Question
Complete the addition-subtraction box.
✓
Answer
We have,
$\frac{2}{3}+\frac{4}{3}=\frac{2+4}{3}=\frac{6}{3}= 2$
$\frac{1}{3}+\frac{2}{3}=\frac{1+2}{3}=\frac{3}{3}= 1$
$\frac{2}{3}-\frac{1}{3}=\frac{2-1}{3}=\frac{1}{3}$
$\frac{4}{3}-\frac{2}{3}=\frac{4-2}{3}=\frac{2}{3}$
and $\frac{1}{3}+\frac{2}{3}=\frac{3}{3} = 1$
Hence, the above-given number box can be completed as:
$\frac{2}{3}+\frac{4}{3}=\frac{2+4}{3}=\frac{6}{3}= 2$
$\frac{1}{3}+\frac{2}{3}=\frac{1+2}{3}=\frac{3}{3}= 1$
$\frac{2}{3}-\frac{1}{3}=\frac{2-1}{3}=\frac{1}{3}$
$\frac{4}{3}-\frac{2}{3}=\frac{4-2}{3}=\frac{2}{3}$
and $\frac{1}{3}+\frac{2}{3}=\frac{3}{3} = 1$
Hence, the above-given number box can be completed as:
| $\frac{2}{3}$ | $\frac{4}{3}$ | $2$ |
| $\frac{1}{3}$ | $\frac{2}{3}$ | $1$ |
| $\frac{1}{3}$ | $\frac{2}{3}$ | $1$ |
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