How do you evaluate #c- ( a + 3) + 4+ b# for #a=1, b=1# and #c=4#?

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Answer 1 · 2017-01-06T17:58:20

Plug the variables into the expression

#5#

Because #a=1, b=1, and c=4#, you would plug those numbers into the expression.

#c-(a+3)+4+b " becomes " 4-(1+3)+4+1#

Then the problem is a case of simplifying.

Calculate inside the parenthesis first #(1+3)# to get #4#

Your expression then becomes:

#4-(4) +4+1#. You would then simplify further:

#= 4-4+4+1#

#=5#.