+0

help

0
170
3
+1195

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

3+0 Answers

#1
+792
+3

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

Jul 23, 2019
#2
+19914
0

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
+23872
+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.

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

Jul 24, 2019