Question 30915
{{{((6+x-x^2)/(x^2-13x+42))divided by((2x^2-5x-3)/(2x^2-13x-7))}}}
when we divide fractions we have to invert the second one and multiply giving us:
{{{((6+x-x^2)/(x^2-13x+42))((2x^2-13x-7)/(2x^2-5x-3))}}} Rewrite this in order of degree of the power
{{{((-x^2+x+6)/(x^2-13x+42))((2x^2-13x-7)/(2x^2-5x-3))}}} now factor if possible
{{{(((-x-2)*(x-3))/((x-6)(x-7)))(((2x+1)(x-7))/((2x+1)(x-3)))}}} Cancel out all factors are the same if one is in the top and one is in the denominator, such as {{{(2x+1)/(2x+1)}}}
{{{(-x-2)/(x-6)}}} Hopes this helps!