Question

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

Answer 1

`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`