Question 737787
2x-3y=-6 gives {{{x=(3/2)y-3}}}

so }}}x^2+y^2=25}}}  becomes {{{((3/2)y-3)^2+y^2=25}}}

so {{{(9/4)y^2-9y+9 +y^2=25}}} 

and {{{(13/4)y^2-9y-16=0}}} 

Using the quadratic formula with a=13/4, b=-9 c=-16

Then (((y=(9 +-sqrt(81-4(13/4)(-16)))/2(13/4)=(9+-sqrt(289))/(13/2)=(9+-17)/(13/2)}}} which gives y=4 or y=-16/13

substitute those into the linear equation to find x.
y=4 gives {{{x=(3/2)(4)-3=6-3=3}}}
y=-16/13 gives {{{x=(3/2)(-16/13)-3=-24/13-3=-(63/13)}}}