If $x - y = 14$ $x-y=14$ and $x^{2} - y^{2} = 7$ $x^2-y^2 = 7$ , what is the value of $x + y$ $x+y$ ?

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Answer 1 · 2018-03-03T07:03:57

$x + y = \frac{1}{2}$ $x+y = 1/2$

$x - y = 14$ $x-y=14$ ----(1)

$x^{2} - y^{2} = 7$ $x^2-y^2=7$

We have an identity for $a^{2} - b^{2}$ $a^2-b^2$ which is equal to $(a - b) (a + b)$ $(a-b)(a+b)$ .

So , $x^{2} - y^{2} = 7$ $x^2-y^2=7$

$\Rightarrow (x - y) (x + y) = 7$ $=> (x-y)(x+y)=7$

Use the value of $(x - y)$ $(x-y)$ from (1).

We get ,

$14 (x + y) = 7$ $14(x+y)=7$

$\Rightarrow x + y = \frac{7}{14}$ $=>x+y=7/14$

$=>x+y=(cancel7^1)/(cancel14^2)$

$=>x+y=1/2$

If x-y=14x−y=14x-y=14 and x^2-y^2 = 7x2−y2=7x^2-y^2 = 7, what is the value of x+yx+yx+y?