How do you find the product of #9/(t-2)*((t+2)(t-2))/3#?

2 Answers
Mar 19, 2017

Answer:

The final answer is 3(t+2) or 3t + 6 if you prefer.

Explanation:

You can cancel full binomials just like you can numerical values. So, the #t-2# term in the denominator of the first factor cancels the one in the numerator of the second factor.

#9/cancel(t-2) * ((t+2)cancel(t-2))/3#

Also, since 9 is 3 x 3, you can cancel the denominator of the second factor:

#(cancel9)color(red)3*(t+2)/cancel3 = 3(t+2)#

Mar 19, 2017

Answer:

#3t+6#

Explanation:

#color(blue)"cancel common factors"# on the numerators/denominators of the fractions.

#rArrcolor(red)(9)/color(blue)((t-2))xx((t+2)color(blue)((t-2)))/color(red)(3)#

#=color(red)cancel(9)^3/(color(blue)cancel((t-2)^1))xx((t+2)color(blue)cancel((t-2)^1))/color(red)cancel(3)^1#

#=(3(t+2))/1#

#=3(t+2)#

#=3t+6#