How do you evaluate #x+(3(y+z)-y)# if w=6, x=4, y=-2, z=6?

1 Answer
Mar 28, 2017

Answer:

#18#

Explanation:

First, substitute the values of #x,y,z# into the equation.

#x+(3(y+z)-y)#

We will simplify #3(y+z)#

Then subtract #y# from whatever we get; #(3(y+z)-y)#

And finally add it to #x#; #x+(3(y+z)-y)#

#4+[3((-2)+6)-(-2)]#

Depending on what country you're studying, you will either have PEMDAS or BODMAS as the order of operation.

P - Parenthesis
E - Exponent
M - Multiplication
D - Division
A - Addition
S - Subtraction

B - Bracket
O - Of i.e also multiplication
D - Division
M - Multiplication
A - Addition
S - Subtraction

Now let's take a look at the problem again. We have parenthesis (or bracket), multiplcation, addition, and subtraction.

Using the order we solve #[3((-2)+6)-(-2)]# first. But that also has inner parenthesis, so we solve those first. i.e: #((-2)+6)# then we multiply by #3# and then subtract #-2# from what we get

#4+[3(4)-(-2)]#

#4+[12-(-2)]#

#4+(12+2)#

#4+14#

#18#