Another mini-middle-school math conundrum

[Minoosh]

A couple of 8th graders I tutor showed me a quiz they had done poorly on. I ran into a problem that bothered me.

There wasn't technically anything wrong with it, I think. But I could see why it would trip up students.

Instructions were to solve for x.

It was, roughly, 5x + 2 = 5x + 2.

Use inverse operations and show your work.

This is a rare case where I would give EXTRA CREDIT for someone *not* showing their work, because it is blindingly obvious they are the exact same expression. Isn't that like a priori true in philosophy, or something?

My student got dinged for using the "wrong inverse operation." Well, she got rid of the addition fine. But if you're left with 5x = 5x, what is the next step?

If you divide by 5, you get x = x, which is fine, except that we've been telling these kiddos that "solving" means "isolating the variable on one side of the equation" and this doesn't meet that test.

If you divide by x, you get 5=5, another identity, obviously, but there is no longer an x = something.

She got there, infinite solutions, but did get confused on that last step. I think she should still get full credit because although she suddenly tried to divide by 4 (I have no idea why) she had the wits to stop, and go with her gut (or her reasoning).

 

[Puppycow]

Well, I agree that the question seems like a trick question unless it has been covered.

Once you simplify it down to x = x, you can use logic to conclude that any x is equal to itself, and therefore x can take any value.

You just have to go back to the definition of a variable.

 

[Manopolus]

x=x

x-x=x-x

x-x=0

There ya go. All x on one side of the equation. Of course, that still boils down to 0=0.

Should we factor it?

x(1-1)=0

x(0)=0

x/0 = ???? But only if you already knew that. It doesn't actually follow. For my next trick, I'll eat some raw cinnamon.

To be quite honest, if I saw it on a quiz, I'd just write "identity statement" and move on. No solving required. But the teacher may have specified something different according to his own tastes in an earlier lecture. If so, he might expect students to remember it.

I always hate that... when you can get the question right and still not do what the teacher wanted you to do. That's basically just an obedience class for dogs jumping through hoops. Has no place in an educational setting. That's called training, not education. With education, the correct answer is the correct answer, no matter how you determined it, mainly because there's almost always more than one way, even with math.

And yet, the early stages of learning are almost always a training course of sorts. They pretty much have to be to some extent. Best not to add any extra unnecessary, overly specific, BS into it, though, because that just punishes creativity.

So yeah. My bet is that the teacher is a jackass. Many of them are.

Oh, and the next step was actually:

x(0)/0=0/0

But dividing by zero is forbidden. Legend states that if you actually accomplish it, the universe will implode.

 

[Wudang]

Helping my daughter with GCSE (high school age 15) maths I had to read her textbooks and notes to understand what some questions were looking for. I have 2 years uni maths but some of the terms and methods were unfamiliar to me. Some were cool (long division algorithm) and some were IMHO bad examples.

 

[Darat]

Helping my daughter with GCSE (high school age 15) maths I had to read her textbooks and notes to understand what some questions were looking for. I have 2 years uni maths but some of the terms and methods were unfamiliar to me. Some were cool (long division algorithm) and some were IMHO bad examples.

One of the biggest uproars at the community quiz was about the answer to a maths equation - a clear demarcation in age (apart from the couple of teachers) I’d say all the over 40s got one answer - the few under 40s got the “right” answer according to the school teacher quiz master. All caused by what precedent order the different operators and brackets have. I told my team the right answer only because of threads on this forum updating me as to the current preferences, but they wouldn’t have it! In the end both answers were awarded a point.

 

[jt512]

If you divide by 5, you get x = x, which is fine, except that we've been telling these kiddos that "solving" means "isolating the variable on one side of the equation" and this doesn't meet that test.

But that is not what solving an equation means. Solving an equation means finding all the solutions to the equation, that is, the set of values of the variables in the equation for which the equation is true. So, the solution to the equation in the problem is "all x" (presumably, all real x). Using inverse operations to isolate the variable to be solved for is a technique to find the solution...a technique that doesn't work when the equation is an identity.

I have no idea why the teacher would instruct the student to solve this problem with a technique that doesn't work for it, unless the teacher specifically explained that x=x implies that the solution is "all x." Well, I do have an idea, and it is not complimentary to the teacher.

 

[Wudang]

Perhaps the answer they're looking for is that the set of valid values for x is isomorphic with the universal set?

See also https://www.smbc-comics.com/comic/2014-12-06

 

[Manopolus]

{x|x=all numbers}

/calculus

Imaginary numbers also count, since 5i+2 = 5i+2, obviously.

I barely remember calculus, though. It has been a while, and I was often hungover in class that year. I can just imagine a math teacher's response if an 8th grader actually put set notation on the test. That would be hilarious.

 

[Ziggurat]

Imaginary numbers also count

So do quaternions. But not vectors, since you can't add a scalar to a vector.

 

[Checkmite]

If you divide by 5, you get x = x, which is fine, except that we've been telling these kiddos that "solving" means "isolating the variable on one side of the equation" and this doesn't meet that test.

I thought "solving" meant finding out what x has to be. The way I was taught, isolating the variable to one side is just "simplifying".

 

[Minoosh]

But that is not what solving an equation means. Solving an equation means finding all the solutions to the equation, that is, the set of values of the variables in the equation for which the equation is true. So, the solution to the equation in the problem is "all x" (presumably, all real x). Using inverse operations to isolate the variable to be solved for is a technique to find the solution...a technique that doesn't work when the equation is an identity.

I have no idea why the teacher would instruct the student to solve this problem with a technique that doesn't work for it, unless the teacher specifically explained that x=x implies that the solution is "all x." Well, I do have an idea, and it is not complimentary to the teacher.

This probably has been covered in their regular math class, but the students I have frequently take a little longer to "get" a concept.

On a practice quiz an equation boiled down to x = -x and the teacher thought that meant no solution. I let him know, tactfully, that there actually was a solution.

I thought "solving" meant finding out what x has to be. The way I was taught, isolating the variable to one side is just "simplifying".

The kind of problems they're working, getting x on one side IS solving because we've been simplifying the other side as we go on. x = 6/3 would be numerically correct but generally not accepted, there would be one more step. You would have a numerical answer, as long as no other variables were in the equation to start.

My students are generally taught that we "simplify" expressions and "solve"
equations, although the equations are also simplified during the process of solving.

 

[Ziggurat]

On a practice quiz an equation boiled down to x = -x and the teacher thought that meant no solution. I let him know, tactfully, that there actually was a solution.

For some reason that reminded me of another problem, which is basically a geometry problem. Consider an idealized spherical earth. From what point on the surface can you walk 1 mile south, 1 mile west, and 1 mile north and end up exactly where you started? No funny business like vertical motion or extra steps involved.

Most people find one particular solution fairly quickly. But there's more than one solution. But there is actually more than one solution.

There are actually infinitely many solutions. And if you tell people a second solution, they can often figure out that a second solution is actually not one point, but a line of solutions. But even that isn't the end of it, because

there's an infinity of sets of solutions which are each different lines.

 

[Yalius]

The north pole, and the line circumscribing the south pole ~(1+1/pi) miles north of the pole. I don't feel like doing the coordinate transformation to get the exact solution for 2d/3d distance along the ground or in 3d space. Where are the infinite sets of lines for solutions?

 

[Ziggurat]

The north pole, and the line circumscribing the south pole ~(1+1/pi) miles north of the pole. I don't feel like doing the coordinate transformation to get the exact solution for 2d/3d distance along the ground or in 3d space. Where are the infinite sets of lines for solutions?

The line you found is the one where you approach the south pole but don't reach it, circle around the south pole once, and then return to your starting point. And that's a line of solutions (in a circle) since any point at the correct latitude will work, regardless of longitude. Get a bit closer to the south pole, and you can circle around twice before returning. So that's a second line of solutions. Get even closer, and you can circle around three times before returning for a third line of solutions. And so on, for all positive integers.

So the distances of these lines from the south pole are approximately 1+1/(n*pi) miles, for all positive integer n. With some math we could calculate the exact distances, but that's not necessary to understand what the solutions look like.

 

[pzkpfw]

One of the biggest uproars at the community quiz was about the answer to a maths equation - a clear demarcation in age (apart from the couple of teachers) I’d say all the over 40s got one answer - the few under 40s got the “right” answer according to the school teacher quiz master. All caused by what precedent order the different operators and brackets have. I told my team the right answer only because of threads on this forum updating me as to the current preferences, but they wouldn’t have it! In the end both answers were awarded a point.

What changed?

(I (53) do remember the method for doing subtraction on paper my daughter was taught in primary school (messing with the lower number) seemed "better" than what I was taught (messing with the upper number). But the precedence etc my kids were taught seemed the same as for me.)

 

[Manopolus]

What changed?

(I (53) do remember the method for doing subtraction on paper my daughter was taught in primary school (messing with the lower number) seemed "better" than what I was taught (messing with the upper number). But the precedence etc my kids were taught seemed the same as for me.)

Yeah, I wish they taught methods as methods from the start and quit pretending that their preferred method is the only way to do math. But that wasn't what you were asking. It just came up.

And then we could all switch to base 12 and live happily ever after (okay, maybe not).

 

[Yalius]

The line you found is the one where you approach the south pole but don't reach it, circle around the south pole once, and then return to your starting point. And that's a line of solutions (in a circle) since any point at the correct latitude will work, regardless of longitude. Get a bit closer to the south pole, and you can circle around twice before returning. So that's a second line of solutions. Get even closer, and you can circle around three times before returning for a third line of solutions. And so on, for all positive integers.

So the distances of these lines from the south pole are approximately 1+1/(n*pi) miles, for all positive integer n. With some math we could calculate the exact distances, but that's not necessary to understand what the solutions look like.

Got it, hadn't thought about integer multiples of circles.

 

[Ziggurat]

Got it, hadn't thought about integer multiples of circles.

That's why I like the problem. People usually get the first solution quickly, but the second one is harder. And even if they get the second one they often don't make the leap to the rest (although once you're past the second one, the pattern becomes obvious pretty quick). So when you ask someone this problem in person and they give you an answer, you get to keep saying, "yes, but there's more."