Question 251531


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



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



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



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



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



{{{(highlight(v)highlight((v+3))(v+1))/(highlight(v)highlight((v+3))(v-1))}}} Highlight the common terms. 



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



{{{(v+1)/(v-1)}}} Simplify. 



So {{{((v^2+3v)/(v^2+2v-3))/((v)/(v+1))}}} simplifies to {{{(v+1)/(v-1)}}}.



In other words, {{{((v^2+3v)/(v^2+2v-3))/((v)/(v+1))=(v+1)/(v-1)}}}