Question 856404
There is more than one way to go about it, but I do not like to have to write all those denominators.
To start, I would eliminate denominators by multiplying times a convenient factor, and the I would simplify:
 
{{{(x-2)/3+(y+1)/5=2}}}-->{{{15(x-2)/3+15(y+1)/5=2*15}}}-->{{{5(x-2)+3(y+1)=30}}}-->{{{5x-10+3y+3=30}}}-->{{{5x+3y=30+10-3}}}-->{{{5x+3y=37}}}
 
{{{(x+2)/7-(y+5)/3=-2}}}-->{{{21(x+2)/7-21(y+5)/3=-2*21}}}-->{{{3(x+2)-7(y+5)=-42}}}-->{{{3x+6-7y-35=-42}}}-->{{{3x-7y=-42-6+35}}}-->{{{3x-7y=-13}}}
 
Then I would set to solve the system
{{{system(5x+3y=37,3x-7y=-13)}}} by elimination.
Multiplying the first equation times 3, and the second times -5,
and adding the resulting equations, I would get
 15x + 9y = 111
-15x +35y = 65
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0x +44y = 176   --> {{{44y=176}}} --> {{{y=176/44}}} --> {{{highlight(y=4)}}}
Then I would substitute 4 for y in {{{5x+3y=37}}} to get
{{{5x+3*4=37}}}-->{{{5x+12=37}}}-->{{{5x=37-12}}}-->{{{5x=252}}}-->{{{x=25/5}}}-->{{{highlight(x=5)}}}