Question

How do you solve the system `3x^2-5y^2+2y=45` and `y=2x+10`?

Answer 1

Solution set is `(-4.2,1.6)` and `(-7.3,-4.6)`

Explanation:

To solve the system of equations `3x^2-5y^2+2y=45` and `y=2x+10`, just substitute value of `y` from latter equation in the former equation.

i.e. `3x^2-5(2x+10)^2+2(2x+10)=45`

or `3x^2-5(4x^2+40x+100)+4x+20=45`

or `-17x^2-196x-525=0` or `17x^2+196x+525=0`

Hence `x=(-196+-sqrt(196^2-4*17*525))/34`

= `(-196+-sqrt(38416-35700))/34`

= `(-196+-sqrt2716)/34=(-196+-52.12)/34`

i.e. `x=-4.2` or `-7.3`

and `y=2xx(-4.2)+10=1.6` or `y=2xx(-7.3)+10=-4.6`

and Solution set is `(-4.2,1.6)` and `(-7.3,-4.6)`

This is intersection of a hyperbola and line.

graph{(3x^2-5y^2+2y-45)(2x+10-y)=0 [-20, 20, -10, 10]}