Question 353653
Don't be afraid that you'll use too many parentheses, they're free, and they can help avoid a lot of confusion.
Let simplify first.
{{{2/(x-3)-3/(x+5)=(2(x+5))/((x-3)(x+5))-(3(x-3))/((x-3)(x+5))}}}
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{{{2/(x-3)-3/(x+5)=(2(x+5)-3(x-3))/((x-3)(x+5))}}}
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{{{2/(x-3)-3/(x+5)=((2x+10)-(3x-9))/((x-3)(x+5))}}}
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{{{2/(x-3)-3/(x+5)=(2x+10-3x+9)/((x-3)(x+5))}}}
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{{{2/(x-3)-3/(x+5)=(-x+19)/((x-3)(x+5))}}}
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{{{7/(x+2)+3/(x+5)=(7(x+5))/((x+2)(x+5))+(3(x+2))/((x+2)(x+5))}}}
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{{{7/(x+2)+3/(x+5)=(7(x+5)+3(x+2))/((x+2)(x+5))}}}
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{{{7/(x+2)+3/(x+5)=((7x+35)+(3x+6))/((x+2)(x+5))}}}
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{{{7/(x+2)+3/(x+5)=(7x+35+3x+6)/((x+2)(x+5))}}}
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{{{7/(x+2)+3/(x+5)=(10x+41)/((x+2)(x+5))}}}
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Dividing by a fraction is the same as multiplying by its reciprocal,
{{{(2/(x-3)-3/(x+5))/(7/(x+2)+3/(x+5))=((-x+19)/((x-3)(x+5)))*(((x+2)(x+5))/(10x+41))}}}
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{{{(2/(x-3)-3/(x+5))/(7/(x+2)+3/(x+5))=((-x+19)/((x-3)cross((x+5))))*(((x+2)cross((x+5)))/(10x+41))}}}
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{{{(2/(x-3)-3/(x+5))/(7/(x+2)+3/(x+5))=highlight(((-x+19)(x+2))/((x-3)(10x+41)))}}}