Question 146709


{{{((x^2-3x+2)/(7x-14))/((x^2-1)/(7x+7))}}} Start with the given expression.



{{{((x^2-3x+2)/(7x-14))((7x+7)/(x^2-1))}}} Multiply the first fraction {{{(x^2-3x+2)/(7x-14)}}} by the reciprocal of the second fraction {{{(x^2-1)/(7x+7)}}}.



{{{(((x-1)*(x-2))/(7x-14))((7x+7)/(x^2-1))}}} Factor {{{x^2-3x+2}}} to get {{{(x-1)*(x-2)}}}.



{{{(((x-1)*(x-2))/(7(x-2)))((7x+7)/(x^2-1))}}} Factor {{{7x-14}}} to get {{{7(x-2)}}}.



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



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



{{{((x-1)*(x-2)7(x+1))/(7(x-2)(x-1)*(x+1))}}} Combine the fractions. 



{{{(highlight(x-1)highlight(x-2)highlight(7)highlight(x+1))/(highlight(7)highlight(x-2)highlight(x-1)highlight(x+1))}}} Highlight the common terms. 



{{{(cross(x-1)cross(x-2)cross(7)cross(x+1))/(cross(7)cross(x-2)cross(x-1)cross(x+1))}}} Cancel out the common terms. 



{{{1}}} Simplify. 



So {{{((x^2-3x+2)/(7x-14))/((x^2-1)/(7x+7))}}} simplifies to {{{1}}}.



In other words, {{{((x^2-3x+2)/(7x-14))/((x^2-1)/(7x+7))=1}}} where.{{{x<>-1}}}, {{{x<>1}}} or {{{x<>2}}}