What happens when you add x to 1 3 </mfrac> x ? I am dealing with an e

If you add 1 (candy bar) to $\frac{1}{3}$ of a (candy bar), how many (candy bars) do you have?
Of course, you have $1$ full candy bar and $\frac{1}{3}$ of another. But we would like to express this in units that are equal! Can we express thirds as wholes? Not in an intuitive way. What about wholes as thirds? Sure! 1 (candy bar) is 3 thirds of a (candy bar). Break it into three pieces and stick them back together, and presto - one whole candy bar made from 3 thirds. Evidently, then, we can say, replacing candy bars with $x$:
$x+\frac{1}{3}x=\frac{3}{3}x+\frac{1}{3}x=\frac{4}{3}x.$
Note that if you take your three candy bar pieces I mentioned before and stick one third of another candy bar in there, you get $\frac{4}{3}$ of a candy bar.
I hope this helps!

Another approach, which at this time does not appear to have been mentioned, is to "clear fractions" from your equation. You can do this by multiplying both sides of the equation by a number that results in no fractions being left. In the case of your equation, multiply both sides by $3$
$\frac{2}{3}b+b=15$
(multiply both sides by $3$)
$3(\frac{2}{3}b+b)=3\left(15\right)$
$2b+3b=45$
$5b=45$
Now solve for $b$ by dividing both sides by $5$ by dividing both sides by $5$ to get $b=9.$
Of course, you could also multiply both sides by $6$ or multiply both sides by $30,$ but $3$ is the most sensible choice because $3$ is the smallest number that does the job.
Note that if we had gotten $6b=45$ at the end, the final answer would involve fractions. However, what this method does is keep the fractions at bay until the very end so you don't have to deal with them until the end.
Other examples:
$\frac{2}{3}b+\frac{1}{4}b=18\text{(multiply both sides by}12)$
$\frac{2}{3}b+\frac{1}{6}b=18\text{(multiply both sides by}6)$
$\frac{2}{3}b+\frac{1}{4}b=\frac{5}{8}\text{(multiply both sides by}24)$
What you want to multiply both sides by is a number that all the denominators will divide into. If you can't think of such a number very quickly, then you can always get such a number by multiplying all the denominators together.
However, this method fails when the coefficients are not fractions or integers, such as
$\sqrt{2}b+b=18$
or
$\mathrm{\pi <!--\; \pi \; -->}b+4b=18$
In these cases, some of the other methods described here can be used (e.g. factor out $b$ and then divide both sides by what $b$ is being multiplied by).