Solve the Simultaneous equations 2x + 4y = 9 and
4x2 + 16y2 = 20x + 4y - 19

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.