# Question #b8408

Oct 11, 2017

$3 \frac{7}{12}$ cups

#### Explanation:

To solve this problem, we need to add together two fractions with different denominators.

In order to add them correctly, we need to change both fractions so they have the same denominator.

One fraction has a denominator of $\textcolor{\lim e g r e e n}{3}$ and the other has a denominator of $\textcolor{b l u e}{4}$. The least common multiple of 3 and 4 is 12, so we need to alter both fractions to have a denominator of 12.

Let's start with $2 \frac{1}{3}$. We can multiply both the top and bottom of the fraction by $\textcolor{b l u e}{4}$ to get a denominator of $\textcolor{red}{12}$:

$2 \textcolor{w h i t e}{\text{-}} \frac{1 \cdot \textcolor{b l u e}{4}}{3 \cdot \textcolor{b l u e}{4}} = 2 \frac{4}{\textcolor{red}{12}}$

Next, let's change $1 \frac{1}{4}$ in the same way by multiplying the top and bottom of the fraction by $\textcolor{\lim e g r e e n}{3}$ to get a denominator of $\textcolor{red}{12}$.

$1 \textcolor{w h i t e}{\text{-}} \frac{1 \cdot \textcolor{\lim e g r e e n}{3}}{4 \cdot \textcolor{\lim e g r e e n}{3}} = 1 \frac{3}{\textcolor{red}{12}}$

Now, we can add the fractions together properly:

$2 \frac{4}{12} + 1 \frac{3}{12}$

$= \left(2 + \frac{4}{12}\right) + \left(1 + \frac{3}{12}\right)$

$= \left(2 + 1\right) + \left(\frac{4}{12} + \frac{3}{12}\right)$

$= 3 + \frac{7}{12}$

$= 3 \frac{7}{12}$