How do you solve the system #10x^2-25y^2-100x=-160# and #y^2-2x+16=0#?

1 Answer
Feb 7, 2018

Answer:

Solution is #(8,0)#. For details please see below.

Explanation:

We solve them as simultaneous equations. Here second equation gives us #y^2=2x-16# and putting this in first equation we get

#10x^2-25(2x-16)-100x=-160#

or #10x^2-50x+400-100x=-160#

or #10x^2-150x+560=0#

or #x^2-15x+56=0#

or #(x-7)(x-8)=0#

i.e. #x=7# or #x=8#

When #x=7#, we have #y^2=2xx7-16=-2#, but this is not possible and hence, no solution.

if #x=8#, we have #y^2=2xx8-16=0# i.e. #y=0#

Hence, only solution is #(8,0)#

Observe that while first equation representsa hyperbola, second equation is of a parabola. As parabola touches one arm of hyperbola at #(8,0)#, that is the only solution.

graph{(10x^2-25y^2-100x+160)(y^2-2x+16)=0 [1.375, 11.375, -1.98, 3.02]}