1. ## [SOLVED] Help w/ an equation please

If x = 6, y = –2, and z = 3, what is the value of the following expression?

xz + xy/z2

2/3 = is the answer (it tells me)

But what I don't get is this:

(6)3 + (6)-2/9

18 + - 12/9

Now wouldn't it make sense to keep 12/9 because of order of operation and then add -12/9 to 18?

I don't get why it's (18 - 12) / 9 because Order of Operation says divide first then add.

Can someone tell me if there's like an exception to the OOO rule please.

Thanks all

2. Originally Posted by Yerallo
If x = 6, y = –2, and z = 3, what is the value of the following expression?

xz + xy/z2

2/3 = is the answer (it tells me)

But what I don't get is this:

(6)3 + (6)-2/9

18 + - 12/9

Now wouldn't it make sense to keep 12/9 because of order of operation and then add -12/9 to 18?

I don't get why it's (18 - 12) / 9 because Order of Operation says divide first then add.

Can someone tell me if there's like an exception to the OOO rule please.

Thanks all

Is this what the equation looks like in your book?

$\frac{xz + xy}{z^{2}}$

because then

$\frac{(6)(3) + (6)(-2)}{(3)^{2}} = \frac{18 - 12}{9} = \frac{2}{3}$

Puting the expression $xz + xy$ over the expression $z^{2}$ (as the numerator and denominator of a fraction, respectively) acts like puting parenthese around it, $(xz + xy)/z^{2}$ making it the first step in the order of opperations: PEMDAS --Parentheses, Exponents, Multiplication and Division, and Addition and Subtraction. And no, there are no exceptions.

3. Originally Posted by sinewave85
Is this what the equation looks like in your book?

$\frac{xz + xy}{z^{2}}$

because then

$\frac{(6)(3) + (6)(-2)}{(3)^{2}} = \frac{18 - 12}{9} = \frac{2}{3}$

Puting the expression $xz + xy$ over the expression $z^{2}$ (as the numerator and denominator of a fraction, respectively) acts like puting parenthese around it, $(xz + xy)/z^{2}$ making it the first step in the order of opperations: PEMDAS --Parentheses, Exponents, Multiplication and Division, and Addition and Subtraction. And no, there are no exceptions.

thanks man you really are smart.

4. Originally Posted by Yerallo
thanks man you really are smart.
Not a man, but thanks. Truthfully, I am stuck on a question of my own, but I am glad to have been of assistance!