If \(x+y=\frac{7}{12}\) and \(x-y=\frac{1}{12}\), what is the value of \(x^2-y^2\)? Express your answer as a common fraction.

Logic Jul 23, 2019

#2**+1 **

Add the two equations together to get

2x = 8/12

x = 4/12

from the first equation then 4/12 + y = 7/12 means y = 3/12

then (4/12)^2 - (3/12)^2 = (1/3)^2 - (1/4)^2 = 1/9 - 1/16 = 16/144 - 9/144 = 7/144

ElectricPavlov Jul 23, 2019

#3**+1 **

If

\(x+y=\dfrac{7}{12} \)

and

\(x-y=\dfrac{1}{12}\),

what is the value of \(x^2-y^2\)?

Express your answer as a common fraction.

\(\begin{array}{|rcll|} \hline (x-y)(x+y) &=& \dfrac{7}{12}\times \dfrac{1}{12} \quad | \quad (x-y)(x+y) = x^2-y^2 \\ \mathbf{x^2-y^2} &=& \dfrac{7}{12}\times \dfrac{1}{12} \\ &=&\mathbf{ \dfrac{7}{144}} \\ \hline \end{array} \)

heureka Jul 24, 2019