Rational Expressions Reasoning with Equations and Inequalities Solving Linear Equations

Solving equations with fractions

Where did the kid get 48 from? How would you help him, err, not get 48 in the future?

9 replies on “Solving equations with fractions”

I can see that he mutliplied the denominators to get 12 as a denominator, but I’m stumped about where the 4 came from on the other side of the equation. From there he multiplied both sides by 12 and voila, x=48. Obviously this student’s understanding of adding fractions with unlike denominators is weak. That’s where I would start to help him out.

The kid multiplied six by two to either a) combine the two factors, or b) because you get rid of a /2 by doubling everything. If it’s b), then that’s why they also doubled the 2 into a 4. If it’s a), then they probably just doubled the 2 into a 4 because they were multiplying 6 times 2, and if you multiply on one side you have to multiply on the other side because that’s the rule, right? (ie. confused application of keeping equation balanced)

Ditto comments above as to where 48 may come from. What concerns me most is the student NOT seeing that 48 is not a reasonable answer. For a problem that is relatively quick to check, the student did not do so.

Do you think this student would know what to do if presented with 48/2 + 48/6? Do you think the student would divide to get 24 + 8, or do you think the student would try (and possibly fail) to create common denominators for the fractions? I foresee the student coming up with 48/12 based on the work in this problem. Either way, you can show the student that none of these equal 2, but if the student can’t get to that conclusion themselves, there’s definitely some work the teacher needs to do with that student.

I also don’t think adding 48/2 + 48/6 would be any easier for this student. I think 96/8 could be an accidental answer by just adding straight across. In which case 96/8 is 12 and could reaffirm his work due to the 12 in the denominator? Regardless, when I have talked to my students about checking their work when they are this far off, they typically don’t know where to start to check their answer.

My suggest might be to back up and discuss what does this problem even represent. Do they understand what they are trying to solve for and why?

I say, “yes, i hate fractions too. The bigger the number on the bottom, the more I hate it.” So, I hate the biggest denominator, and yet I have to be fair. To get rid of the 6 on the bottom, I have to put a 6 on top, to make a special 1, but I have to give everyone a 6 on top, or they complain.
Now I have \dfrac {6} {2} in the first one, which is 3 (with any luck). So now I have 3x+1x=12 using my special 1 and multiplying the other side.
4x=12 is way easier.
If we are learning to use factors, especially if we are irrationally leaping into rational expressions, I will use factors \dfrac {x} {2} +\dfrac {x} {2\cdot 3} = \dfrac {2} {1}
Then I follow the fairness doctrine – who is missing what? You can only multiply by special ones. The first term is missing a 3, so put a 3 on the top and the bottom (and a 1, for the literal). The second term has everything – or multiply by 1 on top and beneath. The last term has only a 1,so we need to multiply by 2 and by 3 on top and bottom.
\dfrac {3x\cdot 1} {3\cdot 2\cdot 1}+\dfrac {x \cdot 1} {3\cdot 2\cdot 1}=\dfrac {2\cdot 3\cdot 2} {1\cdot 3\cdot 2} (omg I have no idea if this will come out right – LaTex challenge!!
Now they are all over the same thing which is not zero. So I can ignore the bottom. Ha! no more fractions.
This is not the most efficient method, but it does lead in to algebraic addition.
If the LaTex is wrong, it will make no sense, but I found out that the blog owner can edit posts in wordpress, so I am relying on mpershan to fix it for me :-)oh well.

We all tend to ignore the fact that we learn most from our mistakes and failures.

A good training does stress the necessity of a check. While it might be true that the student cannot compute 48/2 + 49/6, I think that is not so. What he (or she?) probably did was trying to follow a wrongly remembered path to the solution, and not checking the solution. This is a failure in math education.

I often found problems with my math simply by checking an example. This kind of behavior is what you should teach. And don’t forget to grant the time to check in tests too.

The kid has no clue about the properties of additon: s/he doesn’t understand that only things with exactly the same units can be combined; if the units don’t match, then a “super category” must be created to combine two unlike units. This kid most also believes that you can add X + Y to get XY, and probably lines up decimals because the teacher “told me to do it that way….”

Here’s my take: in a way to reiterate what josh g. kind of said, the kid looks at it like this: There are two x’s, but I need to narrow it down to only one x. I can cancel out x/2 (hence the cross-out) by multiplying everything by 2. Next, perhaps either because the student does not realize that x can have attached coefficients or because the student does not understand the difference between numerator and denominator, to multiply x/6 by 2 means that you get x/12, as 2*6=12. And finally, the 2 on the right side is multiplied by 2 to get 4. Since x/2 was cancelled out, you’re left with x/12=4. To cancel the x/12 out, multiply everything by 12.

I assume that at the point of x/12=4, the student is thinking that to eliminate fractions you multiply by the denominator. The student sees the result as ____=48, or NULL=48 (whatever demonstrates nothingness to you), and with 48 being the final number left, then x=48.

So I ultimately see two problems in understanding: 1) you *eliminate* (as opposed to simplify) fractions by multiplying by the denominator. 2) what the denominator actually means; if I multiply everything *by* the denominator then do I also multiply the denominator (in the case of x/6 becoming x/12)?

I hope that my take on this was clear.

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