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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.

 Jul 23, 2019
 #1
avatar+683 
+3

click here for the answer i do not know the fraction!

 Jul 23, 2019
 #2
avatar+18754 
+2

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

 Jul 23, 2019
 #3
avatar+22896 
+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.

 

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

 

laugh

 Jul 24, 2019

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