Question 118604
I'll do the first two to help you get started.


#33


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





Since the denominators are not equal, we need to get them to a common denominator. Since the LCD is {{{(x-3)(x+3)}}}, we need to get each denominator to {{{(x-3)(x+3)}}}





{{{(((x+3))/((x+3)))((2)/(x-3))+(4)/(x+3)}}} Multiply {{{(2)/(x-3)}}} by {{{((x+3))/((x+3))}}}


 


{{{(x+3)(2)/((x-3)(x+3))+(4)/(x+3)}}} Combine the fractions


 


{{{(x+3)(2)/((x-3)(x+3))+(((x-3))/((x-3)))((4)/(x+3))}}} Multiply {{{(4)/(x+3)}}} by {{{((x-3))/((x-3))}}}


 


{{{(x+3)(2)/((x-3)(x+3))+(x-3)(4)/((x-3)(x+3))}}} Combine the fractions


 


{{{((x+3)(2)+(x-3)(4))/(x-3)(x+3)}}} Since the 2 fractions have the common denominator {{{(x-3)(x+3)}}}, we can combine them. In order to do that, just combine the numerators.



{{{(2x+6+4x-12)/(x-3)(x+3)}}} Distribute



{{{(6x-6)/(x-3)(x+3)}}} Combine like terms



{{{6(x-1)/(x-3)(x+3)}}} Now factor out the GCF 6 from the numerator



So {{{(2)/(x-3)+(4)/(x+3)}}} simplifies to {{{6(x-1)/(x-3)(x+3)}}}




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#35





{{{(4)/(x+1)-(1)/(x-2)}}} Start with the given expression





Since the denominators are not equal, we need to get them to a common denominator. Since the LCD is {{{(x+1)(x-2)}}}, we need to get each denominator to {{{(x+1)(x-2)}}}





{{{(((x-2))/((x-2)))((4)/(x+1))+(1)/(x-2)}}} Multiply {{{(4)/(x+1)}}} by {{{((x-2))/((x-2))}}}


 


{{{(x-2)(4)/((x+1)(x-2))+(1)/(x-2)}}} Combine the fractions


 


{{{(x-2)(4)/((x+1)(x-2))+(((x+1))/((x+1)))((1)/(x-2))}}} Multiply {{{(1)/(x-2)}}} by {{{((x+1))/((x+1))}}}


 


{{{(x-2)(4)/((x+1)(x-2))-(x+1)(1)/((x+1)(x-2))}}} Combine the fractions


 


{{{((x-2)(4)-(x+1)(1))/(x+1)(x-2)}}} Since the 2 fractions have the common denominator {{{(x+1)(x-2)}}}, we can combine them. In order to do that, just combine the numerators.



{{{(4x-8-x-1)/(x+1)(x-2)}}} Distribute




{{{(3x-9)/(x+1)(x-2)}}} Combine like terms




{{{3(x-3)/(x+1)(x-2)}}} Factor out 3 from the numerator



So {{{(4)/(x+1)-(1)/(x-2)}}} simplifies to {{{3(x-3)/(x+1)(x-2)}}}