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

The solutions are the pairs $\left(x , y\right) = \left(- 8 , 0\right)$ and $\left(x , y\right) = \left(10 , 6\right)$

#### Explanation:

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

${y}^{2} = {\left(3 y - 8\right)}^{2} - 64$

Finally after some algebraic calculations you get

${y}^{2} = 6 y \implies y \cdot \left(y - 6\right) = 0 \implies {y}_{1} = 0 \mathmr{and} {y}_{2} = 6$

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

and for

$y = 6$ you get $x = 10$

A graphical solution confirms the results