How do you multiply #\frac { 12r - 16} { 4} \cdot \frac { 1} { 3r - 4}#?

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Answer 1 · 2016-11-07T03:13:06

#(12r-16)/4*1/(3r-4)=1#

Note that #4# can be factored from the numerator of the first fraction.

#(12r-16)/4*1/(3r-4)=(4(3r-4))/4*1/(3r-4)#

The #4# in the numerator will cancel with the #4# denominator, since #4/4=1#:

#(color(red)(cancel(color(black)(4)))(3r-4))/color(red)(cancel(color(black)(4)))*1/(3r-4)=(3r-4)/1*1/(3r-4)#

Multiply straight across:

#(3r-4)/1*1/(3r-4)=(3r-4)/(3r-4)#

Just as before, these will cancel since they are equivalent and we will be left with #1#:

#(3r-4)/(3r-4)=color(red)(cancel(color(black)(3r-4)))/color(red)(cancel(color(black)(3r-4)))=1#