I'm a slow learner person... My teacher so bad at teaching math... And my Life is the worst... And then my teacher send to me on Facebook, and then it changed my life! Thank you so much!!
My math exam is tomorrow and I didn't listen because i always fall asleep when i listen to math. I thought restricted value is really hard. Thanks for making it easy my man
@@breadpitt3369 Just set each denominator equal to zero and solve. You are looking for where the denominator is zero. Let's work through a few: 1) (2x - 5) / (5 - x) Set the denominator equal to zero and solve the equation: 5 - x = 0 -x = -5 x = 5 So this tells you that 5 is restricted from the domain. Why? If you plug in a 5 for x, you end up with 0 as the denominator and you can't divide by zero. {x | x is not equal to 5} is your domain Let's look at #2 2) (x - 2) / (2x - 7) Set 2x - 7 = 0 and solve 2x - 7 = 0 2x = 7 x = 7/2 {x | x is not equal to 7/2}
Just set each denominator equal to zero and solve to find the restricted value(s). Also, I'm not sure if you are asked to simplify? You didn't specify and you should check the instructions. Some teachers want both?
@@breadpitt3369 The answer to what???????????????? You aren't giving instructions. How can you put down an answer with no instructions. Are you finding the restricted values or simplifying? Do you event know? It should state what to do at the top of the homework. If you are finding the restricted values, just set those denominators equal to zero and solve the equations you are producing. Then state the domain by restricted those values. x + 3 = 0 x = -3 4x + 5 = 0 x = -5/4 4a - 7 = 0 a = 7/4 So on and so forth, but you don't list those as the solutions, those are the restricted values and are restricted from the domain.
Same deal, just set the part with the variable equal to zero. 2x - 4 = 0 2x = 4 x = 2 If x was 2, that whole expression would be zero, since you'd be multiplying by zero. 2(2(2) - 4) 2(4 - 4) 2(0) = 0 So the denominator would become zero if x was 2. In general, just set the factors with a variable equal to zero and solve.
If you edit a comment, it doesn't send a notification. You should just reply instead. Your original comment asked where we got 3/4, which I couldn't find. The problem you are looking at is (9x - 1) / (3x - 4) and we are looking for the restricted value. So what can't we plug in for x? Remember you can't divide by zero. So what value makes the denominator zero? Just set up an equation with the denominator equal to zero: 3x - 4 = 0 3x = 4 x = 4/3 So 4/3 is a restricted value, notice if you plug in a 4/3 for x the denominator becomes zero and that is not defined in math. In general for these problems you are setting the denominator equal to zero and then solving the resulting equation.
Just set the denominator equal to zero and solve the equation. Here they made it easy because the denominator is already factored. (p + 4)(p + 5) = 0 p + 4 = 0 p = -4 p + 5 = 0 p = -5 So p can't be equal to -4 or -5 since both of those values would make the denominator zero.
I have a question, what's the point in finding the restricted values and the domains. In other words, why would I want to know what x is not, versus finding what x actually is?
With a rational expression, it's not an equation. The only value for x is when you give it one. We need to find out what values of x are allowed or not allowed. This will have applications for solving rational equations, graphing, and so forth.
@@satorugojo8786 Yes, in the future please post the full problem though. I can't understand why you would just type half of a problem???? Depending on what's in the numerator, the answer could change. Assuming there's nothing fancy in the numerators. If 5x(x + 1) is your denominator then you just need to find what x-values make the denominator zero. You can't divide by zero. 5x(x+1) = 0 Solve by setting each factor equal to zero. 5x = 0 or (x + 1) = 0 5x = 0 x = 0 x + 1 = 0 x = -1 So for the first example, x can't be 0 or -1. You can see if you plug in a zero or -1 you would get zero. And you can't have zero as a denominator. 5(0)(0 + 1) = 0(1) = 0 5(-1)(-1 + 1) = -5(0) = 0 For the second example, we would follow the same logic (x - 5)(x + 4) = 0 x - 5= 0 or x + 4 = 0 x - 5 = 0 x = 5 x + 4 = 0 x = -4 So for the second example x can't be 5 or -4 Again, I don't know what's in the numerators because you didn't want to type out the full problem. There could be more restrictions needed, but there's no way for me to know.
@@satorugojo8786 Please type the full problem. What is the numerator? I can't understand why you aren't typing the full problem. Do you realize you can get the wrong answer by typing half of the problem?
Basically just set the denominator equal to zero, solve the resulting equation, and the solutions are your restricted values. Also let us know what kind of chips. 🥔😋
Is that your denominator? Is it 1/(x^2 - 12) your problem? If you just have x^2 - 12, there isn't a restricted value since your denominator is 1. If x^2 - 12 is in the denominator, then set it equal to zero and solve the resulting equation. x^2 - 12 = 0 x = 2 * sqrt(3), -2 * sqrt(3) So those two values would need to be restricted from the domain. Again this is only if x^2 - 12 is your denominator.
I'm not sure what you are trying to do with this expression? When you say replaceable values? You can replace the x, but you are generally given some value to replace it with. There isn't an equal sign, so it's not an equation to solve. The best I can figure on this example is that you want the restricted value? In that case, set your denominator equal to zero and solve. Also, I'm going to assume this was your problem as the other was just 3/2 - x. 3 /(2 - x) Since your denominator is 2 - x, we can set this equal to zero and solve the resulting equation. 2 - x = 0 2 = x This tells you that x can't be 2 since this makes the denominator 0 and dividing by zero is undefined.
@@jinseiboy5589 So the whole idea with a domain is that it's the set of allowable values for the variable x or in your case n. So think about what can you plug in? If you have a rational expression, let's say: 1/(n +7) Then we know that we can't divide by zero, this is undefined. So we need to find out what value for n makes the denominator zero. To do this, we just set the denominator equal to zero and solve: n + 7 = 0 n = -7 So this means -7 is restricted from the domain. You could write this as: domain: {n | n is not equal to -7} So any value can be plugged in for n except -7 since this makes the denominator zero and we can't divide by zero. :) I'm also assuming you don't have any radicals in the numerator since I can't see the full problem. As long as the numerator is a simple polynomial that's your answer.
@@jinseiboy5589 Again is this the denominator? Why aren't you posting the full problem, you could have something in the numerator with a restriction as well? It's the same deal just set the denominator equal to zero and solve: 2x^2 - 5x + 3 = 0 x = 3/2, 1 So x can't be 3/2 or 1 since that involves division by zero. Again this is only if you have a polynomial in the numerator. No way for me to know since you only post part of the problem. domain: {x | x is not equal to 3/2, 1}
Just set the denominator equal to zero and solve. x - 2 = 0 x = 2 So x can't be 2 since plugging in a 2 for x gives you a denominator of zero which is undefined.
You would factor the numerator and denominator and then cancel common factors. You can watch this video: th-cam.com/video/n6zn3jflZL4/w-d-xo.html If you are still confused leave your example here and I'll show you how to simplify it.
Just set the denominator equal to zero and solve the resulting equation. That's it. If you have a problem you are stuck on, you can reply and I'll try to help :)
There are a zillion things you can do with that expression. Why not put the instructions in? Instead you just give a random problem that could be anything?
@@katenyvin8508 I would start by simplifying. You can factor out a 5 from the top and bottom: (35x - 35) / (25x - 40) 35x - 35 = 5(7x - 7) 25x - 40 = 5(5x - 8) You can cancel the common factor of 5: (7x - 7) / (5x - 8) is the simplified rational expression. A rational expression is not defined where the denominator is zero. This is because division by zero is not defined. What we do is set the denominator equal to zero and solve the resulting equation: 5x - 8 = 0 5x = 8 x = 8/5 This means that if x is 8/5, the rational expression is not defined since we are dividing by zero.
I spent 2 hours reading my book again and again and I still dont understand, my brain finally worked when I watched this video.
I'm so glad it helped, thanks for watching :)
I'm a slow learner person... My teacher so bad at teaching math... And my Life is the worst... And then my teacher send to me on Facebook, and then it changed my life! Thank you so much!!
I'm so glad the video was helpful. Everyone learns at a different pace, keep working hard and you will get where you want to go!
how did you just now get youtube lmao
@@bobjohnson9806 What do you mean?
My math exam is tomorrow and I didn't listen because i always fall asleep when i listen to math. I thought restricted value is really hard. Thanks for making it easy my man
You got this!
Man I got to subscribe, after 5 hours trying to understand, you make it plain and simple!
I’m back to my college homework now! Thanks !
Glad I could help!
Bruh we gonna do this again in college I'm grade 8 right now and it's teaching us this🗿
@@karlishmaelbborja9569it gets worse
you just saved my algebra grade, thank you kind internet man
I'm glad to hear that, let me know if you have any questions or get stuck on a problem. 😎
My teacher is so bad at teaching so i have to go on youtube for me to learn math. This video helped me a lot thank you!!!!
Glad it helped! Good luck with your studies :)
This video is so much better than any textbook, THANK YOU SO MUCHH
Awesome, glad it was helpful!
i needed this video for my highschool class thank you
You are very welcome! :)
The main idea is to set the denominator equal to zero and solve the resulting equation. Hope that helps with homework.
1.) 2x-5/5-x
2.) x-2/2x-7
3.)x²+3x-4/x+3
4.)x²+5x+6/4x+5
How can i answer this? Please help me im confused
5.)5a²-14/4a-7
6.)m²+m+50/3m-12
7.)a² - a - 2/ a² - 13a + 30
8.)x³ - 2x + 3/x² + 12x + 32
@@breadpitt3369 Just set each denominator equal to zero and solve. You are looking for where the denominator is zero.
Let's work through a few:
1) (2x - 5) / (5 - x)
Set the denominator equal to zero and solve the equation:
5 - x = 0
-x = -5
x = 5
So this tells you that 5 is restricted from the domain. Why? If you plug in a 5 for x, you end up with 0 as the denominator and you can't divide by zero.
{x | x is not equal to 5} is your domain
Let's look at #2
2) (x - 2) / (2x - 7)
Set 2x - 7 = 0 and solve
2x - 7 = 0
2x = 7
x = 7/2
{x | x is not equal to 7/2}
@@breadpitt3369 See the comment below.
GreeneMath.com
Good luck with class everyone! :)
1.) x² + 3x -4/x+3
2.) x²+5x+6/4x+5
3.) 5a² - 14/4a - 7
Just set each denominator equal to zero and solve to find the restricted value(s). Also, I'm not sure if you are asked to simplify? You didn't specify and you should check the instructions. Some teachers want both?
@@Greenemath and solution
@@breadpitt3369 The answer to what???????????????? You aren't giving instructions. How can you put down an answer with no instructions. Are you finding the restricted values or simplifying? Do you event know? It should state what to do at the top of the homework. If you are finding the restricted values, just set those denominators equal to zero and solve the equations you are producing. Then state the domain by restricted those values.
x + 3 = 0
x = -3
4x + 5 = 0
x = -5/4
4a - 7 = 0
a = 7/4
So on and so forth, but you don't list those as the solutions, those are the restricted values and are restricted from the domain.
@@Greenemath restricted values
@@breadpitt3369 Alright then just restrict those values I gave you from each domain and you are done.
wow , what a life saver. My head was going to explode before i came across this video LOL
thanks for saving me time and confusion
Glad it helped!
Tandewww berrymuddd... I got all my questions answered!😍😍😍
Glad it helped!
How did the restricted value of the last given became -1/2 ? I Substitute it and it did not because undefined
2(-1/2)^2 - 9 (-1/2) - 5 = 0, This means the denominator is zero and the expression is undefined.
I get this video and thanks for this, but what if the thing is a^2/b ??? thank you
What is the exact problem you are working on?
I understand this much better than the other videos tbh😲
Perfect, glad it was helpful! :)
What if the factor is in parenthesis? Like 2(2x-4) how will u equal it to 0?
Same deal, just set the part with the variable equal to zero.
2x - 4 = 0
2x = 4
x = 2
If x was 2, that whole expression would be zero, since you'd be multiplying by zero.
2(2(2) - 4)
2(4 - 4)
2(0) = 0
So the denominator would become zero if x was 2. In general, just set the factors with a variable equal to zero and solve.
Thank you so much,i don't understand nothing of this before watch this video
You are welcome!
can a restriction be a decimal number
for example
(2x+3) (2x-3) R=-1.5,1.5
Yes, in your case you could write the restricted values as domain: {x | x ≠ 3/2, -3/2} or you can do domain: {x | x ≠ 1.5, -1.5}
@@Greenemath thanks a lot you really helped me understand this :)
@@tessy9672 I'm glad to hear that, let me know if you have any other questions. Happy to try and help if I can. 😎
How do you find your x,like you did in 9x-1
3x-4
5:40
Time marker please? I have no idea what you are referring to.
If you edit a comment, it doesn't send a notification. You should just reply instead. Your original comment asked where we got 3/4, which I couldn't find. The problem you are looking at is (9x - 1) / (3x - 4) and we are looking for the restricted value. So what can't we plug in for x? Remember you can't divide by zero. So what value makes the denominator zero? Just set up an equation with the denominator equal to zero:
3x - 4 = 0
3x = 4
x = 4/3
So 4/3 is a restricted value, notice if you plug in a 4/3 for x the denominator becomes zero and that is not defined in math. In general for these problems you are setting the denominator equal to zero and then solving the resulting equation.
how do you do find the restricted value of
p squared
----
(p+4) (p+5)
Just set the denominator equal to zero and solve the equation. Here they made it easy because the denominator is already factored.
(p + 4)(p + 5) = 0
p + 4 = 0
p = -4
p + 5 = 0
p = -5
So p can't be equal to -4 or -5 since both of those values would make the denominator zero.
I have a question, what's the point in finding the restricted values and the domains. In other words, why would I want to know what x is not, versus finding what x actually is?
With a rational expression, it's not an equation. The only value for x is when you give it one. We need to find out what values of x are allowed or not allowed. This will have applications for solving rational equations, graphing, and so forth.
@@Greenemath ah okay, I thought there was a deeper meaning. thankk you
@@Plasmaaa You are very welcome, good luck with your studies :)
my teacher js didnt explain how to find restrictions besides in basic denominators like 3-4x
Set the denominator equal to zero and solve.
@@Greenemath tysm u and organic chemistry tutor are saving my grades
thank you! this video is so clear and helpful ^^
Glad it was helpful! Good luck with class!
How i can get the restricted value for 5x(x+1) and (x-5)(x+4)??
That's not a rational expression. Are those denominators or something? Can you post the full problem.
@@Greenemath yes they are denominators and i need to find their non permissible values can you help me please?
@@satorugojo8786 Yes, in the future please post the full problem though. I can't understand why you would just type half of a problem???? Depending on what's in the numerator, the answer could change. Assuming there's nothing fancy in the numerators.
If 5x(x + 1) is your denominator then you just need to find what x-values make the denominator zero. You can't divide by zero.
5x(x+1) = 0
Solve by setting each factor equal to zero.
5x = 0 or (x + 1) = 0
5x = 0
x = 0
x + 1 = 0
x = -1
So for the first example, x can't be 0 or -1. You can see if you plug in a zero or -1 you would get zero. And you can't have zero as a denominator.
5(0)(0 + 1) = 0(1) = 0
5(-1)(-1 + 1) = -5(0) = 0
For the second example, we would follow the same logic
(x - 5)(x + 4) = 0
x - 5= 0 or x + 4 = 0
x - 5 = 0
x = 5
x + 4 = 0
x = -4
So for the second example x can't be 5 or -4
Again, I don't know what's in the numerators because you didn't want to type out the full problem. There could be more restrictions needed, but there's no way for me to know.
@@satorugojo8786 Please type the full problem. What is the numerator? I can't understand why you aren't typing the full problem. Do you realize you can get the wrong answer by typing half of the problem?
@@Greenemath
Find their non permissible value
1.numerator 6x+5
denaminator 3x+4
2.numerator x^2 - 1
Denaminator x^2 +4x - 45
Thankssss
*Watched it half way*
*got confused*
*panicked*
*cried*
*ate some chips*
*recuperated*
*got back at **8:29*
*happly* ah record timing!
Basically just set the denominator equal to zero, solve the resulting equation, and the solutions are your restricted values. Also let us know what kind of chips. 🥔😋
How about this problem 3x
6x2
I don't understand
6x * 2 ≠ 0
12x ≠ 0
x ≠ 0/12
6x - 2 ≠ 0
6x ≠ 2
x ≠ 1/3
6x + 2 ≠ 0
6x ≠ -2
x ≠ -1/3
I didn't know if you meant 6x times 2, or if you just forgot the operation (-,+,*) so I did all 3.
Is it (3x)/(6x^2)? In this case x can't be zero.
Do you still help comments? I need help
What's the problem you are working on?
How to find the restricted value of x²-12
Is that your denominator? Is it 1/(x^2 - 12) your problem? If you just have x^2 - 12, there isn't a restricted value since your denominator is 1. If x^2 - 12 is in the denominator, then set it equal to zero and solve the resulting equation.
x^2 - 12 = 0
x = 2 * sqrt(3), -2 * sqrt(3)
So those two values would need to be restricted from the domain. Again this is only if x^2 - 12 is your denominator.
How to solve replaceable values of rational expression
3/2-x what is the answer?
I'm not sure what you are trying to do with this expression? When you say replaceable values? You can replace the x, but you are generally given some value to replace it with. There isn't an equal sign, so it's not an equation to solve. The best I can figure on this example is that you want the restricted value? In that case, set your denominator equal to zero and solve.
Also, I'm going to assume this was your problem as the other was just 3/2 - x.
3 /(2 - x)
Since your denominator is 2 - x, we can set this equal to zero and solve the resulting equation.
2 - x = 0
2 = x
This tells you that x can't be 2 since this makes the denominator 0 and dividing by zero is undefined.
@@Greenemath yes help me on my questions please
@@onetamad3804 I answered above.
I'm confused what's n+7 in domain pleaseeee
Is that the whole expression or is it just the denominator?
@@Greenemath denominator thank you😊😊
@@Greenemath wait can you do 2x²-5x+3
@@jinseiboy5589 So the whole idea with a domain is that it's the set of allowable values for the variable x or in your case n. So think about what can you plug in? If you have a rational expression, let's say:
1/(n +7)
Then we know that we can't divide by zero, this is undefined. So we need to find out what value for n makes the denominator zero. To do this, we just set the denominator equal to zero and solve:
n + 7 = 0
n = -7
So this means -7 is restricted from the domain.
You could write this as:
domain: {n | n is not equal to -7}
So any value can be plugged in for n except -7 since this makes the denominator zero and we can't divide by zero. :)
I'm also assuming you don't have any radicals in the numerator since I can't see the full problem. As long as the numerator is a simple polynomial that's your answer.
@@jinseiboy5589 Again is this the denominator? Why aren't you posting the full problem, you could have something in the numerator with a restriction as well?
It's the same deal just set the denominator equal to zero and solve:
2x^2 - 5x + 3 = 0
x = 3/2, 1
So x can't be 3/2 or 1 since that involves division by zero. Again this is only if you have a polynomial in the numerator. No way for me to know since you only post part of the problem.
domain: {x | x is not equal to 3/2, 1}
How do you do. x/x-2 ?
Just set the denominator equal to zero and solve.
x - 2 = 0
x = 2
So x can't be 2 since plugging in a 2 for x gives you a denominator of zero which is undefined.
3 days before exam
Will update after my result
Good luck.
@@Greenemath thanks
how to know the restricted value of x?
Set the denominator equal to zero and solve the equation. The solutions are the restricted values.
How do you simplify it?
You would factor the numerator and denominator and then cancel common factors. You can watch this video: th-cam.com/video/n6zn3jflZL4/w-d-xo.html
If you are still confused leave your example here and I'll show you how to simplify it.
what about
3
__
5x ??
Set 5x = 0 and solve.
5x = 0
x = 0
This means x can't be zero.
@@Greenemath Thank you! Your video was very helpful, in fact i learn faster by watching your videos! I also aced test so Thank you again!
@@Greenemath Thank you! Your video was very helpful, in fact i learn faster by watching your videos! I also aced my test so Thank you again!
@@friedegg7994 I'm glad you passed! :)
Thank you 😭math In Philippines is hell
It's hard everywhere, math is a difficult subject to learn.
i still don't get it :
Just set the denominator equal to zero and solve the resulting equation. That's it. If you have a problem you are stuck on, you can reply and I'll try to help :)
Who is watching this in 2022
Everyone.
Omgg i really get it now tyy
That's so awesome to hear!
I answered all of my schoolworks except for this.. 35x - 35/25x-40 pls help 🙏✨
There are a zillion things you can do with that expression. Why not put the instructions in? Instead you just give a random problem that could be anything?
@@Greenemathi need to find its restricted value :(
@@katenyvin8508 I would start by simplifying. You can factor out a 5 from the top and bottom:
(35x - 35) / (25x - 40)
35x - 35 = 5(7x - 7)
25x - 40 = 5(5x - 8)
You can cancel the common factor of 5:
(7x - 7) / (5x - 8)
is the simplified rational expression.
A rational expression is not defined where the denominator is zero. This is because division by zero is not defined. What we do is set the denominator equal to zero and solve the resulting equation:
5x - 8 = 0
5x = 8
x = 8/5
This means that if x is 8/5, the rational expression is not defined since we are dividing by zero.
Legend🔥🔥🔥🔥
Thanks! 😎
Woah thank you
You are welcome!
Thank you
You're welcome :)
I still cabt answer it
Can't answer what?
@@Greenemath i finally answerd it now i need to simplify rational expressions
@@Greenemath thank u i finally answerd it
@@lean5777 It's just factoring and canceling common factors. You can always drop the example in the comments and I'll try to help! :)
@@lean5777 I'm glad to hear that! :)
wow
😎
Kl9
Not sure what that means but thanks for watching and commenting! Good luck with your studies :)