Add and Subtract Mixed Numbers - 4th Grade Math

Add the whole numbers and add the fractional parts. This separates the problem into simpler parts to solve.

$5\ \frac{6}{7}$. $2+3=5$ and $\frac{4}{7}+\frac{2}{7}=\frac{6}{7}$.

$7\ \frac{3}{8}$. $4+2=6$ and $\frac{5}{8}+\frac{6}{8}=\frac{11}{8}=1\frac{3}{8}$, so $6+1\frac{3}{8}=7\frac{3}{8}$.

$8\ \frac{2}{10}$. $6+1=7$ and $\frac{3}{10}+\frac{9}{10}=\frac{12}{10}=1\frac{2}{10}$, so $7+1\frac{2}{10}=8\frac{2}{10}$.

Subtract whole numbers and subtract fractional parts. This works when the first fraction is larger than the second.

$3\ \frac{2}{9}$. $5-2=3$ and $\frac{6}{9}-\frac{4}{9}=\frac{2}{9}$.

$3\ \frac{2}{4}$. Borrow: $7\frac{1}{4}=6\frac{5}{4}$, then $6-3=3$ and $\frac{5}{4}-\frac{3}{4}=\frac{2}{4}$.

$4\ \frac{3}{5}$. Borrow: $9\frac{2}{5}=8\frac{7}{5}$, then $8-4=4$ and $\frac{7}{5}-\frac{4}{5}=\frac{3}{5}$.

$3\ \frac{2}{8}$. Borrow: $10\frac{1}{8}=9\frac{9}{8}$, then $9-6=3$ and $\frac{9}{8}-\frac{7}{8}=\frac{2}{8}$.

$2 \frac{7}{8}$. $\frac{3}{8}+\frac{4}{8}=\frac{7}{8}$, so result is $2 \frac{7}{8}$.

$4\ \frac{3}{12}$. $1+2=3$ and $\frac{7}{12}+\frac{8}{12}=\frac{15}{12}=1\frac{3}{12}$, so $3+1\frac{3}{12}=4\frac{3}{12}$.

$2\ \frac{3}{6}$. $5=4\frac{6}{6}$, then $4\frac{6}{6}-2\frac{3}{6}=2\frac{3}{6}$.

$4\ \frac{1}{4}$. Divide: $17÷4=4$ remainder $1$, so $\frac{17}{4}=4\frac{1}{4}$.

$6\ \frac{4}{6}$. Borrow: $12\frac{3}{6}=11\frac{9}{6}$, then $11-5=6$ and $\frac{9}{6}-\frac{5}{6}=\frac{4}{6}$.

$\frac{23}{7}$. Multiply whole number by denominator and add numerator: $3×7+2=23$.

$3\ \frac{3}{9}$. $\frac{5}{9}-\frac{2}{9}=\frac{3}{9}$, whole number stays $3$.

$11$. $8+2=10$ and $\frac{1}{6}+\frac{5}{6}=\frac{6}{6}=1$, so $10+1=11$.

$2\frac{7}{5}$. Borrow $1=\frac{5}{5}$: $3\frac{2}{5}=2+\frac{5}{5}+\frac{2}{5}=2\frac{7}{5}$.

$4\frac{9}{8}$. Borrow $1=\frac{8}{8}$: $5\frac{1}{8}=4+\frac{8}{8}+\frac{1}{8}=4\frac{9}{8}$.