Question 121336
#1


{{{4x/(x-5)-20/(x-5)}}} Start with the given expression



{{{(4x-20)/(x-5)}}} Since the denominators are equal, we can combine the fractions



{{{(4(x-5))/(x-5)}}}   Factor {{{4x-20}}} to get {{{4(x-5)}}} 




{{{4cross((x-5))/cross((x-5))}}} Cancel like terms



{{{4}}} Simplify



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Answer:


So {{{4x/(x-5)-20/(x-5)}}} simplifies to {{{4}}}. In other words {{{4x/(x-5)-20/(x-5)=4}}}




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#2



{{{((x^2-36y^2)/(5x^2-30xy))/((x^2+6xy)/1)}}} Start with the given expression



{{{((x^2-36y^2)/(5x^2-30xy))((1)/(x^2+6xy))}}} Multiply the first fraction by the reciprocal of the second fraction




{{{(((x+6y)(x-6y))/(5x^2-30xy))((1)/(x^2+6xy))}}}   Factor {{{x^2-36y^2}}} to get {{{(x+6y)(x-6y)}}} 


{{{(((x+6y)(x-6y))/(5x(x-6y)))((1)/(x^2+6xy))}}}   Factor {{{5x^2-30xy}}} to get {{{5x(x-6y)}}} 


{{{(((x+6y)(x-6y))/(5x(x-6y)))((1)/(x(x+6y)))}}}   Factor {{{x^2+6xy}}} to get {{{x(x+6y)}}} 



{{{(x+6y)(x-6y)/5x(x-6y)x(x+6y)}}} Combine the fractions




{{{cross((x+6y))cross((x-6y))/(5x*cross((x-6y))x*cross((x+6y)))}}} Cancel like terms




{{{1/5x^2}}}Simplify




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Answer:


So {{{((x^2-36y^2)/(5x^2-30xy))/((x^2+6xy)/1)}}} simplifies to {{{1/5x^2}}}. In other words {{{((x^2-36y^2)/(5x^2-30xy))/((x^2+6xy)/1)=1/5x^2}}}