Answer:
[tex]\frac{1}{169}[/tex]
Step-by-step explanation:
It is given:
[tex]x+y=\frac{7}{13}[/tex]
and
[tex]x-y=\frac{1}{91}[/tex]
We are asked to find the value of "[tex]x^2-y^2[/tex]"
We remember a rule/formula from algebra:
[tex]x^2-y^2=(x+y)(x-y)[/tex]
Using this formula we can write:
[tex]x^2-y^2=(x+y)(x-y)=(\frac{7}{13})(\frac{1}{91})=\frac{1}{13}*\frac{1}{13}=\frac{1}{169}[/tex]
The answer is:
[tex]\frac{1}{169}[/tex]
