Had a quick search but can't find what I need, so I'll make a new thread instead.

I've got a quick problem here that I can't get my head around, and it's the only simultaneous equation out of 10 that I can't solve, I keep getting answers that satisfy one or the other!

$\displaystyle 3x+2y=8$ (1)

$\displaystyle 4x-5y=6$ (2)

(solving $\displaystyle x$ from 1)

$\displaystyle 3x = 8-2y$ (subtract 2y from both sides)

$\displaystyle x = \frac{8-2y}{3}$ (divide by 3 to find x)

(substituting $\displaystyle x$ into into 2)

$\displaystyle 4(\frac{8-2y}{3})-5y=6$

$\displaystyle \frac{32-8y}{3}-5y=6$ (expanded brackets)

$\displaystyle 32-8y-15y=18$ (multiply both sides by 3 to remove denominator)

$\displaystyle -23y=-14$ (subtract 32 from both sides, $\displaystyle -8y-15y=-23y$)

Which leaves me with $\displaystyle y=\frac{14}{23}$ which of course then does not satisfy the equation very well... I've gone over my working and isolated the problem to this point, I hope it's not something glaring, but what have I done wrong?