Question

Complete the addition and subtraction box.

Answer 1

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$a)\frac{1}{2}+\frac{1}{3}=\frac{3+2}{6}=\frac{5}{6}$

$\frac{1}{3}+\frac{1}{4}=\frac{4+3}{12}=\frac{7}{12}$

$\frac{1}{6}+\frac{1}{12}=\frac{1}{4}$

$\frac{1}{2}-\frac{1}{3}=\frac{3-2}{6}=\frac{1}{6}$

$\frac{1}{3}-\frac{1}{4}=\frac{4-3}{12}=\frac{1}{12}$

$\frac{5}{6}-\frac{7}{12}=\frac{10-7}{12}=\frac{3}{12}=\frac{1}{4}$

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$b)\frac{2}{3}+\frac{4}{3}=\frac{6}{3}=2$

$\frac{1}{3}+\frac{2}{3}=\frac{3}{3}=1$

$\frac{1}{3}+\frac{2}{3}=1$

$\frac{2}{3}-\frac{1}{3}=\frac{1}{3}$

$\frac{4}{3}-\frac{2}{3}=\frac{2}{3}$

$2-1=1$