Question:
$$ \text{It is given that } y= \frac{3a+2}{2a-4} \text{and }x= \frac{a+3}{a+8} \\ $$
$$ \text{Express } y \text{ in terms of } x. $$
From using $x$ to solve for $a$, I discovered that $$ a = \frac{8x-3}{1-x} $$
Then I proceeded to substitute $a$ into $y$. I did this twice to ensure no mistakes are made, and my final answer for both was
$$ \frac{22x-7}{20x-10} $$
There's a problem, the correct answer is $$ \frac{7-22x}{10-20x} $$
This makes me want to cry, more so because I checked it twice and I was very careful about my working out, here it is:
$$ \frac{2+ 3(\frac{8x-3}{1-x})}{ 2(\frac{8x-3}{1-x}) -4 } $$
$$\rightarrow{}$$
$$ \frac{(\frac{2(1-x) + 3(8x-3)}{1-x})}{(\frac{-4(1-x)+2(8x-3)}{1-x})} $$
$$\rightarrow{}$$
$$ \frac{2(1-x) + 3(8x-3)}{1-x} * \frac{1-x}{-4(1-x)+2(8x-3)} $$
$$\rightarrow{}$$
$$ \frac{22x-7}{1-x} * \frac{1-x}{20x-10} $$
The $(1-x)$'s cancel out $$\rightarrow{}$$
$$ \frac{22x-7}{20x-10} $$
Can someone please tell me as to what I did incorrectly in the process? Thank you in advance!