Originally Posted by

**mark** hi i've been trying to solve a simultaneous equation and i'm sure i've gone wrong somewhere and would be grateful if someone could point out where:

$\displaystyle 5x + 3y = 4$

$\displaystyle 5x^2 - 3y^2 = 8$

i started by making x the determinant in the first equation then subbing it into equation 2 like so

$\displaystyle x = \frac{4-3y}{5}$(1)

$\displaystyle {5}(\frac{4-3y}{5}) - 3y^2 = 8$

then multiplied everthing by 5 to remove the fraction

$\displaystyle 25(4 - 3y)^2 - 15y^2 = 40$ then expanded the squared bracketed term:

$\displaystyle 25(16 - 24y + 9y^2) - 15y^2 = 40$ then carried on with

$\displaystyle 400 - 600y + 225y^2 - 15y^2 = 40$ then

$\displaystyle 360 - 600y + 210y^2 = 0$