How do you solve #y^2=x^2-64# and #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)# and #(x,y)=(10,6)#

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

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