Question 274297
The key to this problem is factoring



Let's factor {{{2xy+5x-2y-5}}}



{{{2xy+5x-2y-5}}} Start with the given expression.



{{{(2xy+5x)+(-2y-5)}}} Group the terms.



{{{x(2y+5)+(-2y-5)}}} Factor out the GCF 'x' from the first group.



{{{x(2y+5)-(2y+5)}}} Factor out the GCF -1 from the second group.



{{{(x-1)(2y+5)}}} Factor out the GCF {{{3y+4}}} from the entire expression.



So {{{2xy+5x-2y-5}}} factors to {{{(x-1)(2y+5)}}}


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Now let's factor {{{3xy+4x-3y-4}}}



{{{3xy+4x-3y-4}}} Start with the given expression.



{{{(3xy+4x)+(-3y-4)}}} Group the terms.



{{{x(3y+4)+(-3y-4)}}} Factor out the GCF 'x' from the first group.



{{{x(3y+4)-(3y+4)}}}  Factor out the GCF -1 from the second group.



{{{(x-1)(3y+4)}}} Factor out the GCF {{{3y+4}}} from the entire expression.



So {{{3xy+4x-3y-4}}} factors to {{{(x-1)(3y+4)}}}


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Now that we have the factorizations, we can now simplify:



{{{(2xy+5x-2y-5)/(3xy+4x-3y-4)}}} Start with the given expression.



{{{((x-1)(2y+5))/(3xy+4x-3y-4)}}} Factor the numerator (see the first factorization above)



{{{((x-1)(2y+5))/((x-1)(3y+4))}}} Factor the denominator (see the second factorization above)



{{{(highlight((x-1))(2y+5))/(highlight((x-1))(3y+4))}}} Highlight the common terms.



{{{(cross((x-1))(2y+5))/(cross((x-1))(3y+4))}}} Cancel out the common terms.



{{{(2y+5)/(3y+4)}}} Simplify.



So {{{(2xy+5x-2y-5)/(3xy+4x-3y-4)}}} simplifies to {{{(2y+5)/(3y+4)}}}



In other words, {{{(2xy+5x-2y-5)/(3xy+4x-3y-4)=(2y+5)/(3y+4)}}}



Note: One thing to point out is that the value of 'x' does not affect the final outcome of the expression since the 'x' terms cancel out. This isn't so obvious when you look at the original expression, but it becomes clear when we reach the last step.