well (no, not again) I have to use substitution method for this one.
Alright here goes nothing.
2x+4y=9 and 4x^2 +16y^2=20x+4y-19
2x+4y=9
2x=9-4y
x=9-4y/2 ......this is equation 1
substituting value of x in 4x^2 + 16y^2=20x+4y-19 ,
4(9-4y/2)^2+ 16(9-4y/2)^2 = 20(9-4y/2)^2+4y-19
=324+64y^2-144y + 324+64y^2-144y = 71-36y
=648+128y^2-288y=71-36y
= 648-71+128y^2-288y+36y
= 577+128y^2-252y
solving the above equation using quadratic formula , we get y=2.56
substituting value of y in equation 1 , we get
2x+4(2.56)=9
2x= 9-4(2.56)
x=0.62 or -0.62
Thank you very much lol , this problem was so huge that it made my day.
