## Elementary Algebra

We use the rule $\frac{a^{m}}{a^{n}}=a^{m-n}$ to help us cancel out the common factors of $y$ in the numerator and denominator: $\frac{12xy^{3}}{-16y^{2}}$ =$\frac{12}{-16}\times x\frac{y^{3}}{y^{2}}$ =$\frac{3}{-4}\times x\times y^{3-2}$ =$-\frac{3}{4}\times x\times y$ =$-\frac{3xy}{4}$ Since this answer matches the one in the statement, the statement is true.