Question

Question 83 (xviii)

Multiply the following :

$(\frac{3}{4}x-\frac{4}{3}y),(\frac{2}{3}x+\frac{3}{2}y)$

Answer 1

We have, $(\frac{3}{4}x-\frac{4}{3}y)and(\frac{2}{3}x+\frac{3}{2}y)$
$\therefore (\frac{3}{4}x-\frac{4}{3}y)(\frac{2}{3}x+\frac{3}{2}y)=\frac{3}{4}x(\frac{2}{3}x+\frac{3}{2}y)-\frac{4}{3}y(\frac{2}{3}x+\frac{3}{2}y)=\frac{3}{4}\times \frac{2}{3}{x}^2+\frac{3}{4}\times \frac{3}{2}xy-\frac{4}{3}\times \frac{2}{3}yx-\frac{4}{3}\times \frac{3}{2}{y}^2=\frac{1}{2}{x}^2+\frac{9}{8}xy-\frac{8}{9}xy-2y^2=\frac{1}{2}{x}^2+(\frac{9}{8}-\frac{8}{9})xy-2y^2=\frac{1}{2}{x}^2+\left(\frac{81-64}{72}\right)xy-2y^2=\frac{1}{2}{x}^2+\frac{17}{72}xy-2y^2$