Question 862167
{{{(x/(a+y))/b=a+b}}}, your first equation;


{{{x/(a+y)^2/b^2=2}}}, your second equation.


Those already look crazy.  Here is another try to understand:

{{{x/a+y/b=a+b}}}, your first equation;
Simplifying, {{{bx+ay=ab(a+b)}}};
{{{x/a^2+y/b^2=2}}}, your second equation.
Simplifying, {{{b^2*x+a^2y=2a^2b^2}}}.
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Continuing with this, these equations are both linear in x and y.  That is assuming if a and b are taken as constants.
System is:
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bx+ay=ab(a+b)
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b^2x+a^2y=a(ab)^2
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Maybe use Elimination Method.
{{{b*bx+b*ay=ab^2(a+b)}}}, as first equation.
{{{(b^2x+a^2y)-(b^2x+aby)=a(ab)^2-ab^2(a+b)}}}, subtracting first from second.
{{{a^2y-aby=a^3*b^2-ab^2(a+b)}}}, the x term has been "eliminated".
{{{(a^2-ab)y=a^3b^2-ab^2(a+b)}}}
{{{y=(a^3b^2-ab^2(a+b))/(a^2-ab)}}}
{{{highlight(y=(a^2b^2-b^2(a+b))/(a-b))}}}
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Next, to solve for x, you could either try elimination of y to solve for x, or you could try substituting what was just found for y into either system equation and solve for x.  
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Here is an attempt substituting.
{{{b^2x+a^2y=a(ab)^2}}}
{{{x*b^2=a^3b^2-a^2y}}}
{{{x=(a^3b^2-a^2y)/(b^2)=a^3-(a/b)^2*y}}}
{{{x=a^3-(a/b)^2((a^2b^2-b^2(a+b))/(a-b))}}}... and then a bit of further work to simplify this.
{{{x=a^3-a^2(a^2-ab)/(a-b)}}}
{{{x=a^3-a^3(a-b)/(a-b)}}}
{{{x=a^3-a^3*1}}}
{{{highlight(x=0)}}}-----Interesting result.