Question 178319
Yikes, there are a lot of problems! I'm going to do the first three to get you started. If you're still stuck, then post again or email me.



# 1


{{{((4x)/(x-1))((x^2-1)/(3x+3))}}} Start with the given expression.



{{{((4x)/(x-1))(((x-1)(x+1))/(3x+3))}}} Factor the second numerator



{{{((4x)/(x-1))(((x-1)(x+1))/(3(x+1)))}}} Factor the second denominator



{{{(4x(x-1)(x+1))/(3(x-1)(x+1))}}} Combine the fractions.



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



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



{{{(4x)/3}}} simplify




So {{{((4x)/(x-1))((x^2-1)/(3x+3))=(4x)/3}}} where {{{x<>-1}}} or {{{x<>1}}}



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# 2


{{{((x-3)/(x+3))/((3x)/(x^2-9))}}} Start with the given expression.



{{{((x-3)/(x+3))*((x^2-9)/(3x))}}} Multiply the first fraction by the reciprocal of the second fraction.



{{{((x-3)/(x+3))*(((x+3)(x-3))/(3x))}}} Factor the second numerator



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



{{{((x-3)highlight((x+3))(x-3))/(3x*highlight((x+3)))}}} Highlight the common terms



{{{((x-3)cross((x+3))(x-3))/(3x*cross((x+3)))}}} Cancel out the common terms



{{{((x-3)(x-3))/(3x)}}} Simplify



{{{((x-3)^2)/(3x)}}} Condense the terms.




So {{{((x-3)/(x+3))/((3x)/(x^2-9))=((x-3)^2)/(3x)}}} where {{{x<>0}}}, {{{x<>-3}}}, or {{{x<>3}}}



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# 3


Since "the length of a rectangular garden is represented by x square +2x over x square + 2x -15 and its width is represented by 2x-6 over x+4", this means that {{{L=(x^2+2x)/(x^2+2x-15))}}} and {{{W=(2x-6)/(x+4)}}}



{{{A=LW}}} Start with the area of a rectangle formula



{{{A=((x^2+2x)/(x^2+2x-15))((2x-6)/(x+4))}}} Plug in {{{L=(x^2+2x)/(x)}}} and {{{W=(2x-6)/(x+4)}}}



{{{A=((x(x+2))/(x^2+2x-15))((2x-6)/(x+4))}}} Factor the first numerator.



{{{A=((x(x+2))/((x+5)(x-3)))((2x-6)/(x+4))}}} Factor the first denominator.



{{{A=((x(x+2))/((x+5)(x-3)))((2(x-3))/(x+4))}}} Factor the second numerator.



{{{A=(2x(x+2)(x-3))/((x+5)(x-3)(x+4))}}} Combine the fractions.



{{{A=(2x(x+2)highlight((x-3)))/((x+5)highlight((x-3))(x+4))}}} Highlight the common terms



{{{A=(2x(x+2)cross((x-3)))/((x+5)cross((x-3))(x+4))}}} Cancel out the common terms



{{{A=(2x(x+2))/((x+5)(x+4))}}} Simplify



{{{A=(2x^2+4x)/((x+5)(x+4))}}} Distribute



{{{A=(2x^2+4x)/(x^2+9x+20)}}} FOIL




So {{{((x^2+2x)/(x^2+2x-15))((2x-6)/(x+4))=(2x^2+4x)/(x^2+9x+20)}}}



This means that the area is {{{A=(2x^2+4x)/(x^2+9x+20)}}}