Distributive Property Problem

J

Jason76

Senior Member
Joined
Oct 19, 2012
Messages
1,180
With this:

Simplify

\(\displaystyle 4(x + 9) - 6\) - Use distributive property

\(\displaystyle 4x + 36 -6\)

\(\displaystyle 4x + 30\)

The answer is straightforward

But what about this?

Simplify

\(\displaystyle 4(\dfrac{4x - 2}{2}) - 5\)

What's the first move? :confused:
 
serds said:
Or this: For example

View attachment 2962
Click to expand...
\(\displaystyle \dfrac{10x + 6}{2} \ = \ \dfrac{1}{2}(10x + 6) \ = \ \dfrac{1}{2}\bigg[2(5x + 3)\bigg] \ = \ 5x + 3\) \(\displaystyle \ \ \ \ \ \ \)In your third expression, there is no multiplication going on between \(\displaystyle "\dfrac{1}{2}" \ \) and \(\displaystyle \ "2(5x + 3)"\) without the inclusion of more grouping symbols in the appropriate place(s).
 
\(\displaystyle 4(\dfrac{4x - 2}{2}) - 5\)

\(\displaystyle \dfrac{16x - 8}{2} - 5\) :confused: This is what it seems to be to me.

\(\displaystyle 8x -4 -5\)

\(\displaystyle 8x - 9\)

This is nothing like what you guys wrote. Am I doing something wrong? :confused:
 
Jason76 said:
\(\displaystyle 4(\dfrac{4x - 2}{2}) - 5\)

\(\displaystyle \dfrac{16x - 8}{2} - 5\) :confused: This is what it seems to be to me.

\(\displaystyle 8x -4 -5\)

\(\displaystyle 8x - 9\)

This is nothing like what you guys wrote. Am I doing something wrong? :confused:
Click to expand...
They were using an example, as serds explicitly explained, not your problem. You need to read posts in their entirety.
 
Jason76 said:
\(\displaystyle 4(\dfrac{4x - 2}{2}) - 5\)

\(\displaystyle \dfrac{16x - 8}{2} - 5\) :confused: This is what it seems to be to me.

\(\displaystyle 8x -4 -5\)

\(\displaystyle 8x - 9\)
Click to expand...
Jason76, you might as well take advantage and divide out a factor of 2 between the numerator and denominator, thereby eliminating the fraction right away:
\(\displaystyle 4(\dfrac{4x - 2}{2}) - 5 \ = \)

\(\displaystyle 2(4x - 2) - 5 \ = \)

\(\displaystyle 8x - 4 - 5 \ = \)

\(\displaystyle 8x - 9\)
 
lookagain said:
Jason76, you might as well take advantage and divide out a factor of 2 between the numerator and denominator, thereby eliminating the fraction right away:
\(\displaystyle 4(\dfrac{4x - 2}{2}) - 5 \ = \)

\(\displaystyle 2(4x - 2) - 5 \ = \)

\(\displaystyle 8x - 4 - 5 \ = \)

\(\displaystyle 8x - 9\)
Click to expand...

Post Edited after Mistake:

Here is the most logical way to do it, I think:

- Due to Order of Operations (PEMDAS) We do inside the parenthesis, and see two fractions in one fraction. Next, simplify the fractions (which gets rid of them). Next, do the distributive property. Finally, subtract the two numbers that don't have an x.

\(\displaystyle 4(\dfrac{4x}{2} -\dfrac{2}{2}) - 5 \ = \)

\(\displaystyle 4(2x - 1) - 5 =\)

\(\displaystyle 8x - 4 - 5 =\)

\(\displaystyle 8x - 9 =\)
 
Last edited:
Denis said:
If you can't see that this is the 1st move that makes the most sense:
2(4x - 2) - 5
then explain why...
Click to expand...

\(\displaystyle 4(\dfrac{4x - 2}{2}) - 5\)

The first move was seeing the fraction as two fractions (and then simplifying the fractions before doing the distributive property). But you can still see it as one fraction (from the start) and get a correct answer. In that method you would notice that the \(\displaystyle 4\) on the left side could be divided by the \(\displaystyle 2\) in the denominator (which is actually multiplying \(\displaystyle 4\) by \(\displaystyle \dfrac{1}{2}\)).

\(\displaystyle 2(4x - 2) - 5 \)

\(\displaystyle 8x - 4 - 5\)

\(\displaystyle 8x - 9\)
 
Last edited:
lookagain said:
\(\displaystyle \dfrac{10x + 6}{2} \ = \ \dfrac{1}{2}(10x + 6) \ = \ \dfrac{1}{2}\bigg[2(5x + 3)\bigg] \ = \ 5x + 3\) \(\displaystyle \ \ \ \ \ \ \)In your third expression, there is no multiplication going on between \(\displaystyle "\dfrac{1}{2}" \ \) and \(\displaystyle \ "2(5x + 3)"\) without the inclusion of more grouping symbols in the appropriate place(s).
Click to expand...

I do not agree with your claim that, in the absense of additional grouping symbols, multiplication of 'one-half times two' is not implied in the typeset-notation \(\displaystyle \dfrac{1}{2} 2\).

Reading it in context, as serds presented, I parse that expression as 'multiplication and division -- from left to right, as the operations occur'.

In other words, vertically-typeset ratios can themselves sometimes act in place of formal grouping symbols. In this thread, I think that serd's intent to multiply 1/2*2 is clear, as posted.

Cheers :cool:
 
mmm4444bot said:
I do not agree with your claim that, in the absense of additional grouping symbols, multiplication of 'one-half times two' is not implied It's not merely a "claim," it's a mathematical fact. "implied" doesn't cut. Either write it in one of the correct forms, or don't bother to write it incorrectly. in the typeset-notation \(\displaystyle \dfrac{1}{2} 2\).

Reading it in context, Again, it doesn't matter "in context." Stating something as allowable, because it is "in context" isn't logical. It's more along the lines of emotionalism. as serds presented, I parse that expression as 'multiplication and division -- from left to right, as the operations occur'.

In other words, vertically-typeset ratios can themselves > > sometimes < < <--- That "sometimes" doesn't apply here. act in place of formal grouping symbols. In this thread, I think that > > serd's intent << to multiply 1/2*2 is clear <---- That is a totally irrelevant point. It doesn't change the fact that what serd is posted is wrong (and meaningless), as posted.
Click to expand...


It's wrong no matter who types it, so I am not isolating serds about it. It should always be stated/told that it's wrong, too.And it'll be a reduction of points on any exams/quizzes/homework/projects, at least for the initial infraction. \(\displaystyle \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \) Edit: The intended use of that "expression" is just as wrong in the same sense, as, say, intending \(\displaystyle \ \dfrac{1}{2}\dfrac{1}{2} \ \) to equal \(\displaystyle \ \dfrac{1}{4}.\)
 
Last edited:
I agree with lookagain → \(\displaystyle \dfrac{1}{2}2\) does not have an "implied" multiplication sign (as 2x does) - at least not for beginning algebra class.

We do have \(\displaystyle 2\dfrac{1}{2}\) with an implied "addition" sign. This can create real confusion (I have seen it!) for beginners.
 
I am of the opinion -

If computer/calculator will not take it - don't write it that way.

Saves lot of headache later while trying to "input" equations/functions in computer/calculator.
 
You must log in or register to reply here.
Top