Answer
$\frac{13}{12}$ or $1 \frac{1}{12}$
Work Step by Step
$ x= \frac{1}{3}$ $y=\frac{3}{4}$
$ x+y= \frac{1}{3}+ \frac{3}{4}$
The least common multiplier of the denominators is 12. Adjust the fractions based on the least common multiplier. For the first fraction, multiply the numerator and denominator by 4.
$ \frac{1}{3}\times \frac{4}{4}= \frac{4}{12}$
For the second fraction, multiply the numerator and denominator by 3.
$ \frac{3}{4} \times \frac{3}{3}= \frac{9}{12}$
So, $x+y= \frac{4}{12}+ \frac{9}{12}= \frac{13}{12}= 1 \frac{1}{12}$
