What if you have a rational expression where both the numerator and the denominator were factorable in themselves? Would you factor both the numerators and denominators or just the denominators? Thank you.
You would in that case factor everything and try to cancel before you find the LCD. In most cases, they won't give you questions like that but it could happen. Let's say you had: 2/8 and 1/2 If you don't cancel first, the LCD is 8, but in reality the 2/8 is really 1/4 after you simplify and so the real LCD is 4. It's the same concept with rational expressions.
The LCD is 30, since you have denominators of 10 and 15. So now you need to rewrite the fraction 3/10 as an equivalent fraction with the LCD 30 as the denominator: 3/10 * 3/3 = 9/30 3/3 is just 1 and multiplying by 1 does not change the number.
hi! i just discovered your video today and it helped me a lot with my assignment! im down with most of it except one. could you perhaps help me with it? i'd really appreciate it! the problem is: find the LCD of x²/x-3 and 4/3-x its really confusing, ty!
The denominator (x - 3) occurs one time and the denominator (3 - x) occurs one time, so the LCD is (x - 3)(3 - x). Now for this problem, you may notice that -(3 - x) = x - 3, so what you could do is multiply the second fraction by -1/-1 and then write the second fraction as -4/(x - 3). Now, you have a common denominator of (x - 3), which would be the LCD. I guess it just depends on what the teacher wants.
You want the LCD? You have a denominator of x and a denominator of x + 1. What's in common? Nothing so the answer is the product of the denominators: LCD: x(x + 1)
So for the numbers you have: 6, 12, and 36, this leads to a number part of 36. 6 = 2 x 3 12 = 2 x 2 x 3 36 = 2 x 2 x 3 x 3 LCM(6, 12, 36) = 2 x 2 x 3 x 3 = 36 For the variable part it is much easier, we have a y in each denominator, so we just go with the largest exponent which is 5. LCM(y^2, y^5, y^4) = y^5 If you put these two results together, you get your LCD: 36y^5
At 15:02 - Why do you select the terms you do- i.e. why do you decide to find the LCM of the second term from the denominator of the first fraction, and the first term from the second term as opposed to the other way around? Could you do it the other way well(first term from first fraction, second term from second fraction) or would that not work? . Is there a rule to this, or do you just pick the ones that will be easier to factor? I'm having trouble understanding this one part.
You can find the LCM in any order that you want. I was trying to show that (x + 3) is common to both, as many students will miss this. When you build the LCM it involves multiplication, which can be done in any order.
15:16? What if the denominator didn't have exponent will I still do the same? Oops never mind you had one in the next example How would I do it if the denominator didn't had variables?
can someone tell me why he divided 3/10 with 3/3 instead of 2/2 and vice versa with 7/15? Will it make a big difference 3/10 was divided by 2/2 and if 7/15 was divided by 3/3? hehe, It just seems like a mistake i'll make if so edit: plus where did the "x+3" for 5x² come from in 15:25?
The whole point is we are factoring the denominators to find the LCD. So one denominator is 5x^2 + 15x = 5x(x + 3). In other words we can factor out the 5x part to get 5x(x + 3). As for your first question, we aren't dividing, we are multiplying. 3/10 is multiplied by 3/3 to get a denominator of 30. 7/15 is multiplied by 2/2 to get a denominator of 30. You need a common denominator, that's the whole point.
You can simplify each of those to make your task a bit easier: (4x + 12) / (x^2 + 5x + 6) Let's factor the numerator and the denominator: 4(x + 3) / (x + 2)(x + 3) We can cancel a common factor of (x + 3) 4 / (x + 2) The second one is also able to be simplified: (5x + 15) / (10x + 20) 5(x + 3) / 10(x + 2) You can cancel a common factor of 5: (x + 3) / 2(x + 2) So now your problem is what is the LCD of (4) / (x + 2), (x + 3) / 2(x + 2) The denominators each have an (x + 2), then the second one has a 2. Since the (x + 2) is repeated, we only need 1: LCD = 2(x + 2) I just want to put a note here, if you are asked for the restricted values, use the original denominators, not the simplified ones.
The 3 in 3/10 is given. The 3 from factoring the number 15 as 5 * 3? I'm unclear on which 3 you mean, again please leave a time marker so I can answer your question.
So for the first one, you would just multiply everything by the LCD which is 5x: After you simplify you will get: 10 - x^2 - x = -4x Which can be changed into: x^2 - 3x - 10 = 0 which has solutions: x = 5, -2
For the second one, you would multiply everything by the LCD, which is 2(x + 3): 4 + x(x - 3) = -(x - 3) This can be changed into: x^2 - 2x + 1 = 0 (x -1)^2 = 0 x = 1
Dear sir you are explaining the LCM and LCD ONLY in one method either we have other easy ways to find. Can you please guide me about another easy way of this.
Take all your denominators and find the LCM: (x - 3), 2, 2 So you will multiply just one copy of 2 by (x - 3) LCD will be: 2(x - 3) If you want to solve the equation, multiply both sides by the LCD: 4 + x^2 - 3x = -x + 3 This will simplify to: x^2 - 2x + 1 = 0 Which can be factored as: (x - 1)^2 = 0 So x is just equal to 1. x = 1 Hope this helps :)
So your three denominators are p^2, p^3, and 1. Your LCD is the LCM of the denominators. Since one of them is 1, you don't need to even worry about it. You basically have p^2 and p^3. With variables for the LCM you go with the highest exponential power. In this case it's a 3, so your LCM/LCD is going to be p^3.
What are those slashes in front of the 12? You can add the left side by getting a common denominator. 2/3 + 1/6 = 4/6 + 1/6 = 5/6 I don't know what you are trying to do on the right side. Did you miss anything with the problem? Variables or question marks? It's not very clear what you are trying to do. Let me know and I'll show you how to get a solution.
I dont understand, why do you have to choose the larger exponent 5 when you calculate lcd of 21x^3 and 12x^5. It doesnt make sense, i think there should be more explanation why this is how you should write the expression. Also if theres any method to find the prime of a term with a variable raised to an exponent, it would be helpful.
Let's start with the first question. Why is the LCM of x^3 and x^5 given as x^5? With numbers, the LCM is the smallest that is divisible by all numbers in the group. So for example with 2 and 3, the LCM is 6, since 6/2 = 3 and 6/3 = 2. The same logic applies when we work with variables. What's the smallest power on x that will be divisible by both x^5 and x^3. It will always be the variable raised to the largest power in the group. So x^5/x^5 = 1 and x^5/x^3 = x^2. I think this lesson may help you: www.greenemath.com/Algebra1/45/LCDofRationalExpressionsLesson.html It's from our website, which is free. Method to find the prime of a term raised to an exponent? Just break it up and find the prime factorization of the number and the prime factorization of the variable part separately, then put them back together. Hope this helps you and good luck in your studies.
can someone please tell me how x + 100,000/x became x^2+10,000 when simplified? this LCD as well is included in our lesson but i totally forgot how to do this. thank you so much 🥲
No, not quite, if you had two x's next to each other like that it implies multiplication so xxy^2 is really x^2y^2. In your case, if you are trying to find the LCM of those terms, you would only have one single x. You only put the largest number of repeats of any given factor. 3x -> 3 * x 5 -> 5 4xy^2 ->4 * x * y * y So you want to make a list of every factor that occurs but if there is a duplicate, only put in the largest number of repeats. See how x is in 3x and also in 4xy^2, so you only have 1 that goes in your list: LCM (3x, 5, 4xy^2) = 3 * 5 * 4 * x * y^2 = 60xy^2
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i learned more by watching this video than by our online class teacher i-
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I have been stuck on a question for 2 hours but i watch this for 2 minutes and I understand thank you
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This helps so much thanks, I've been stuck on this assignment and this really helped.
Glad it helped!
Oh my god. You are so great. I couldnt listen well in my math class because i am at the back and i was going to cry.
You can do it!
excellent teacher deserves the best and deserves a big hype!!!!!!
cool thank you :)
i have an exam tomorrow, thank you for making this
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You just save my PRE-COLLEGE MATH class thanks a lot
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I'm in Calculus 2, we're doing basic limits and somehow I completely forgot how to do this. Thank you! :D
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I remember that... Partial fraction technique
Thank u so much! My module is so confusing and my head is wrecked. U helped me understand every thing, thank u sm. 😭✋
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Thank you very much!!! You just saved a life.
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This really helped me before my test. Thank you.
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Thank you so much! This really helped me. You teach clearly.
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What if you have a rational expression where both the numerator and the denominator were factorable in themselves? Would you factor both the numerators and denominators or just the denominators? Thank you.
You would in that case factor everything and try to cancel before you find the LCD. In most cases, they won't give you questions like that but it could happen.
Let's say you had:
2/8 and 1/2
If you don't cancel first, the LCD is 8, but in reality the 2/8 is really 1/4 after you simplify and so the real LCD is 4. It's the same concept with rational expressions.
thank you so muchhh i finally get it I have my math final this afternoon and I couldn't figure it out at all until now
I hope you are able to pass :)
@@Greenemath thank you :)
@@ls-ip6ed You are welcome, good luck with your studies :)
Sir, u saved my life
I'm glad to hear the video was helpful!
At 9:35 in the video, you multiplied -9/20 by 6/6. Why is that? Where did we get the 6/6 from?
You need a common denominator.
Where you able to learn that because I am stuck on that too
Thank you so much and btw @ 4:13 how did you get 3/3 from?
The LCD is 30, since you have denominators of 10 and 15.
So now you need to rewrite the fraction 3/10 as an equivalent fraction with the LCD 30 as the denominator:
3/10 * 3/3 = 9/30
3/3 is just 1 and multiplying by 1 does not change the number.
hi! i just discovered your video today and it helped me a lot with my assignment! im down with most of it except one. could you perhaps help me with it? i'd really appreciate it!
the problem is:
find the LCD of x²/x-3 and 4/3-x
its really confusing, ty!
The denominator (x - 3) occurs one time and the denominator (3 - x) occurs one time, so the LCD is (x - 3)(3 - x). Now for this problem, you may notice that -(3 - x) = x - 3, so what you could do is multiply the second fraction by -1/-1 and then write the second fraction as -4/(x - 3). Now, you have a common denominator of (x - 3), which would be the LCD. I guess it just depends on what the teacher wants.
Never thought to do a factor ‘tree’ thank you!
Happy to help!
Thanks this really helped alot
Please can you help me with a question it's
2/x + 5/x+1
You want the LCD? You have a denominator of x and a denominator of x + 1. What's in common? Nothing so the answer is the product of the denominators:
LCD: x(x + 1)
1.Give the least common denominator of a/36y^2 , b/12y^5 and c/6y^4..thank you
So for the numbers you have: 6, 12, and 36, this leads to a number part of 36.
6 = 2 x 3
12 = 2 x 2 x 3
36 = 2 x 2 x 3 x 3
LCM(6, 12, 36) = 2 x 2 x 3 x 3 = 36
For the variable part it is much easier, we have a y in each denominator, so we just go with the largest exponent which is 5.
LCM(y^2, y^5, y^4) = y^5
If you put these two results together, you get your LCD:
36y^5
What is the application that you use to write out your steps?
I use Microsoft One Note and Microsoft Journal and Camtasia for recording and editing.
Thank you for making this video. Very helpful
You're very welcome!
can you do more about this please also include some variable i still cant understand it........
We have more videos on the topic at greenemath.com
At 15:02 - Why do you select the terms you do- i.e. why do you decide to find the LCM of the second term from the denominator of the first fraction, and the first term from the second term as opposed to the other way around? Could you do it the other way well(first term from first fraction, second term from second fraction) or would that not work? . Is there a rule to this, or do you just pick the ones that will be easier to factor? I'm having trouble understanding this one part.
You can find the LCM in any order that you want. I was trying to show that (x + 3) is common to both, as many students will miss this. When you build the LCM it involves multiplication, which can be done in any order.
is it possible to find lcd if one of the numbers are negative?
Yes, what's the example you are stuck on?
15:16?
What if the denominator didn't have exponent will I still do the same?
Oops never mind you had one in the next example
How would I do it if the denominator didn't had variables?
What is the specific problem you are struggling with? If you post it I can try to guide you in the right direction.
long but VERY HELPFUL!!! Thank You!
Thank you, all my videos are kind of long. I try to be detailed and give a few examples. Good luck in your studies :)
@@sadubentertainment Exactly and they always choose an easy example that doesn't help.
You’re a life saver!
Awesome, glad it helped. :)
why do people unlike this video do they hate math or what ? its helphul for goodness sake
Lots of reasons, they are probably frustrated and the video didn't solve their exact problem.
can someone tell me why he divided 3/10 with 3/3 instead of 2/2 and vice versa with 7/15? Will it make a big difference 3/10 was divided by 2/2 and if 7/15 was divided by 3/3? hehe, It just seems like a mistake i'll make if so
edit: plus where did the "x+3" for 5x² come from in 15:25?
The whole point is we are factoring the denominators to find the LCD. So one denominator is 5x^2 + 15x = 5x(x + 3). In other words we can factor out the 5x part to get 5x(x + 3). As for your first question, we aren't dividing, we are multiplying. 3/10 is multiplied by 3/3 to get a denominator of 30. 7/15 is multiplied by 2/2 to get a denominator of 30. You need a common denominator, that's the whole point.
@@Greenemath ohhh!! It makes so much sense now! thank you so much ❤
@@janifhewasbig I'm glad it was helpful, let me know if you have any other questions or get stuck on a problem :)
Hi sir. I have a question. What is the LCD of 1/n! , 1/n+7! , 1/((n-1)+9)! And may you give an explanation as well? Thank you very much.
Did you mean to put the exclamation points in? Those are for factorials and would make your expressions way more complicated.
what is the LCD of ab^3 and a^3b^2? thanks!
You find the LCD of rational expressions. Those are not rational expressions.
Can I get the LCD of 2/(x²-1) - 1/x-1 = 1/2?
(x^2 - 1) = (x + 1)(x - 1)
(x - 1)
2
So your LCD would be 2(x + 1)(x - 1)
can i ask what the LCD of 4x+12/ x^2 + 5x + 6 and 5x+15/10x+20 is?
You can simplify each of those to make your task a bit easier:
(4x + 12) / (x^2 + 5x + 6)
Let's factor the numerator and the denominator:
4(x + 3) / (x + 2)(x + 3)
We can cancel a common factor of (x + 3)
4 / (x + 2)
The second one is also able to be simplified:
(5x + 15) / (10x + 20)
5(x + 3) / 10(x + 2)
You can cancel a common factor of 5:
(x + 3) / 2(x + 2)
So now your problem is what is the LCD of (4) / (x + 2), (x + 3) / 2(x + 2)
The denominators each have an (x + 2), then the second one has a 2. Since the (x + 2) is repeated, we only need 1:
LCD = 2(x + 2)
I just want to put a note here, if you are asked for the restricted values, use the original denominators, not the simplified ones.
@@Greenemath Thanks!
@@ls-ip6ed You are welcome :)
I confused on where did you get the number 3 on first example where there is an addition😅😅😅
The 3 in 3/10 is given. The 3 from factoring the number 15 as 5 * 3? I'm unclear on which 3 you mean, again please leave a time marker so I can answer your question.
transform to quadratic equations 2/x - x+1/5 = - 4/5 and 2/x-3 + x/2 = - 1/2 thank you
So for the first one, you would just multiply everything by the LCD which is 5x:
After you simplify you will get:
10 - x^2 - x = -4x
Which can be changed into:
x^2 - 3x - 10 = 0
which has solutions:
x = 5, -2
For the second one, you would multiply everything by the LCD, which is 2(x + 3):
4 + x(x - 3) = -(x - 3)
This can be changed into:
x^2 - 2x + 1 = 0
(x -1)^2 = 0
x = 1
Dear sir you are explaining the LCM and LCD ONLY in one method either we have other easy ways to find. Can you please guide me about another easy way of this.
There isn't some trick to this, just find the LCM of the denominators and that is the LCD for your group of rational expressions.
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@@Greenemath your a filipino :0
@@10sogogo34 Nah, I just used Google Translate :)
What is the LCD of:
3/x-2 and 1/(x-2)²
Thanks;)
You have (x - 2) and (x - 2)^2, so it's just (x - 2)^2.
Thanks
@@khenruzel3690 You are welcome
Ay 9:30 why do you multiply 11/24 by 5/5 and 9/20 by 6/6?????
You need a common denominator. 24 * 5 = 120 and 20 * 6 = 120.
Did he explain how we get 2/2?
If you could give me a time marker for your question, I'd be happy to answer.
What if you have different variables like
4/x, x-2/y
In that case, you would just multiply them together:
x * y or just xy is the LCD.
what is the lcd of this ? 2/x-3 + x/2 = - 1/2
Take all your denominators and find the LCM:
(x - 3), 2, 2
So you will multiply just one copy of 2 by (x - 3)
LCD will be: 2(x - 3)
If you want to solve the equation, multiply both sides by the LCD:
4 + x^2 - 3x = -x + 3
This will simplify to:
x^2 - 2x + 1 = 0
Which can be factored as:
(x - 1)^2 = 0
So x is just equal to 1.
x = 1
Hope this helps :)
transform to quadratic equations 2/x - x+1/5 = - 4/5 and 2/x-3 + x/2 = - 1/2 thank you
@@nizichu4761 Please leave a new comment and I'll help you out.
Pls solve this
Lcd of
x2-5x-6
and
x2-12x+36
You find the LCD for rational expressions, what you have are two polynomials.
Whats the lcd of 8/c and 8-c/c+2?
Did you get this yet?
@@Greenemath not yet huhu i cant understand the ppt presented by my teacher:(
@@jerzx You have c as one denominator and c + 2 as the other, nothing is common, so the lcd is just c(c + 2)
What is the LCD of x² - 4, x² - 5x + 6
You don't have any denominators in that problem.
I'm sure u solved It already but you have to simply the denominator like X²-4 is equal to (x+2)(x-2) and x²-5x+6 is eqaul to (X-2)(X-3)
LCD: (X-2)(X+2)(X-3)
what is the lcd of this denominator, (x+2),(x-6),(x+2),(x-6)?
Assuming each expression separated by a comma is a denominator, then you would just have: (x + 2)(x - 6)
thank you, such a big help♥️😭
@@jhaymarkofiana2363 You are welcome! :)
What’s the LCD of:
3/p^2 + 3/p^3 - 4/1?
So your three denominators are p^2, p^3, and 1.
Your LCD is the LCM of the denominators. Since one of them is 1, you don't need to even worry about it. You basically have p^2 and p^3. With variables for the LCM you go with the highest exponential power. In this case it's a 3, so your LCM/LCD is going to be p^3.
@@Greenemath thanks!!
@@Adam-ck2gc You are welcome! :)
i have a ? what is 2/3+1/6= /12+ /12
What are those slashes in front of the 12?
You can add the left side by getting a common denominator.
2/3 + 1/6 = 4/6 + 1/6 = 5/6
I don't know what you are trying to do on the right side.
Did you miss anything with the problem? Variables or question marks? It's not very clear what you are trying to do. Let me know and I'll show you how to get a solution.
LCD of the denominators x-2 and x^2-2x pls
Since you don't have a rational expression, I can't give you an LCD.
You have x - 2 and x^2 - 2x
x - 2, x(x - 2)
What's the LCM?
x(x - 2)
What is the LCD of 2/x - x+1/5 =4/5
One denominator is x and the other two are 5, so it's just going to be 5x.
Wow thx 🙏🏻 helped so much
Glad it helped!
How did 7/15 become 2/2?
If you could give me a time marker for your question, I'd be happy to answer.
Thank you. Very helpful
You are very welcome, thanks for watching :)
14:40 why (x+3)?
It comes from factoring each denominator. 7x^2 + 21x can be factored as 7x(x + 3), also 5x^2 + 15x can be factored as 5x(x + 3).
@@Greenemath ohhhh Now I Know. Thank You❤💖💖
@@earlerrolsucilan3463 That's great to hear, good luck with your studies :)
I dont understand, why do you have to choose the larger exponent 5 when you calculate lcd of 21x^3 and 12x^5. It doesnt make sense, i think there should be more explanation why this is how you should write the expression. Also if theres any method to find the prime of a term with a variable raised to an exponent, it would be helpful.
Let's start with the first question. Why is the LCM of x^3 and x^5 given as x^5? With numbers, the LCM is the smallest that is divisible by all numbers in the group. So for example with 2 and 3, the LCM is 6, since 6/2 = 3 and 6/3 = 2. The same logic applies when we work with variables. What's the smallest power on x that will be divisible by both x^5 and x^3. It will always be the variable raised to the largest power in the group. So x^5/x^5 = 1 and x^5/x^3 = x^2.
I think this lesson may help you: www.greenemath.com/Algebra1/45/LCDofRationalExpressionsLesson.html
It's from our website, which is free.
Method to find the prime of a term raised to an exponent? Just break it up and find the prime factorization of the number and the prime factorization of the variable part separately, then put them back together. Hope this helps you and good luck in your studies.
gg you are better than my teacher
I'm glad the video was helpful for you. Thanks for the nice comment; good luck with your studies :)
Thank you
You're welcome, let me know if you have questions 😎
can someone please tell me how x + 100,000/x became x^2+10,000 when simplified?
this LCD as well is included in our lesson but i totally forgot how to do this. thank you so much 🥲
What's the time marker? I can't re-watch the whole video, if you give me a time marker, I'll let you know what's going on in the video.
Thank you!!
You're welcome! :)
Where did he the 6
Which 6?
Hello welcome to algebra 2 homework lol
Awesome, hope it's easy for you :)
Pls give many example not many talk ha
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I know you're here
?
Awesome
Perfect, thanks for watching :)
10:00
The chapters are listed in the description and on the video.
What if it's 3x, 5, 4xy² does x and xy² gonna be together like "x xy²" ?
No, not quite, if you had two x's next to each other like that it implies multiplication so xxy^2 is really x^2y^2. In your case, if you are trying to find the LCM of those terms, you would only have one single x. You only put the largest number of repeats of any given factor.
3x -> 3 * x
5 -> 5
4xy^2 ->4 * x * y * y
So you want to make a list of every factor that occurs but if there is a duplicate, only put in the largest number of repeats. See how x is in 3x and also in 4xy^2, so you only have 1 that goes in your list:
LCM (3x, 5, 4xy^2) = 3 * 5 * 4 * x * y^2 = 60xy^2