time speed distance Two horses start simultaneously towards each other and meet after $3h 20 min$. How much time will it take the slower horse to cover the whole distance if the first arrived at the place of departure of the second $5 hours$ later than the second arrived at the departure of the first.
MY TRY::
Let speed of 1st be a kmph and 2nd be b kmph
Let the distance between A and B be d km
d = 10a/3 + 10b/3
and
d/a - d/b = 5 
now i cant solve it. :(
Spoiler: The answer is $10$ hours.

First, let's identify what you actually want to solve for, which is $\frac{d}{b}$. Solve for $a$ in your first equation: $a = 3/10 d - b$ and substitute into the second equation
$$
\frac{d}{\frac{3}{10} d - b} - \frac{d}{b} = 5\\
db- d\left(\frac{3}{10} d - b\right) = 5b\left(\frac{3}{10} d - b\right)\\
d\left(2b - \frac{3}{10}d \right) = \frac{3}{2}bd - 5b^2 \\
 \frac{3}{10}d^2- \frac{1}{2}d b -5b^2 = 0
$$
then, dividing by $b^2$
$$
\frac{3}{10}\left(\frac{d}{b}\right)^2 - \frac{1}{2}\frac{d}{b} - 5  =0 \\
3\left(\frac{d}{b}\right)^2 - 5\frac{d}{b} - 50  =0 
$$
which is a quadratic in the variable you want to solve for.
