Question
9(y+2) ÷ 3 x 2 + 1
Since I am self-studying and my textbook only provides answers to half of the questions, I am not able to find the answer to this question and have some doubts. Does multiplying out the bracket represent the same priority as any other multiplication?
There are several important ideas to deal with.
First, "order of operations" and "multiplying out brackets" are in different categories; they don't really interact at all. The former tells us how to interpret the
meaning of an expression; "9(y+2) ÷ 3 x 2 + 1" means that to find the
value of the expression, we multiply 9 by the sum of y and 2, then divide the result by 3, then multiply the result by 2, then add 1. This is because the expression in parentheses is treated as a single quantity ("P"), and the multiplications and divisions are done from left to right by convention ("MD") , and then the addition is performed ("AS").
Given that meaning, we can apply properties of operations to
transform the expression to another that has the same value; for example, if we wished, it turns out that we could think of it as [MATH]\frac{9(y+2)}{3}\cdot 2 + 1[/MATH], and cancel to change that to [MATH]3(y+2)\cdot 2 + 1[/MATH]. PEMDAS is not a straightjacket but a lens through which to see.
What you are told to do here is similar to that: You are to
simplify the expression, which means that you want to write a new expression that will always have the same value, but is simpler. In doing this, you must
retain the meaning; so in particular, you must ensure that 9(y+2), in its entirety, is divided by 3. When you write 9y + 18 ÷ 3 x 2 + 1, you have violated the requirement not to change the meaning, because now the 18 ÷ 3 is done before adding 9y. In order to retain the meaning, you need to add parentheses: (9y + 18) ÷ 3 x 2 + 1.
By the way, you added "= 0" to the expression, and changed the problem from simplifying to
solving an equation that you were not told is true (unless you failed to show us the actual problem you were working on!)
I don't understand the rule which says we should retain the brackets when multiplying out.
At what point can we add other terms with the contents of the parentheses?
Do we consider multiplying out the parentheses an operation of 'P' or 'M'?
I guess we should always assume an invisbible '1' prefixes the parentheses after it has been multiplied out...
If I make an example up of:
10(9y + 3c + 2) ÷ 3 x 4 + 10 x 10 x 2(7y + c) + 2
My attempt
10(9y + 3c + 6) ÷ 3 x 4 + 10 x 10 x 2(7y + c) + 2 = 0
(90y + 30c + 60) ÷ 3 x 4 + 10 x 10 x 2(7y + c) + 2 = 0
(90y + 30c + 60) ÷ 3 x 4 + 10 x 10 x (14y + 2c) + 2 = 0
(30y +10c + 20) x 4 +10 x 10 x (14y + 2c) + 2 = 0
(120y + 40c + 80) + 10 x 10 x (14y + 2c) + 2 = 0
(120y + 40c + 80) + 100 x (14y + 2c) + 2 = 0
(120y + 40c + 80) + 100 x (14y + 2c) + 2 = 0
(120y + 40c + 80) + (140y + 200c) + 2 = 0
260y + 240c + 82 = 0
Do you see now why "we should retain the brackets when multiplying out"? This is not part of PEMDAS proper, but rather part of not changing the meaning when you rearrange an expression. You just have to be sure that the new expression
treats as a single quantity whatever
is a single quantity. And rather than thinking of magic invisible 1's, you might want to add extra parentheses wherever that expresses the correct meaning, like (((9(y+2)) ÷ 3) x 2) + 1, if you need to go that far. Those parentheses express what PEMDAS tells you about the meaning.
In this new example (apart from the fact that you again changed it from an expression to equation), you are right until the last line, where you multiplied 100*14 incorrectly. (By the way, please avoid using "x" to mean multiplication in an algebraic expression, as it looks too much like a variable.)
