eq #1 says that 7y^2=5x^2-20x+3, so plug that into eq #2 and you have
5x^2+3(5x^2-20x+3) = 209
20x^2-60x-200 = 0
x^2-3x-10 = 0
(x-5)(x+2) = 0
And I think you can take it from there, no?
Complete the following for each system that involves one or more quadratic equations:
Solve the system algebraically
7y^2-5x^2+20x=3
5x^2+21y^2=209
2 answers
yes, thanks i can :)