1. ## Simultaneous Equation trouble

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?

2. You did not do anything wrong.

Using the first equation and your value $\displaystyle y = \frac{14}{23}$:

$\displaystyle 3x + 2y = 8$
$\displaystyle 3x + 2(\frac{14}{23}) = 8$
$\displaystyle 3x + \frac{28}{23} = \frac{184}{23}$ (Multiplying and putting 8 in a different form)
$\displaystyle 3x = \frac{156}{23}$
$\displaystyle x = \frac{156}{69}$
$\displaystyle x = \frac{52}{23}$ (Putting the factor into lowest terms)

Substituting $\displaystyle x = \frac{52}{23}$ and $\displaystyle y = \frac{14}{23}$ into your equations gives the correct answers.

So your problem was not something glaring, as there was no problem at all. If there is something not making sense to you still, let me know.