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To do:

We have to complete the addition-subtraction boxes.

Solution:

(a) $\frac{2}{3}+\frac{4}{3}=\frac{2+4}{3}$

$=\frac{6}{3}$

$=2$

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

$=\frac{3}{3}$

$=1$

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

$=\frac{1}{3}$

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

$=\frac{2}{3}$

$2-1=1$

(b) $\frac{1}{2}+\frac{1}{3}=\frac{1\times3+1\times2}{6}$

$=\frac{5}{6}$

$\frac{1}{3}+\frac{1}{4}=\frac{1\times4+1\times3}{12}$

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

$=\frac{7}{12}$

$\frac{1}{2}-\frac{1}{3}=\frac{1\times3-1\times2}{6}$

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

$=\frac{1}{6}$

$\frac{1}{3}-\frac{1}{4}=\frac{1\times4-1\times3}{12}$

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

$=\frac{1}{12}$

$\frac{5}{6}-\frac{7}{12}=\frac{5\times2-7\times1}{12}$

$=\frac{10-7}{12}$

$=\frac{3}{12}$

$=\frac{1}{4}$

Updated on 10-Oct-2022 13:33:02