Question

$(\frac{3}{5}x+\frac{1}{2}y)$ by $(\frac{5}{6}x+4y)$.

Answer 1

Let us find the product of $(\frac{3}{5}x+\frac{1}{2}y)$ and $(\frac{5}{6}x+4y)$ as shown below:

$(\frac{3}{5}x+\frac{1}{2}y)(\frac{5}{6}x+4y)=(\frac{3}{5}x\times \frac{5}{6}x)+(\frac{3}{5}x\times 4y)+(\frac{1}{2}y\times \frac{5}{6}x)+(\frac{1}{2}y\times 4y)=\frac{1}{2}{x}^2+\frac{12}{5}xy+\frac{5}{12}xy+2y^2=\frac{1}{2}{x}^2+2y^2+(\frac{12}{5}xy+\frac{5}{12}xy)\left(Combiningliketerms\right)$

$=\frac{1}{2}{x}^2+2y^2+(\frac{12}{5}+\frac{5}{12})xy=\frac{1}{2}{x}^2+2y^2+\left(\frac{144+25}{60}\right)xy=\frac{1}{2}{x}^2+2y^2+\frac{169}{60}xy$

Hence, $(\frac{3}{5}x+\frac{1}{2}y)(\frac{5}{6}x+4y)=\frac{1}{2}{x}^2+2y^2+\frac{169}{60}xy$.