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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
+2602
+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
+25
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