Question

Find (x + y) ÷ (x + y), if
(i) $x=\frac{2}{3},y=\frac{3}{2}$
(ii) $x=\frac{2}{5},y=\frac{1}{2}$
(iii) $x=\frac{5}{4},y=\frac{-1}{3}$

Answer 1

(i) x = $\frac{2}{3},y=\frac{3}{2}$
Then, (x+y) = $\frac{2}{3}+\frac{3}{2}=\frac{2\times 2}{3\times 2}+\frac{3\times 3}{2\times 3}=\frac{4}{6}+\frac{9}{6}=\frac{13}{6}$
$(x-y)=\frac{2}{3}-\frac{3}{2}=\frac{4}{6}-\frac{9}{6}=\frac{-5}{6}$
Then, $(x+y)÷(x-y)=\frac{13}{6}÷\frac{-5}{6}=\frac{13}{6}\times \frac{6}{-5}=\frac{-13}{5}$.

(ii) x = $\frac{2}{5},y=\frac{1}{2}$
Then, (x+y) = $\frac{2}{5}+\frac{1}{2}=\frac{2\times 2}{5\times 2}+\frac{1\times 5}{2\times 5}=\frac{4}{10}+\frac{5}{10}=\frac{9}{10}$
$(x-y)=\frac{2}{5}-\frac{1}{2}=\frac{4}{10}-\frac{5}{10}=\frac{-1}{10}$

Then, $(x+y)÷(x-y)=\frac{9}{10}÷\frac{-1}{10}=-9$

(iii) x = $\frac{5}{4},y=\frac{-1}{3}$
Then, (x+y) = $\frac{5}{4}+\frac{-1}{3}=\frac{5\times 3}{4\times 3}+\frac{-1\times 4}{3\times 4}=\frac{15}{12}+\frac{-4}{12}=\frac{11}{12}$
$(x-y)=\frac{5}{4}-\frac{-1}{3}=\frac{5}{4}+\frac{1}{3}=\frac{19}{12}$
Then, $(x+y)÷(x-y)=\frac{11}{12}÷\frac{19}{12}=\frac{11}{12}\times \frac{12}{19}=\frac{11}{19}.$