Solve the simultaneous equations x+y=a+b;`a/x+b/y=2` :

From the first equation let y=a+b-x
Multiply the second equation on both sides by xy to get:

ay+bx=2xy

Substituting for y:

a(a+b-x)+bx=2x(a+b-x)

`a^2+ab-ax+bx=2ax+2bx-2x^2`

`2x^2-(3a+b)x+(a^2+ab)=0`  Use the quadratic equation:

`x=( (3a+b) +- sqrt((3a+b)^2-4(2)(a^2+ab)))/(2(2))`

`=((3a+b)+-sqrt(a^2-2ab+b^2))/4`

`=> x=(3a+b+a-b)/4==>x=(4a)/4=a` Then y=b

`=>x=(3a+b-(a-b))/4 ==>x=(a+b)/2` Then `y=(a+b)/2`

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The solutions are `(a,b);((a+b)/2 , (a+b)/2 )`

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