Question
Complete the addition$-$subtraction box.
✓
Answer
$a.$ 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,
$\frac{2}{3}-\frac{1}{3}=\frac{2-1}{3}=\frac{1}{3}$
$\frac{4}{3}-\frac{2}{3}=\frac{4-2}{3}=\frac{2}{3}$
and $2-1=1$
Hence, the complete addition$-$subtraction box is:
$b.$ Solving rows first,
$\frac{1}{2}+\frac{1}{3}=\frac{1\times3}{2\times3}+\frac{1\times2}{3\times2}=\frac{3}{6}+\frac{2}{6}=\frac{3+2}{6}=\frac{5}{6}$
$\frac{1}{3}+\frac{1}{4}=\frac{1\times4}{3\times4}+\frac{1\times3}{4\times3}=\frac{4}{12}+\frac{3}{12}=\frac{4+3}{12}=\frac{7}{12}$
Solving the columns,
$\frac{1}{2}-\frac{1}{3}=\frac{1\times3}{2\times3}-\frac{1\times2}{3\times2}=\frac{3}{6}-\frac{2}{6}=\frac{3-2}{6}=\frac{1}{6}$
$\frac{1}{3}-\frac{1}{4}=\frac{1\times4}{3\times4}-\frac{1\times3}{4\times3}=\frac{4}{12}-\frac{3}{12}=\frac{4-3}{12}=\frac{1}{12}$
and $\frac{5}{6}-\frac{7}{12}=\frac{5\times2}{6\times2}-\frac{7\times1}{12\times1}=\frac{10}{12}-\frac{7}{12}=\frac{10-7}{12}=\frac{3}{12}=\frac{1}{4}$
Hence, the complete addition$-$subtraction box is:
$\frac{2}{3}+\frac{4}{3}=\frac{2+4}{3}=\frac{6}{3}=2$
$\frac{1}{3}+\frac{2}{3}=\frac{1+2}{3}=\frac{3}{3}=1$
Solving the columns,
$\frac{2}{3}-\frac{1}{3}=\frac{2-1}{3}=\frac{1}{3}$
$\frac{4}{3}-\frac{2}{3}=\frac{4-2}{3}=\frac{2}{3}$
and $2-1=1$
Hence, the complete addition$-$subtraction box is:
$b.$ Solving rows first,
$\frac{1}{2}+\frac{1}{3}=\frac{1\times3}{2\times3}+\frac{1\times2}{3\times2}=\frac{3}{6}+\frac{2}{6}=\frac{3+2}{6}=\frac{5}{6}$
$\frac{1}{3}+\frac{1}{4}=\frac{1\times4}{3\times4}+\frac{1\times3}{4\times3}=\frac{4}{12}+\frac{3}{12}=\frac{4+3}{12}=\frac{7}{12}$
Solving the columns,
$\frac{1}{2}-\frac{1}{3}=\frac{1\times3}{2\times3}-\frac{1\times2}{3\times2}=\frac{3}{6}-\frac{2}{6}=\frac{3-2}{6}=\frac{1}{6}$
$\frac{1}{3}-\frac{1}{4}=\frac{1\times4}{3\times4}-\frac{1\times3}{4\times3}=\frac{4}{12}-\frac{3}{12}=\frac{4-3}{12}=\frac{1}{12}$
and $\frac{5}{6}-\frac{7}{12}=\frac{5\times2}{6\times2}-\frac{7\times1}{12\times1}=\frac{10}{12}-\frac{7}{12}=\frac{10-7}{12}=\frac{3}{12}=\frac{1}{4}$
Hence, the complete addition$-$subtraction box is:
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