Question 158977

{{{((t^2-2t-3)/(t^2-1))((3t-3)/(t^2-4t+3))}}} Start with the given expression.



{{{(((t+1)*(t-3))/(t^2-1))((3t-3)/(t^2-4t+3))}}} Factor {{{t^2-2t-3}}} to get {{{(t+1)*(t-3)}}}.



{{{(((t+1)*(t-3))/((t-1)*(t+1)))((3t-3)/(t^2-4t+3))}}} Factor {{{t^2-1}}} to get {{{(t-1)*(t+1)}}}.



{{{(((t+1)*(t-3))/((t-1)*(t+1)))((3(t-1))/(t^2-4t+3))}}} Factor {{{3t-3}}} to get {{{3(t-1)}}}.



{{{(((t+1)*(t-3))/((t-1)*(t+1)))((3(t-1))/((t-1)*(t-3)))}}} Factor {{{t^2-4t+3}}} to get {{{(t-1)*(t-3)}}}.



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



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



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



{{{(3)/(t-1)}}} Simplify. 



So {{{((t^2-2t-3)/(t^2-1))((3t-3)/(t^2-4t+3))}}} simplifies to {{{(3)/(t-1)}}}.



In other words, {{{((t^2-2t-3)/(t^2-1))((3t-3)/(t^2-4t+3))=(3)/(t-1)}}} where {{{t<>-1}}}, {{{t<>1}}}, or {{{t<>3}}}