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

Question

3814 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2018-03-03T07:03:57

#x+y = 1/2#

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

#x^2-y^2=7#

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

So , #x^2-y^2=7#

#=> (x-y)(x+y)=7#

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

We get ,

#14(x+y)=7#

#=>x+y=7/14#

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

#=>x+y=1/2#