Common factors? Simplify 8^2 - 16 ÷ 2^2 × 4 - 3

[Squish93]

The question states to simplify the expression, I thought I understood canceling common factors but I don’t understand why you would cancel both 4s on one side of the division sign. I’m completely lost at this point. Wouldn’t canceling 4x4 cancel out the -16 and and make it -1? I get 60.

 

[Otis]

I don’t understand why you would cancel both 4s on one side of the division sign.

Hi Squish. The Order of Operations tells us to do multiplication AND division in the order they appear, reading from left to right.

16 ÷ 4 × 4

The division is done first, then the multiplication. In effect, the multiplication by 4 cancels out the division by 4, bringing us back to 16.

16 ÷ 4 = 4

4 × 4 = 16

Another way to look at it:

\(\displaystyle \frac{16}{\cancel{4}_1} \cdot \frac{\cancel{4}^1}{1} = 16\)


[imath]\;[/imath]

 

[Harry_the_cat]

Do the division before the multiplication (because it appears first when reading from left to right):
\(\displaystyle 64 - 16\div4 \times4 - 3\)
\(\displaystyle =64 - 4 \times4 - 3\)
\(\displaystyle =64 - 16 - 3\)
\(\displaystyle =45\)

@Otis SNAP!

 

[Squish93]

Thank you both! Rereading my old math books and practicing for fun I can’t believe how dumb this makes me feel. On the bright side, I’m smiling because I understand and now I’m not frustrated ?

 

[lookagain]

The question states to simplify the expression, ...

Squish93, on your second line, you need to make your second subtraction line longer, because it started looking like a
multiplication dot.

I see you have no parenthetical/bracketed expressions.
All of your exponentiation should be taken care of in the same step. Then, multiplication/division will be done after that
from left to right, depending on which operation shows up first on the left. I will break down the last step of Harry_the_cat
into two steps, and I will use the "times" sign instead of a multiplication dot.

\(\displaystyle 8^2 - 16 \div 2^2 \times 4 - 3 \ = \)

\(\displaystyle 64 - 16 \div 4 \times 4 - 3 \ = \)

\(\displaystyle 64 - 4 \times 4 - 3 \ = \)

\(\displaystyle 64 - 16 - 3 \ = \)

\(\displaystyle 48 - 3 \ = \)

\(\displaystyle 45 \)

 

[Steven G]

If you have 16÷4x4 you can do the multiplication 1st, ie do 16x4 and then do the division by 4.
I will say that the order of operations says to do the multiplication/division as they come up from left to right. This is not wrong but you can do whichever one you like 1st.

 

[Otis]

do the multiplication/division as they come up from left to right. This is not wrong but you can do whichever one you like 1st

Indeed! I'd done so, myself, upon first recognizing those adjacent operations.

For other readers: When we view division by 4 to be the same as multiplication by ¼, then the Commutative Property of Multiplication tells us that we may multiply the three factors in any order we like. (The green version below is how I'd mentally parsed the 16÷4×4 chunk in the op image.)

(16)(¼)(4)
(16)(4)(¼)
(¼)(16)(4)
(¼)(4)(16)
(4)(16)(¼)
(4)(¼)(16)


[imath]\;[/imath]

 

[Steven G]

Indeed! I'd done so, myself, upon first recognizing those adjacent operations.

For other readers: When we view division by 4 to be the same as multiplication by ¼, then the Commutative Property of Multiplication tells us that we may multiply the three factors in any order we like. (The green version below is how I'd mentally parsed the 16÷4×4 chunk in the op image.)

(16)(¼)(4)
(16)(4)(¼)
(¼)(16)(4)
(¼)(4)(16)
(4)(16)(¼)
(4)(¼)(16)


[imath]\;[/imath]

Personally I like the last one
I never did consider in this situation that dividing by m is the same as multiplying by 1/m. I just realized that when you have a string of just multiplication and division it just doesn't matter which you do first or 2nd or ...

 

[Harry_the_cat]

If you have 16÷4x4 you can do the multiplication 1st, ie do 16x4 and then do the division by 4.
I will say that the order of operations says to do the multiplication/division as they come up from left to right. This is not wrong but you can do whichever one you like 1st.

Yeah you can, but you must apply the multiplication to the 16 not the 4, ie \(\displaystyle 16\div4\times4 =16\times4\div4\) NOT \(\displaystyle 16\div(4\times4)\).

I would be reluctant to say "you can do whichever one you like 1st" without further clarification.

 

[Otis]

I never did consider … dividing by m is the same as multiplying by 1/m

Oh, I misunderstood what you'd done (I'd thought you were mentally multiplying factors).
[imath]\;[/imath]