need help on this fraction equation 2 <mrow class="MJX-TeXAtom-ORD"> / </mrow> 5

Hint: Start by clearing the denominators. Do you know how to do that?

You said you wanted a step-by-step explanation so here is one: As others have pointed out, you need to find the common denominator (why? If you do not know why, then I'd suggest reviewing the material of whatever book you are using or asking a close friend/teacher). A very easy way to obtain the common denominator is to simply multiply all of the denominators of your expression together.
Your expression is
$\frac{2}{5}=\frac{2}{3}-<!--\; -\; -->\frac{r}{5}.$
How would you get the common denominator here? As mentioned above, you multiply all of the denominators of your expression together: $5\times <!--\; \times \; -->3=15$ . Aha! Your common denominator, then, is 15. How do you make use of the common denominator? You multiply both sides of your original expression by it (why? Again, consult your text or talk to a friend or teacher if you do not know why you should do this). Let's multiply both sides of your original expression by 15 and see what happens:
$15\times <!--\; \times \; -->\left(\frac{2}{5}\right)=15\times <!--\; \times \; -->(\frac{2}{3}-<!--\; -\; -->\frac{r}{5})⟺<!--\; ⟺\; -->\frac{15\times <!--\; \times \; -->2}{5}=\frac{15\times <!--\; \times \; -->2}{3}-<!--\; -\; -->\frac{15\times <!--\; \times \; -->r}{5}.$
You should now perform the actual multiplications and simplify this expression. When you do this, you will get the expression
$6=10-<!--\; -\; -->3r.$
Wow! Suddenly we are actually very close to figuring out what r is. What value of r will make the equation $6=10-<!--\; -\; -->3r$ true? The easiest way to figure this out is to rearrange the equation $6=10-<!--\; -\; -->3r$as follows:
$\begin{array}{}\text{(add }3r\text{ to both sides)}& 6=10-<!--\; -\; -->3r& ⟺<!--\; ⟺\; -->3r+6=10\text{(subtract }6\text{ from both sides)}& & ⟺<!--\; ⟺\; -->3r=4\text{(divide both sides by }3\text{)}& & ⟺<!--\; ⟺\; -->r=\frac{4}{3}.\end{array}$
Thus, we can see that
$\frac{2}{5}=\frac{2}{3}-<!--\; -\; -->\frac{r}{5}$
is true when $r=\frac{4}{3}$