Question

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

Answer 1

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]}