SOLUTION: How do I use the substitution method to solve the following problem: 3x^2-20y^2-12x+80y-96=0 and 3x^2+20y^2=80y+48?

Question 665440: How do I use the substitution method to solve the following problem: 3x^2-20y^2-12x+80y-96=0 and 3x^2+20y^2=80y+48?
Answer by mananth(16946) (Show Source): You can put this solution on YOUR website!
3x^2-20y^2-12x+80y-96=0
3x^2-20y^2-12x+(80y+48) -144=0
Substitute 80y+48 by 3x^2+20y^2
3x^2-20y^2-12x+3x^2+20y^2-144=0
6x^2+12x-144=0
/6
x^2+2x-24=0
(x+6)(x-4)=0
x=-6 or 4
Plug the values of x to get the values of y
and 3x^2+20y^2=80y+48?
x=4
3(16)+20y^2=80y+48
48+20y^2=80y+48
20y^2=80y
/y
20y=80
/20
y=4
Similarly find the other value of y
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