Finally after some algebraic calculations you get

`y^2 = 6 y=>y*(y-6)=0=>y_1=0 or y_2=6`

Hence for `y=0` you get `x=-8`

and for

`y=6` you get `x=10`

A graphical solution confirms the results