How do you evaluate #(3a-2b)/(c-1)# for #a=9,b=3,c=2#?

1 Answer
Makan · EZ as pi
Dec 23, 2016

The answer is #21#

Explanation:

In a problem like this, you just need to plug in your given values into the appropriate variables.

#1.# #(3(9)-2(3))/((2)-1) = (27-6)/(2-1)#

#2.# The answer on the numerator is #21#, whereas the answer on the denominator is just #1#. And #21/1=21#.

#3.# Therefore you get #21# as the final answer.

Hope this helped (c:

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