Solving Rational Equations | Three Mistakes

Being a teacher for many years I can tell you I know where students make their mistakes. In this video I will show you the top three mistakes I see students making when solving rational equations








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#Brianmclogan #rationalexpressions

[thewalnutwoodworker]

Still active on YouTube after over 10 years! What a trooper.

[itsNaloxone]

i love how you still manage to teach better than my math teacher

[nyashabrown]

2:04 "That I see students to make"

[its.yara.]

Yea I prefer this math teacher teaching me instead of my others

[MarcleYTClashRoyale]

Is there a way to clone him so he can be my teacher instead

[jasonkataria]

I literally did all of this on my precalculus exam last week. I always learn so much watching your videos.

[steph]

Mr. McLogan I cannot thank you enough! I found you on YouTube one night after wrecking my brain on how to help my 5th grader add/subtract using a mixed fraction. I’ve been out of school for a couple of decades and of course need a refresher here and there. You helped so much! I have subscribed and will keep following along to help my daughter who does not like math to get going after it and be the best math student she can be! Thank you again!

[haruh]

Man
can u believe this is my fav youtuber

[indianbatman]

The most impossible thing for me on earth to get heart from (Brain:mclogan)
Sir your teaching style is so cool

[johnnolen]

In the problem @3:20 you can still cross multiply, you just have to be smart about how you do it. First, factor the denominator on the LHS. Second, add the fractions on the RHS. Third, cancel any factors common to both denominators. Finally carry out the cross multiplication. In this case with the result that 1 = 4(x-1) + 3(x-2).

[Bigstanky]

This is exactly what I needed help with today because my teacher only shoes us her doing the problem instead of giving us notes or anything to refer to. I am so bad at algebra and this has helped me so much thank you.

[sahistabodi]

Hi please can you teach us how to do fraction in ascending order

[MarieK]

Me: still in Algebra 1...
Also me: 🤩🤭
(What grade/Math class would this be in?)

[hiy]

Can you assist me with dimensional analysis

[angelmendez-rivera]

1. Multiplying by the least common multiple is something that one should actually avoid teaching, for two reasons. The first reason is that, as stated later in the video, this tends to get used incorrectly because the method is not consistent. The second reason is that doing so is prone to introducing extraneous solutions, which should be avoided. A better method that is far more general and consistent, avoids introducing extraneous solutions, and is still sufficiently simple for most students to understand, is to first tell them to ensure that they manipulate the equation so that one side is 0, and this can be accomplished by adding the appropriate fractions to both sides of the equation. Once this is accomplished, it will result in a sum of fractions, which you can always rewrite as a single fraction, using the concept of the least common multiple, and then simplifying by using polynomial division on the common factors of the numerator and denominator. Once this is done, you can simplify the equation by eliminating the denominator altogether and only working with the numerator. For example, I would tell students to solve the equation 3/(x – 5) = x/(x + 1) by rewriting the equation as x/(x + 1) – 3/(x – 5) = 0, which you obtain by adding –3/(x – 5) to both sides of the equation. Now that I have a sum of fractions equal to 0, I can turn the sum of fraction into a single fraction. How? The least common multiple of x + 1 and x – 5 is (x + 1)·(x – 5), so x/(x + 1) – 3/(x – 5) = x·(x – 5)/[(x + 1)·(x – 5)] – 3·(x + 1)/[(x + 1)·(x – 5)] = [x·(x – 5) – 3·(x + 1)]/[(x + 1)·(x – 5)] = (x^2 – 8·x – 3)/[(x + 1)·(x – 5)]. Therefore, (x^2 – 8·x – 3)/[(x + 1)·(x – 5)] = 0, and now the left side has been rewritten as a single fraction, as promised. Now, the student should ask themselves if the numerator and denominator have any common factors, and divide those away until no common factors are shared. In this case, x^2 + 8·x – 3 = x^2 + 8·x + 16 – 13 = (x + 4)^2 – 13 = [x + 4 + sqrt(13)]·[x + 4 – sqrt(13)], so there already are no shared factors. Therefore, one can proceed to simply eliminate the denominator altogether, and solve [x + 4 + sqrt(13)]·[x + 4 – sqrt(13)] = 0, which is now easy. This is nice, because these steps work for ANY rational equation, and the steps are all justified by polynomial algebra, not by mindless algorithms like "cross-multiplying" which only work for a very specific type of equation, something that students are never taught to realize because they are never taught why the algorithm works, they are simply told to memorize it.

2. The same method works for 1/(x^2 – 3·x + 2) = 4/(x – 2) + 3/(x – 1). Instead of multiplying by the least common multiple mindlessly, the student add –1/(x^2 – 3·x + 2) to both sides of the equation. This results in 4/(x – 2) + 3/(x – 1) – 1/(x^2 – 3·x + 2) = 0. The left is a sum of fractions, which the student should rewrite as a single fraction. How? Again, with the least common multiple. Notice that x^2 – 3·x + 2 = (x – 1)·(x – 2), so 4/(x – 2) + 3/(x – 1) – 1/[(x – 2)·(x – 1)]. Rewriting this is now easy, because 4/(x – 2) + 3/(x – 1) – 1/[(x – 2)·(x – 1)] = 4·(x – 1)/[(x – 2)·(x – 1)] + 3·(x – 2)/[(x – 2)·(x – 1)] – 1/[(x – 2)·(x – 1)] = [4·(x – 1) + 3·(x – 2) – 1]/[(x – 2)·(x – 1)] = (7·x – 11)/[(x – 2)·(x – 1)]. Thus (7·x – 11)/[(x – 2)·(x – 1)] = 0. Since the numerator and denominator have no common factors, the student can simply conclude 7·x – 11 = 0, and this is because any fraction can only ever be equal to 0 if the numerator is 0, and the denominator is not 0 at the same time. Again, no extraneous solutions occur here.

3. Checking for extraneous solutions here is also unnecessary, because yet again, the same method still works, and you immediately get "No solutions" after the very first step. 1 + 4/(x + 1) = 4/(x + 1) can be rewritten as 1 = 0 by adding –4/(x + 1) to both sides of the equation. Now there is a sum of fractions in the left side, and 0 in the right side, but notice how 1 = 0 is trivially false. Therefore, the equation has no solutions. This is also why multiplying by the least common multiple right away is actually a bad idea.

[stopmopic]

Hello so I have test tomorrow and I was wondering if you could tell me what exactly is expected on algebra l

[guccichumba]

This video was controversial, I’m unsubscribing

[ahmedmustafa]

R u wearing under wear under this dress?