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

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Answer 1 · 2016-03-10T16:23:49

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

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

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

Finally after some algebraic calculations you get

$y^{2} = 6 y \Rightarrow y \cdot (y - 6) = 0 \Rightarrow y_{1} = 0 or y_{2} = 6$ $y^2 = 6 y=>y*(y-6)=0=>y_1=0 or y_2=6$

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

and for

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

A graphical solution confirms the results

enter image source here

How do you solve y2=x2−64y^2=x^2-64 and 3y=x+83y=x+8 using substitution?