What does #(3a-b)/(2a+b)# equal when #a=4# and #b=3#?

1 Answer
Apr 17, 2017

Answer:

When #a=4# and #b=3#, we have #(3a-b)/(2a+b)=9/11=0.stackrel_81#.

Explanation:

In order to evaluate this expression properly, we must remember two things: (1) the order of operations, and (2) how to treat variables when they have coefficients.

The expression is #(3a-b)/(2a+b)#. The order of operations tells us we need to get a value for #3a-b#, and get a value for #2a+b#, and then divide #(3a-b) divide (2a+b)#.

So how do we get a value for #3a-b#? We are given #a=4# and #b=3#, so we simply substitute these values in. But, the #a# has a coefficient—a number on its left. Coefficients tell you how many of a variable is represented by the term. In this case, the coefficient of #3# says we have three #a#'s. All this means is, after plugging in our value for #a#, we need to multiply that value by #3#, and then subtract #b#. #3a-b=(3 times a) - b#.

Here's how the substitution looks:

#color(white)(=" ") 3a-b#
#=3(4)-3#

And now simplifying:

#=12-3#
#=9#

The same procedure is done to find the value of #2a+b#:

#color(white)(=" ")2a+b#
#=2(4)+3#
#=8+3#
#=11#

Normally, these two simplifications would be done at the same time, since neither one affects the other until the division needs to be done. So, we would simplify the whole expression like this:

#(3a-b)/(2a+b)=(3(4)-3)/(2(4)+3)=(12-3)/(8+3)=9/11#

That's as far as you should need to go. However, if you prefer a decimal expansion, this answer can be written as:

#9/11=0.stackrel_(81)=0.818181...#