Question 178412

{{{((6x^2-x-12)/(9x^2-16))/((2x^2-7x+6)/(3x^2-x-4))}}} Start with the given expression.



{{{((6x^2-x-12)/(9x^2-16))((3x^2-x-4)/(2x^2-7x+6))}}} Multiply the first fraction {{{(6x^2-x-12)/(9x^2-16)}}} by the reciprocal of the second fraction {{{(2x^2-7x+6)/(3x^2-x-4)}}}.



{{{(((3x+4)(2x-3))/(9x^2-16))((3x^2-x-4)/(2x^2-7x+6))}}} Factor {{{6x^2-x-12}}} to get {{{(3x+4)(2x-3)}}}.



{{{(((3x+4)(2x-3))/((3x-4)(3x+4)))((3x^2-x-4)/(2x^2-7x+6))}}} Factor {{{9x^2-16}}} to get {{{(3x-4)(3x+4)}}}.



{{{(((3x+4)(2x-3))/((3x-4)(3x+4)))(((x+1)(3x-4))/(2x^2-7x+6))}}} Factor {{{3x^2-x-4}}} to get {{{(x+1)(3x-4)}}}.



{{{(((3x+4)(2x-3))/((3x-4)(3x+4)))(((x+1)(3x-4))/((2x-3)(x-2)))}}} Factor {{{2x^2-7x+6}}} to get {{{(2x-3)(x-2)}}}.



{{{((3x+4)(2x-3)(x+1)(3x-4))/((3x-4)(3x+4)(2x-3)(x-2))}}} Combine the fractions. 



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



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



{{{(x+1)/(x-2)}}} Simplify. 



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Answer:



So {{{((6x^2-x-12)/(9x^2-16))/((2x^2-7x+6)/(3x^2-x-4))}}} simplifies to {{{(x+1)/(x-2)}}}.



In other words, {{{((6x^2-x-12)/(9x^2-16))/((2x^2-7x+6)/(3x^2-x-4))=(x+1)/(x-2)}}} where {{{x<>-4/3}}}, {{{x<>4/3}}}, {{{x<>-3/2}}}, {{{x<>3/2}}}, {{{x<>2}}}, or {{{x<>-1}}}.