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

1 Answer
Apr 22, 2018

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]}