Dividing instead of Multiplying--why??

[Steven G]

I am not a big fan of going from 9x=90 to x=10 by dividing by 9.

So many teachers, tutors, helpers do this and I just do not understand why.

Almost all (if not all) helpers on this forum and most students who come here for help have studied/memorized/practiced times tables. So why when they have a problem like 9x=90 are they told to divide by 9?? Students should know, and most probably do know, that 9*10 = 90 making x=10. Converting a multiplication problem to a division problem in my mind is ludicrous, yet so many use it. If I am tired or not in the mood to think what 90/9 is, I ask myself 9 times what equal 90--which was the original problem!

I have the same problem in problems like 7+x=9. Students know that 7+ 2 = 9, so why tell them to subtract 2 from both sides. I always hear that my way does not show the student's work. OK, so I did ONE very simple computation in my head. Now if you subtract 2 from both sides you are NOT showing TWO simple computations, namely that x+ 2 - 2 = x and 9-2 =7. So in the end using my method I showed more work.

I truly believe that students need to be pushed into thinking more than they are currently being pushed by their teachers.

I remember right here on this forum getting into a dispute with a helper (no names given) who said that students should not be able to easily see that 97*8 + 3*8 =800 but should know that 97x+3x=100x.

I would love to hear your opinion on this.

 

[pka]

So take this equation: 6=(15*x)/(15+x)
Now apparently the answer is x=10, but i require an explanation on how u get there. Also, is there a name for this type of equation? I think it would help if i could google it and get more practice

I am not a big fan of going from 9x=90 to x=10 by dividing by 9.
I would love to hear your opinion on this.

@Jomo, I really cannot understand your concerns. My concern is that some here seem to think that there is one best solution.
That is an absolutely ill informed position for anyone who teaches courses in basic mathematics to hold.
Here is my solution to the OP.
$\displaystyle 6=\frac{15x}{15+x}\\2=\dfrac{5x}{15+x}\\30+2x=5x\\x=10$
Now if you, Jomo, or the student either have a problem with that solution then I say tough.
Maybe the student has chosen the wrong course of study given the student's clear weakness in basic mathematics.
Jomo unless you were a tenured professor, I suspect you would be seeking another position.

 

[Harry_the_cat]

Given that 9*x =90, what is x? Sure we can see that x=10.

But what about 9*x =80. We need to know the process to get x =80/9 (ie divide both sides by 9).

And what about, given that a*b=c, what if you are asked to make 'a' the subject.

I think the processes we teach are preparing the students for algebra at a higher level.

 

[HallsofIvy]

I am puzzled by what you think division is. If I see the equation 9x= 90 then, yes, in this very simple problem which was clearly selected just to make your point I would recognize immediately that 90 is 9 times 10. If I didn't know that wouldn't be able to divide 90 by 9! What if, instead the problem were 103x= 7725? My reasoning would be (which I would do in a split second) that since x is multiplied by 103 to get back to 103, I need to do the opposite of "multiply" which is "divide". Since I know that I have to do the same thing to both sides of an equation in order to keep it a true equation, I would divide 103x by 103 to get x and divide 7725 by 105, either "by hand" or using a calculator to get 75. The solution is x= 75. Are you seriously suggesting that I should have been taught to solve this equation by immediately "seeing" that 7725 is 103 times 75 without dividing?

 

[Dr.Peterson]

I am not a big fan of going from 9x=90 to x=10 by dividing by 9.
...
I truly believe that students need to be pushed into thinking more than they are currently being pushed by their teachers.

I'd say that 9x = 90 is a problem in which we need to find the missing factor. There is a name for that: division. There are also several ways to accomplish that, one of which is "inspection" (recognizing a known multiplication); another is long division; another is to use a calculator. No matter how we do it, we are dividing.

But if your point is that students shouldn't treat this as a rote activity in which they automatically write 9 under a fraction bar on both sides, without thinking at all, then you're absolutely right. They too often bypass the thinking I just described. (But I've seen plenty of students who do think just as I said -- you just can't see what they are thinking because they don't need to write it all out!)

Students should not be required to show every step of their work; an algebra textbook I've used has a specific section devoted to showing students how to do part of the work in their heads! They have to do this because they have until then showed all the details in what appears to be a rote manner, because otherwise they can't show the thinking behind the steps; and because until then they have asked students to write all the steps so the teacher can see that they are thinking correctly. These are training wheels, and must be removed as soon as possible! A good point can be made, too, that students' first exposure to solving equations should not use rote methods at all, but allow the student to think individually, being creative. (This is, in fact, the modern Common Core way to teach these ideas, which many people decry because it looks more complicated than the rote procedures they learned!)

 

[JeffM]

Someone (tkhunny ?) says correct answers do not care how you find them. BUT, as our sole feline helper cogently indicated, a general method for solving linear equations in one unknown is to simplify them to the form of

[MATH]a \ne 0 \text { and } ax = b \implies[/MATH]
[MATH]\dfrac{ax}{a} =\dfrac{b}{a} \text { by the division axiom of algebra, which in turn} \implies x = \dfrac{b}{a}.[/MATH]
Now it is certainly true that in the admittedly broad special case of

[MATH]a \ne 0 \text { and } ax = ac \implies x = c.[/MATH]
But if you are asked to justify that process, you must either introduce an extra axiom or revert to the division axiom.

And obviously, in practice, I would "see"

[MATH]3x = 12 \implies x = 4[/MATH]
without the need for any justificatory steps because I memorized my "threes table." (Like Halls, I never committed my "one hundred and threes table" to memory.)

But what I "see" after more than 60 years of using algebra and what a beginning student is "sure to see" will frequently differ. I want every student to ingrain the following basic axioms of elementary algebra. Then it does not matter what is or is not immediately obvious to the student.

[MATH]x = y \implies x + a = y + a,\ x - a = y - a, \ a * x = a * y, \text { and, if } a \ne 0,\ \dfrac{x}{a} = \dfrac{y}{a}.[/MATH]

 

[Mr. Bland]

Students should know, and most probably do know, that 9*10 = 90 making x=10.

I certainly hope anyone involved in such an equation can easily see that [MATH]x = 10[/MATH]. I hope that it is immediately obvious to anyone who sees it.

That being said, the fundamental concept being studied here is not the arithmetic, but the algebra. It's not about the fact that [MATH]9 * 10 = 90[/MATH], but rather that [MATH]ax = b[/MATH]. The solution is fundamentally division, in that [MATH]x = \frac{b}{a}[/MATH].

If a student can intuitively see that [MATH]\frac{90}{9} = 10[/MATH] without even framing it in the context of division, then that's good. If not (somehow), or if the numbers are different and non-obvious, then understanding the underlying process is more important than knowing the solution for a specific case.

 

[Steven G]

Almost everyone missed the point I was getting at. Sure 9x=90 was made to show my point (actually it was from a recent post). What I am getting at is that students need to be told often that in solving 6x=45 they are trying to find a number that when multiplied by 6 gives 45. I had students who could easily solve these type of problems yet when I asked them 6 times what equals 45 they would say that there is no answer (even though the answer was literally in front of them). I feel that this is a big problem. Teaching rote math does not work as you all know. IMO students need to know exactly what they are solving, ie what the meaning is to the equation. In 2x+5=11 I have been know to say that I am thinking of a number that when you double it and then add 5 to it you get 11, so what is the number. Sure some students will never get this but other can but never will unless you point this out to them. Student, as we all know, memorize way too much math and all I am saying is that they need to go away from that mentality and we as teachers/tutors/mentors/helpers must help them move away from that.

 

[Steven G]

@Jomo, I really cannot understand your concerns. My concern is that some here seem to think that there is one best solution.
That is an absolutely ill informed position for anyone who teaches courses in basic mathematics to hold.
Here is my solution to the OP.
$\displaystyle 6=\frac{15x}{15+x}\\2=\dfrac{5x}{15+x}\\30+2x=5x\\x=10$
Now if you, Jomo, or the student either have a problem with that solution then I say tough.
Maybe the student has chosen the wrong course of study given the student's clear weakness in basic mathematics.
Jomo unless you were a tenured professor, I suspect you would be seeking another position.

I always think that thinking is the best solution. It makes math more fun (compared to doing some boring dry steps that maybe they do not even understand). As a mathematician I was surprised that you do not see that. Having said that I do not think that there is a best solution after thinking if the answer is obvious fails. I 100% agree that students should find their own way of doing problems.

 

[Steven G]

Given that 9*x =90, what is x? Sure we can see that x=10.

But what about 9*x =80. We need to know the process to get x =80/9 (ie divide both sides by 9).

And what about, given that a*b=c, what if you are asked to make 'a' the subject.

I think the processes we teach are preparing the students for algebra at a higher level.

By teaching them rote math? They need to understand the problem they are working on. In solving 2x=6 I am sure that every student who can solve this knows that 2*3=6 and that a VERY high percentage of them would simply write down x=3 if they were taught to think about what the problem is asking.

 

[Steven G]

I am puzzled by what you think division is. If I see the equation 9x= 90 then, yes, in this very simple problem which was clearly selected just to make your point I would recognize immediately that 90 is 9 times 10. If I didn't know that wouldn't be able to divide 90 by 9!

That is exactly my point! I am not sure why you are not getting my point. If you know that 9*10=90, then why bother trying to figure out what 90/9 is? After all, as you said, if you do not know what to multiply 9 by to get 90 then you will not know what 90/9 equals. Students need to know what is being asked in a given problem. A linear equation is a puzzle and the students would benefit by knowing what the puzzle is asking.

 

[Steven G]

I certainly hope anyone involved in such an equation can easily see that [MATH]x = 10[/MATH]. I hope that it is immediately obvious to anyone who sees it.

That being said, the fundamental concept being studied here is not the arithmetic, but the algebra. It's not about the fact that [MATH]9 * 10 = 90[/MATH], but rather that [MATH]ax = b[/MATH]. The solution is fundamentally division, in that [MATH]x = \frac{b}{a}[/MATH].

If a student can intuitively see that [MATH]\frac{90}{9} = 10[/MATH] without even framing it in the context of division, then that's good. If not (somehow), or if the numbers are different and non-obvious, then understanding the underlying process is more important than knowing the solution for a specific case.

As long as a student knows that ax=b is asking what number do you need to multiply a by to get b then I agree with everything you said.

 

[pka]

As long as a student knows that ax=b is asking what number do you need to multiply a by to get b then I agree with everything you said.

If the student does not know then the student has no business in engineering school.

 

[Mr. Bland]

As long as a student knows that ax=b is asking what number do you need to multiply a by to get b then I agree with everything you said.

I often chuckle at my personal struggles to perform subtraction in my head, even though I don't struggle with division. Internally, I always frame a division in the context of "what multiplication produces this result?" instead of trying to perform the mechanical division. It could be said that I don't do well with division either, since I habitually rephrase divisions as multiplications in this way. (For subtraction, I always seem to try to add backwards and for whatever reason I keep confusing myself.)

In my experience, however, there has always been a problem with this approach: for less familiar numbers, it often leads into trial-and-error. "I know the answer is between this and that, so what if I multiply by this? Nope, too high. How about this? Now it's too low." On the other hand, performing division will directly yield the correct result, every time. Using multiplication to solve this kind of problem may be feasible in some (many?) cases, but when intuition and memory fall short, the best one can hope for is to test educated guesses. In contrast, division always succeeds (even if I have to write it down).

The case of [MATH]9x = 90[/MATH] demonstrates a situation where intuition is more useful than arithmetic. It makes sense to conceptualize this problem in the form of "what multiplication produces this result?" Another problem in the same form, such as [MATH]28.3924x = 1083.98[/MATH], may not work through intuition and may require a mechanical division instead. In one of the two cases, intuition will work. In both cases, division will work. For this reason, it's more practical to consider it a problem that is solved by division.

My predominant exposure to algebra is in software engineering, where equations must be solved without knowing any of the variables. Even if the computer receives "easy" numbers, it has no intuition or experience and must consequently perform a division to reach the correct answer. When I see [MATH]9x = 90[/MATH], my training tells me it's a question of division, even if my intuition knows that [MATH]x = 10[/MATH].

 

[firemath]

The subtraction reverse is a problem I have too, @Mr. Bland. I think that the process of the basic arithmetic should be more enforced in earlier work.
For example, most students can figure out that x in the equation 9x=90 equals 10.
But can any of them apply the same reasoning to 708x=28,954.23?

But the students who were taught the ax=b method would be more prepared for the variable switches later in their studies.

 

[JeffM]

That is exactly my point! I am not sure why you are not getting my point. If you know that 9*10=90, then why bother trying to figure out what 90/9 is? After all, as you said, if you do not know what to multiply 9 by to get 90 then you will not know what 90/9 equals. Students need to know what is being asked in a given problem. A linear equation is a puzzle and the students would benefit by knowing what the puzzle is asking.

I have two responses, which, I admit, repeat what I said in my previous post.

If I were talking to a student experienced in algebra, I would go

[MATH]9x = 90 \implies x = 10[/MATH] without any intermediate steps.

If I were talking to a student just starting off, who needs to see all intermediate steps, I would go

[MATH]9x = 90 \implies \dfrac{9x}{9} = \dfrac{90}{9} \implies x = 10[/MATH]
because it is a general method rather than the less general and more roundabout method of

[MATH]9x = 90 \implies 9x = 9 * 10 = \dfrac{9x}{9} = \dfrac{9 * 10}{9} \implies x = 10.[/MATH]
Second, I am not so disturbed by rote rules. Is memorizing the nines table anything but rote learning? Is it in fact necessary in a day of calculators to learn the nines table? Actually, I have no objection to learning a few basic things by rote. For algebra, these include

[MATH]x = y \implies x + a = y + a;[/MATH]
[MATH]x = y \implies x - a = y - a;[/MATH]
[MATH]x = y \implies x * a = y * a; \text { and}[/MATH]
[MATH]x = y \implies \dfrac{x}{a} = \dfrac{y}{a} \text { if } a \ne 0.[/MATH]
I see no reason to add another rule to learn by rote:

[MATH]c * x = c * y \implies x = y[/MATH]

 

[Otis]

… most students who come here for help have [memorized the] times tables …

At the moment, I seem to be at a loss for words.

 

[Steven G]

If I were talking to a student just starting off, who needs to see all intermediate steps, I would go

[MATH]9x = 90 \implies \dfrac{9x}{9} = \dfrac{90}{9} \implies x = 10[/MATH]

This is precisely where I disagree with you. I think that a student should know that 9x=90 is asking 9 times what number is 90. If a students decides to solve this by dividing by 9 that is fine by me. I would wonder why they chose to do that, but as long as they get the answer I will accept that. My problem is when the beginning student is not told that 9x=90 is asking 9 times what number is 90.

 

[Harry_the_cat]

Students are taught about expressions (eg 9x means "9 times a number") before they are taught about equations (eg 9x =90), so that would be a natural progression.

 

[JeffM]

This is precisely where I disagree with you. I think that a student should know that 9x=90 is asking 9 times what number is 90. If a students decides to solve this by dividing by 9 that is fine by me. I would wonder why they chose to do that, but as long as they get the answer I will accept that. My problem is when the beginning student is not told that 9x=90 is asking 9 times what number is 90.

And that is precisely where I disagree with you. First, an equation does not "ask" anything; it is a statement rather than a question. Second, and much more importantly, mathematics is about generalization. If I want to know what number when multiplied by a given number results in a second given number, the general method of solution is division, not memorizing times tables.