How do you solve 13 2/9 - 7 1/3 and simplify the answer?

Dec 10, 2017

Convert to improper fractions, make the denominator common, subtract, then convert to the asked fraction form (perhaps a mixed fraction) to obtain $5 \frac{8}{9}$.

Explanation:

Before starting, I would turn any subtraction into negative addition to reduce ambiguity and reduce the possibility of errors:

$= 13 \frac{2}{9} + \left(- \left(7 \frac{1}{3}\right)\right)$

Now, let's convert each to improper fractions. Mixed fractions really are a sum of a whole number and a proper fraction:

$= \left(13 + \frac{2}{9}\right) + \left(- \left(7 + \frac{1}{3}\right)\right)$

I'll take the whole number, $13$, from $13 \frac{2}{9}$, and make it a fraction with a denominator of $9$. To do that, first divide $13$ by $1$ (which shouldn't change anything):

$13 = \frac{13}{1}$

Then multiply by $1$ (doesn't change anything either):

$= \frac{13}{1} \cdot 1$

Still not changing anything, we can turn that $1$ into a number divided by itself. Choose $9$:

$= \frac{13}{1} \cdot \frac{9}{9}$

And multiply:

$= \frac{13 \cdot 9}{1 \cdot 9} = \frac{117}{9}$

$13 + \frac{2}{9} = \frac{117}{9} + \frac{2}{9} = \frac{119}{9}$

Alright, that's our first improper fraction, now we have:

$= \left(13 + \frac{2}{9}\right) + \left(- \left(7 + \frac{1}{3}\right)\right) = \frac{119}{9} + \left(- \left(7 + \frac{1}{3}\right)\right)$

Time to tackle the second! However, we need to make sure both fractions end up having common denominators. So let's make both $7$ and $\frac{1}{3}$ have a denominator of $9$ (without changing the value!) by the same method:

$7 + \frac{1}{3}$

$= \left(\frac{7}{1}\right) + \left(\frac{1}{3}\right)$

$= \left(\frac{7}{1} \cdot 1\right) + \left(\frac{1}{3} \cdot 1\right)$

$= \left(\frac{7}{1} \cdot \frac{9}{9}\right) + \left(\frac{1}{3} \cdot \frac{3}{3}\right)$

$= \frac{63}{9} + \frac{3}{9} = \frac{66}{9}$

Putting this back into our problem:

$\frac{119}{9} + \left(- \left(7 + \frac{1}{3}\right)\right) = \frac{119}{9} + \left(- \left(\frac{66}{9}\right)\right) = \frac{119}{9} - \frac{66}{9}$

Now solve:

$\frac{119}{9} - \frac{66}{9} = \frac{53}{9}$

I assume you need a mixed fraction... Well, split $53$ into a multiple of $9$ and another number... maybe $45$ and $8$?

$\frac{53}{9} = \frac{45}{9} + \frac{8}{9}$

Then divide $\frac{45}{9}$:

$= 5 + \frac{8}{9} = 5 \frac{8}{9}$