Question 755197
Those equations have x and y to degree 1.  Each equation can be simplified but the degrees on x and y will still be 1.  


[(1+b)/b](x) + [(1+a)/a](y) = b-a
As rendered should be as
{{{((1+b)/b)(x) + ((1+a)/a)(y) = b-a}}}
Multiply both sides by {{{a*b}}} to clear the fractions.


{{{x/a - 4y/b = 5}}}
See the l.c.d. is also {{{a*b}}}, so multiply both sides by {{{a*b}}} to clear the fractions.


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I'll jump ahead some, since you may already know how to clear the fractions and simplify each of these simultaneous equations.  You will find you have this system linear in x and y, exponent on x and y being 1:


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{{{a(1+b)x+b(1+a)y=ab^2-ba^2}}}
AND
{{{bx-4ay=5ab}}}
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I would recommend solving the second equation for either x or y and subsitute into the first equation and solve for the single variable there; and then solve for the other variable.   I'll give one of those pathways here:


{{{bx-4ay+4ay=5ab+4ay}}}
{{{bx=5ab+4ay}}}
{{{x=(5ab+4ay)/b}}}
and substitute this for x in the other equation.


{{{a(1+b)x+b(1+a)y=ab^2-ba^2}}}
{{{a(1+b)(1/b)(5ab+4ay)+b(1+a)y=ab^2-ba^2}}}
.
Many very detailed simplification steps too difficult to show in typed text form, so done on paper, yielding this not necessarily finished solution for only y:
.
.
{{{y=b(ab^2-ba^2-5a^2(1+a))/(4a^2(1+b)+b^2(1+a))}}}

Seems to have no special useful factorizations there so the best simplification is 
{{{highlight(y=(ab^3-a^2b^2-5a^2b-5a^3b)/(4a^2+4a^2b+ab^2+b^2))}}}


Best way to find x is to go back to the second equation and solve that for y and substitute into the first equation and go through the long steps to solve for a formula for x.