I'm trying to solve the following equation for a, my work is as follows.
$d_1=\frac{d_2-\frac{1}{2}at_2^2}{t_2}t_1+\frac{1}{2}at_1^2$
$d_1=\frac{d_2t_1-\frac{1}{2}at_2^2t_1}{t_2}+\frac{1}{2}at_1^2$
$d_1-\frac{d_2t_1}{t_2}=\frac{-\frac{1}{2}at_2^2t_1}{t_2}+\frac{1}{2}at_1^2$
$d_1-\frac{d_2t_1}{t_2}=-\frac{1}{2}at_2t_1+\frac{1}{2}at_1^2$
$d_1=-\frac{1}{2}at_2t_1+\frac{1}{2}at_1^2+\frac{d_2t_1}{t_2}$
$d_1=t_1(-\frac{1}{2}at_2+\frac{1}{2}at_1+\frac{d_2}{t_2})$
$\frac{d_1}{t_1}=-\frac{1}{2}at_2+\frac{1}{2}at_1+\frac{d_2}{t_2}$
$\frac{d_1}{t_1}-\frac{d_2}{t_2}=-\frac{1}{2}at_2+\frac{1}{2}at_1$
$\frac{d_1}{t_1}-\frac{d_2}{t_2}=a(-\frac{1}{2}t_2+\frac{1}{2}t_1)$
$a=\frac{\frac{d_1}{t_1}-\frac{d_2}{t_2}}{-\frac{1}{2}t_2+\frac{1}{2}t_1}$
As you can see I end up with a really messy fraction at the end. Can I rewrite the above as the following?
$a=\frac{d_1-d_2}{t_1t_2(-\frac{1}{2}t_2+\frac{1}{2}t_1)}$
No. You have proceeded correctly thus far. We have $$a = \frac{\frac{d_1}{t_1} - \frac{d_2}{t_2}}{-\frac{t_2}{2}+\frac{t_1}{2}}$$
Consider the numerator: $$\frac{d_1}{t_1} - \frac{d_2}{t_2} = \frac{d_1t_2-d_2t_1}{t_1t_2}$$
The denominator is: $$\frac{t_1-t_2}{2}$$
Now simplify the expression to get: $$a = \frac{2[d_1t_2-d_2t_1]}{t_1t_2[t_1-t_2]}$$