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

Mar 28, 2017

$18$

#### Explanation:

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

$x + \left(3 \left(y + z\right) - y\right)$

We will simplify $3 \left(y + z\right)$

Then subtract $y$ from whatever we get; $\left(3 \left(y + z\right) - y\right)$

And finally add it to $x$; $x + \left(3 \left(y + z\right) - y\right)$

$4 + \left[3 \left(\left(- 2\right) + 6\right) - \left(- 2\right)\right]$

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
S - Subtraction

B - Bracket
O - Of i.e also multiplication
D - Division
M - Multiplication
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 $\left[3 \left(\left(- 2\right) + 6\right) - \left(- 2\right)\right]$ first. But that also has inner parenthesis, so we solve those first. i.e: $\left(\left(- 2\right) + 6\right)$ then we multiply by $3$ and then subtract $- 2$ from what we get

$4 + \left[3 \left(4\right) - \left(- 2\right)\right]$

$4 + \left[12 - \left(- 2\right)\right]$

$4 + \left(12 + 2\right)$

$4 + 14$

$18$