+0  
 
0
113
2
avatar

If $x+y=\frac{15}{7}$ and $x-y=\frac7{9}$, find the value of $x^2-y^2$, expressed as a common fraction.

 Feb 22, 2023
 #1
avatar+2667 
+1

Adding the two equations gives us \(2x = {184 \over 63}\), meaning \(x = {92 \over 63}\).

 

This means that \(y = {43 \over 63}\), so \(x^2 - y^2 = ({92 \over 63})^2 - ({43 \over 63})^2 = \color{brown}\boxed{5 \over 3}\)

 Feb 22, 2023
 #2
avatar+27 
0

Alternative Solution: 

 

Notice that \((x+y)(x-y) = x^2-y^2\), so simply multiplying the two together gives: 

 

\((x+y)(x-y) = x^2-y^2 = \frac{15}{7} \cdot \frac{7}{9} = \frac{15}{9} = \boxed{\frac{5}{3}}\)

 Feb 22, 2023

2 Online Users

avatar
avatar