Question
Find $(x + y) ÷ (x − y)$, if $\text{x}=\frac{2}{7},\text{y}=\frac{4}{3}$
✓
Answer
$\text{x}=\frac{2}{7},\text{y}=\frac{4}{3}$
$\therefore\text{x}+\text{y}=\frac{2}{7}+\frac{4}{3}$
$=\frac{6+28}{21}=\frac{34}{21}$
$\text{x}-\text{y}=\frac{2}{7}-\frac{4}{3}$
$=\frac{6-28}{21}=\frac{-22}{21}$
$\therefore(\text{x}+\text{y})\div(\text{x}-\text{y})=\frac{34}{21}\div\frac{-22}{21}$
$=\frac{34}{21}\times\frac{21}{-22}$
$=\frac{17\times1}{1\times(-11)}=\frac{17}{-11}$
$=\frac{17\times(-1)}{-11\times(-1)}=\frac{-17}{11}$
$\therefore\text{x}+\text{y}=\frac{2}{7}+\frac{4}{3}$
$=\frac{6+28}{21}=\frac{34}{21}$
$\text{x}-\text{y}=\frac{2}{7}-\frac{4}{3}$
$=\frac{6-28}{21}=\frac{-22}{21}$
$\therefore(\text{x}+\text{y})\div(\text{x}-\text{y})=\frac{34}{21}\div\frac{-22}{21}$
$=\frac{34}{21}\times\frac{21}{-22}$
$=\frac{17\times1}{1\times(-11)}=\frac{17}{-11}$
$=\frac{17\times(-1)}{-11\times(-1)}=\frac{-17}{11}$
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