Question

Complete the addition-subtraction box.

(a)

(b)

Answer 1

(a) Solving rows first,
$\frac{2}{3}+\frac{4}{3}=\frac{2+4}{3}=\frac{6}{3}=2$

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

Solving the columns,
$\frac{2}{3}-\frac{1}{3}=\frac{2-1}{3}=\frac{1}{3}$

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

$2-1=1$

Hence, the complete addition-subtraction box is



(b) Solving rows first,
$\frac{1}{2}+\frac{1}{3}=\frac{1\times 3}{2\times 3}+\frac{1\times 2}{3\times 2}=\frac{3}{6}+\frac{2}{6}=\frac{3+2}{6}=\frac{5}{6}$

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

Solving the columns,
$\frac{1}{2}-\frac{1}{3}=\frac{1\times 3}{2\times 3}-\frac{1\times 2}{3\times 2}=\frac{3}{6}-\frac{2}{6}=\frac{3-2}{6}=\frac{1}{6}$

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

$\frac{5}{6}-\frac{7}{12}=\frac{5\times 2}{6\times 2}-\frac{7}{12}=\frac{10}{12}-\frac{7}{12}=\frac{10-7}{12}=\frac{3}{12}=\frac{1}{4}$

Hence, the complete addition-subtraction box is