How do you evaluate #12(n+11)-4n#, when #n=1/4#?

1 Answer
Jim G.
Nov 28, 2016

#134#

Explanation:

The first step is to substitute #n=1/4# into the expression.

#rArr12(1/4+11)-4xx1/4#

There are 2 approaches we could take here to evaluate the expression.

#color(blue)"Approach 1"#

distribute the bracket and simplify the multiplication.

#cancel(12)^3xx1/cancel(4)^1+(12xx11)-cancel(4)^1xx1/cancel(4)^1#

#=3+132-1=134#

#color(blue)"Approach 2"#

add the terms inside the bracket and proceed as in Approach 1.

#rArr12(11 1/4)-4xx1/4=12xx45/4-4xx1/4#

#=cancel(12)^3xx45/cancel(4)^1-1#

#=(3xx45)-1=135-1=134#

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