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

Feb 7, 2018

Solution is $\left(8 , 0\right)$. For details please see below.

#### Explanation:

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

$10 {x}^{2} - 25 \left(2 x - 16\right) - 100 x = - 160$

or $10 {x}^{2} - 50 x + 400 - 100 x = - 160$

or $10 {x}^{2} - 150 x + 560 = 0$

or ${x}^{2} - 15 x + 56 = 0$

or $\left(x - 7\right) \left(x - 8\right) = 0$

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

When $x = 7$, we have ${y}^{2} = 2 \times 7 - 16 = - 2$, but this is not possible and hence, no solution.

if $x = 8$, we have ${y}^{2} = 2 \times 8 - 16 = 0$ i.e. $y = 0$

Hence, only solution is $\left(8 , 0\right)$

Observe that while first equation representsa hyperbola, second equation is of a parabola. As parabola touches one arm of hyperbola at $\left(8 , 0\right)$, 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]}