Question 107453
{{{((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 by the reciprocal of the second



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



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


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



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



{{{(cross((x-2))cross((x-1))/cross(7)cross((x-2)))*(cross(7)cross((x+1))/cross((x-1))cross((x+1)))}}} Cancel out like terms. Notice every term in the numerator has a corresponding term in the denominator. So everything cancels out



{{{1}}}  So everything simplifies to 1



In other words {{{((x^2-3x+2)/(7x-14))/((x^2-1)/(7x+7))=1}}}