Question

Add. Write each sum in the lowest terms.

$$\begin{gathered} \frac{1}{6} \\ +\frac{7}{12} \\ \_\_\_\_\_ \end{gathered}$$

Answer 1

If the denominators are different, we change to equivalent fractions that have the same denominator.

Also, if one denominator is a multiple of the other, the greater number can be used as a common denominator.

$$\begin{gathered} \frac{1}{6}+\frac{7}{12}=\frac{2}{12}+\frac{7}{12}(Changing\frac{1}{6}toequivalentfractionthathavesamedenominator) \\ =\frac{2+7}{12} \\ =\frac{9}{12}(Dividingnumeratoranddenominatorbycommonfactor3) \\ =\frac{3}{4} \end{gathered}$$

Therefore the required sum in the lowest term is $\frac{3}{4}$