Damian Kerr

2022-02-03

How do you simplify $3\frac{{x}^{2}}{4}\left(xy-z\right)+4x\frac{y}{3}\left({x}^{2}+2\right)-2\frac{{x}^{2}}{3}\left(xy-z\right)+3x\frac{y}{5}\left({x}^{2}+2\right)?$

Zayden Sims

Step 1
Let’s first simplify the individual terms and then combine them finally.
$13\left(\frac{{x}^{2}}{4}\right)\left(xy-z\right)=\left(\frac{13{x}^{2}}{4}\right)\left(xy-z\right)=\frac{13{x}^{3}y}{4}-\frac{13{x}^{2}z}{4}$
$\left(4x\right)\left(\frac{y}{3}\right)\left({x}^{2}+2\right)=\left(\frac{13xy}{3}\right)\left({x}^{2}+2\right)=\frac{13{x}^{3}y}{3}+\frac{26xy}{3}$
$2\left(\frac{{x}^{2}}{3}\right)\left(xy-z\right)=\left(7\frac{{x}^{2}}{3}\right)\left(xy-z\right)=\frac{7{x}^{3}y}{3}-\frac{7{x}^{2}z}{3}$
$\left(3x\right)\left(\frac{y}{5}\right)\left({x}^{2}+2\right)=\left(\frac{16xy}{5}\right)\left({x}^{2}+2\left(=\frac{16{x}^{3}y}{5}+\frac{32xy}{5}$
Combining all the terms,
$\frac{13{x}^{3}y}{4}-\frac{13{x}^{2}z}{4}+\frac{13{x}^{3}y}{3}+\frac{26xy}{3}-\frac{7{x}^{3}z}{3}+\frac{7{x}^{2}z}{3}+\frac{16{x}^{3}y}{5}+\frac{32xy}{5}$
$=-\frac{13{x}^{2}z}{4}+\frac{7{x}^{2}z}{3}+\frac{13{x}^{3}y}{4}+\frac{13{x}^{3}y}{3}-\frac{7{x}^{3}y}{3}+\frac{16{x}^{3}y}{5}+\frac{26xy}{3}+\frac{32xy}{5}$

Do you have a similar question?