I am dealing with an equation that requires me to add $x$ to $\frac{1}{3}x$:

$x + \frac{1}{3}x$ = ??

I know this might be simple to any of you on this site, because you are all asking questions with symbols I have never seen, but this is confusing to me.

I guess one way of thinking about this is - You are adding $x$ to $yx$, right? Or just adding another $\frac{1}{3}$?

The complete equation that I am working on is [- don't laugh at its simplicity ;)]:

$\frac{2}{3}b + 5 = 20 - b$

So, when worked out... I got:

$\frac{2}{3}b + b = 15$

And this is where I get stuck.


7 Answers 7


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!

  • 2
    $\begingroup$ @CarlosCarlsen Glad I could help! Don't accept this answer yet, though. See if an even better one comes along, perhaps in an hour or so. $\endgroup$
    – The Count
    Jun 29, 2016 at 13:48
  • 7
    $\begingroup$ +1 for an answer that intuits and addresses the source of the OP's confusion, rather then simply providing a mathematically correct answer.. $\endgroup$ Jun 29, 2016 at 13:50
  • 1
    $\begingroup$ @CarlosCarlsen Of course, if that time passes, you might consider accepting if you like it, but then again, I am biased. Your profile made me very happy. I wish all young students (including my past self) were like you. $\endgroup$
    – The Count
    Jun 29, 2016 at 13:51
  • 1
    $\begingroup$ The best part of this answer is clearly the username. Very good catch on The Count's part $\endgroup$ Jun 29, 2016 at 16:52
  • 1
    $\begingroup$ @Casquibaldo Ah! Ah! Ah! $\endgroup$
    – The Count
    Jun 29, 2016 at 17:03

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 $3$)

$$3\left(\frac{2}{3}b \; + \; b\right) \; = \; 3\left( 15 \right)$$

$$2b \; + \; 3b \; = \; 45$$

$$5b \; = \; 45$$

Now solve for $b$ 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

$$\sqrt{2}\,b \; + \; b \; = 18$$


$$\pi \, b \; + \; 4b \; = 18$$

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).


$$\frac{2}{3}b+b = \frac{2}{3}b+1b= \bigg(\frac{2}{3}+1\bigg)b = \bigg(\frac{2+3}{3}\bigg)b = \frac{5}{3}b$$

Does this help?

  • 1
    $\begingroup$ $\dfrac 23b + b = \dfrac 23b + 1b = \cdots$ $\endgroup$ Jun 29, 2016 at 13:44
  • $\begingroup$ @StevenGregory, added to the answer. Thanks. $\endgroup$
    – fosho
    Jun 29, 2016 at 13:45
  • $\begingroup$ @Dman - yes, thanks - in retrospect, it was very simple, yet I couldnt grasp the whole process of adding a variable to a fraction - so your answer drew it out nicely for me - thanks! $\endgroup$ Jun 29, 2016 at 13:49
  • $\begingroup$ "yet I couldnt grasp the whole process of adding a variable to a fraction" ... well, you aren't actually adding a variable to a fraction; you are adding a variable to a fraction of the variable. Subtle difference. $\endgroup$
    – fleablood
    Jul 1, 2016 at 19:36
  • $\begingroup$ Best answer on the site !! :) $\endgroup$
    – mick
    Feb 15, 2017 at 22:59

$$ x + \frac{1}{3} x = 1 \cdot x + \frac{1}{3} x = \left( 1 + \frac{1}{3} \right) x = \frac{4}{3} x $$ Your full equation is slighty different, but the same rules (distributivity) apply:

$$ 15 = \frac{2}{3} b + b = \left( \frac{2}{3} + 1 \right) b = \frac{5}{3} b \iff \\ b = \frac{3}{5} \cdot 15 = 9 $$

  • $\begingroup$ Thanks for the help - I can see its process now! $\endgroup$ Jun 29, 2016 at 13:48

You know that you can:

  • Add fractions as long as they have the same denominator
  • Multiply both the nominator and the denominator with the same value and the value of the fraction remains the same

Having that in mind we are going to add $x$ to $\frac{x}{3}$.We know that $x=\frac{3x}{3}$ (multiply both nominator and denominator by 3)

Now we can add them together $\frac{3x+x}{3}=\frac{4x}{3}$


You need to grasp what is known as the distributive law: $$a(b+c)=ab+ac$$ In your case it is backwards, in other words $$xb+yb=(x+y)b$$ Can you see how that might help. Put $x=\frac{2}{3},y=1$ and you have $$\frac{2}{3}b+1\cdot b=(\frac{2}{3}+1)b$$ I guess that also uses $1\cdot b=b$, which I am sure you are happy with.

Why is that law true? Well first think of $b$ as "apples" we are saying that $x$ apples + $y$ apples equals $x+y$ apples. Then think of $b$ as a (box of a) dozen apples. Then we are saying $x$ dozen apples plus $y$ dozen apples is $x+y$ dozen apples. Drop the apples and we have: $$x\times 12+y\times 12=(x+y)\times 12$$

The final part is what to do about $\frac{2}{3}+1$. Do you need help there? You can express 1 as $\frac{3}{3}$ and then the two fractions (or as math guys like to say "rational numbers") have the same "denominator" (= bottom part), so we can add the top parts to get $$\frac{2}{3}+1=\frac{2}{3}+\frac{3}{3}=\frac{5}{3}$$ and hence finally $$\frac{2}{3}b+b=\frac{5}{3}b$$


Technically, this is called the "distributive law": a(b+ c)= ab+ ac but you probably have never seen it and don't need it for basic 8th or 9th grade algebra. What you can do think of the "b" as representing some object, say a cup of flour. If you put 2/3 of a cup of flour in a bowl and add one cup of flour, how much flour do you now have in the bowl? Well, here, the fact that it is flour doesn't affect the arithmetic: 2/3+ 1= 2/3+ 3/3= (2+ 3)/3= 5/3. You would have 5/3 cups of flour. In fact, 2/3 of anything plus 1 of the samething is 5/3 of that thing. If that "thing" is b, then you have (5/3)b.


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