If $x+y=\frac{15}{7}$ and $x-y=\frac7{9}$, find the value of $x^2-y^2$, expressed as a common fraction.
+2668 Adding the two equations gives us \(2x = {184 \over 63}\), meaning \(x = {92 \over 63}\).
This means that \(y = {43 \over 63}\), so \(x^2 - y^2 = ({92 \over 63})^2 - ({43 \over 63})^2 = \color{brown}\boxed{5 \over 3}\)
+27 Alternative Solution:
Notice that \((x+y)(x-y) = x^2-y^2\), so simply multiplying the two together gives:
\((x+y)(x-y) = x^2-y^2 = \frac{15}{7} \cdot \frac{7}{9} = \frac{15}{9} = \boxed{\frac{5}{3}}\)
