# even simpler algebra problem

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Either I am missing something very obvious or I've discovered a new unsolveable math problem.....I'm pretty certain I know which statement is true. Helping my son with some middle school math and came across this simple one:

Solve for x:

3(4x-2) = 12x

start by multipling whats in the parenthesis by 3

12x - 6 = 12x

then subtract 12x from each side

-6 = 0x

You see the problem: Can't divide by zero by zero to get x by itself.

I can't remember if these type of problems are common in algebra but my recollection is that you should be able to solve for x in any simple algebra problem. Plus I doubt that Holt Pre-Algebra wanted to spring an undefined answer on the students (and their parents).

Any thoughts?

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Either I am missing something very obvious or I've discovered a new unsolveable math problem.....I'm pretty certain I know which statement is true. Helping my son with some middle school math and came across this simple one:

Solve for x:

3(4x-2) = 12x

start by multipling whats in the parenthesis by 3

12x - 6 = 12x

then subtract 12x from each side

-6 = 0x

You see the problem: Can't divide by zero by zero to get x by itself.

I can't remember if these type of problems are common in algebra but my recollection is that you should be able to solve for x in any simple algebra problem. Plus I doubt that Holt Pre-Algebra wanted to spring an undefined answer on the students (and their parents).

Any thoughts?

Simplifying

3(4x + -2) = 12x

Reorder the terms:

3(-2 + 4x) = 12x

(-2 * 3 + 4x * 3) = 12x

(-6 + 12x) = 12x

Add '-12x' to each side of the equation.

-6 + 12x + -12x = 12x + -12x

Combine like terms: 12x + -12x = 0

-6 + 0 = 12x + -12x

-6 = 12x + -12x

Combine like terms: 12x + -12x = 0

-6 = 0

Solving

-6 = 0

Couldn't find a variable to solve for.

This equation is invalid, the left and right sides are not equal, therefore there is no solution.

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Either I am missing something very obvious or I've discovered a new unsolveable math problem.....

You are absolutely correct.

x is undefined.

(The answer, by the way, is 42 ... not to this problem, but to the big question.)

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You can build any sort of untrue identity by starting with a false premise:

4=0

4+ 12x = 12x

4(1+3x) = 12x

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that reminds me of the following (I'll simplify it a bit though it misses all the magic involved)

2x - 6 = 5(x-3)

2(x-3) = 5(x-3)

divide both sides by (x-3)

2 = 5(x-3) / (x-3)

this leaves

2 = 5

............

years ago I worked with an actuary and he suggested the following for an audit. put the value you want into memory.

then, in front of the auditor start pucnhing different keys and the last step "recall memory"

then, simply say, an alternative way would have been to do the following.

punch a bunch of different keys, and for the last step, hit recall memory...you got it... same answer.

in fact you could actually do it a third way....

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that reminds me of the following (I'll simplify it a bit though it misses all the magic involved)

2x - 6 = 5(x-3)

2(x-3) = 5(x-3)

divide both sides by (x-3)

2 = 5(x-3) / (x-3)

this leaves

2 = 5

............

years ago I worked with an actuary and he suggested the following for an audit. put the value you want into memory.

then, in front of the auditor start pucnhing different keys and the last step "recall memory"

then, simply say, an alternative way would have been to do the following.

punch a bunch of different keys, and for the last step, hit recall memory...you got it... same answer.

in fact you could actually do it a third way....

too funny!

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Another way to solve it:

3(4x-2) = 12x

divide both sides of the equation by 3 and you are left with:

4x-2 = 4x

Although at that point is should be obvious that the equation isn't equal, you can take it a step further by subtracting 4x from each side to get to"

-2=0

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Thanks for all the replies. I didn't realize that math teachers accept "there is no answer" as an answer Just my faulty recollection that you could always solve for X. But you cannot obviously. Maybe thats why they call it "X"?

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12x - 6 = 12x

then subtract 12x from each side

-6 = 0x

If you subtract 12x from each side, you would actually get -6 = 0, not 0x because 12x - 12x = 0

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12x - 6 = 12x

then subtract 12x from each side

-6 = 0x

If you subtract 12x from each side, you would actually get -6 = 0, not 0x because 12x - 12x = 0

Yep.

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get -6 = 0, not 0x because 12x - 12x = 0

I wouldn't say that 0x is wrong, although it's not the way we usually expect to see it.

It's like having 12 apples from which you take 12 apples, so you are left with 0 apples. Of course, having zero apples is about the same as having zero solar systems or zero x's. They all equal 0.

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get -6 = 0, not 0x because 12x - 12x = 0

I wouldn't say that 0x is wrong, although it's not the way we usually expect to see it.

It's like having 12 apples from which you take 12 apples, so you are left with 0 apples. Of course, having zero apples is about the same as having zero solar systems or zero x's. They all equal 0.

But in this case, "x" isn't a label such as inches or apples or solar systems. It ostensibly stands for an unknown quantity. In the case of the original problem, "x" is undefined.

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Nice to know that my little trick is still remembered!!! ##### Share on other sites

Actuarial Exam, Part 1: answer (e) = undefined