## Elementary Algebra

$= \frac{2z^{3}}{3x^{2}y}$
Simplify $\frac{-32}{-48}$ $= \frac{32}{48}$ $= \frac{16}{24}$ $= \frac{8}{12}$ $= \frac{4}{6}$ $= \frac{2}{3}$ Simplify$\frac{xy^{2}z^{4}}{x^{3}y^{3}z}$ 1) Cancel $x$ from the numerator and denominator to result in $1$ in the numerator and $x^{2}$ in the denominator. 2) Cancel $y^{2}$ from the numerator and denominator to result in $1$ in the numerator and $y$ in the denominator. 3) Cancel $z$ in the denominator and numerator to result in $z^{3}$ in the numerator and $1$ in the denominator. $= \frac{(1)y^{2}z^{4}}{x^{2}y^{3}z}$ $= \frac{(1)(1)z^{4}}{x^{2}yz}$ $= \frac{(1)(1)z^{3}}{x^{2}y(1)}$ $= \frac{z^{3}}{x^{2}y}$ Multiply the two components to get the result $= \frac{2}{3}(\frac{z^{3}}{x^{2}y})$ $= \frac{2z^{3}}{3x^{2}y}$