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## Algebra (all content)

### Course: Algebra (all content) > Unit 13

Lesson 6: Solving rational equations- Rational equations intro
- Rational equations intro
- Equations with one rational expression (advanced)
- Rational equations (advanced)
- Equations with rational expressions
- Equations with rational expressions (example 2)
- Rational equations
- Equation with two rational expressions (old example)
- Equation with two rational expressions (old example 2)
- Equation with two rational expressions (old example 3)

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# Equations with one rational expression (advanced)

Sal solves the following equation by first simplifying the rational expression: x^2-(x^2-4)/(x-2)=4. Created by Sal Khan and Monterey Institute for Technology and Education.

## Want to join the conversation?

- Are there any practice problems for rational equations?(9 votes)
- Yes, you can practice from this link if you want : https://www.khanacademy.org/math/algebra2/rational-expressions/solving-rational-equations/e/linear_equations_4(2 votes)

- The interesting thing about this problem is that, if you don't spot the difference of squares and factor out (x-2), as Sal did, and instead multiply through by (x-2) to get it out of the denominator of the second term, you will be left with a third degree polynomial with three roots, one of which happens to be 2, the excluded value. Attempting to solve it this way inadvertently adds an extra root. Factoring out the (x-2), as Sal did, turns the problem into a quadratic, with only two roots. It wasn't immediately clear to me where the extra root "went" or why we needed to worry about excluding it, until I tried to solve it the other way. Hope this helps someone.(4 votes)
- This isn't about the video, but I was wondering how to solve this: (2/m+2) - (m+2/m-2) = 7/3 and if there were any videos on this that could help? I have tried solving it myself and ended up with m^2+19m+8 ; which I'm unsure of how to simplify, if it can be simplified, or even if it is correct.(1 vote)
- Here's how I would solve this.

First of all, I think your parentheses probably need adjusting so that the problem reads :

2 / ( m + 2 ) - ( m + 2 ) / ( m - 2 ) = 7 / 3

(Otherwise, after removing parentheses you'd be left with 5 terms on the lefthand side

2/m + 2 - m - 2/m + 2

which would simplify to

4 - m = 7 / 3 which seems too simple at this level.)

Second, once the problem has the correct parentheses, get a common denominator for both sides of the equation. Then set the numerators equal to each other.

Third, get all terms onto the lefthand side of the equation.

Fourth, factor the terms as the product of two binomials.

There will be 2 real roots.

Do let me know how you get on and if I can help you further.(3 votes)

- I have to solve the problem 2/c-2=2-4/c (the denominator c only belongs to 4). I can't solve it because I don't know what the LCD is. How do you figure out the LCD in this problem?(2 votes)
- at1:40...why did he subtract 4 to get a 0 on the one side instead of just adding 2 + 4 so you would end up with x^2 - x = 6?

Wouldn't x^2 - x^1 = just x? Therefore, x=6?

I don't understand how he knew that you would have to get a zero on the one side and then factor it out??

Thanks for any help and clarification!(1 vote)- He subtracts 4 from both sides because he wants to turn it into a quadratic equation. (ax^2+bx+c). When you see x raised to the 2 power, x^2, you will want to turn it into a quadratic equation that is equal to zero. Once that is done, now you can factor to solve the equation and find the answers.

Or, using the quadratic equation, you can use the quadratic formula to solve the equation, if that is easier than factoring. Quadratic formula: -b+-sqrt b^2-4(a)(c)/2a. Or you could complete the square to solve it. Just know, it needs to be in the form ax^2+bx+c=0 when you have a x^2 term and that there are 3 ways to solve it once you've converted it into a quadratic equation

ax^2+bx+c=0:

Complete the square

Factor

Quadratic Formula

HInt:

x^2 term= turn into quadratic equation equal to zero

ax^2+bx+c=0

If you don't know what I"m talking about, you should watch the videos in the quadratics section of algebra.(3 votes)

- I have a question concerning a problem on Khan Academy that I need help understanding. It goes like this.

36y^2 - 1/ 30y + 5 = 7

Before you even start, you set the denominator as equal to zero, ie.

30y + 5 = 0

We then end up with y = -1/6.

Then we disregard -1/6 as extraneous if it ever is a possible answer. But, -1/6 makes the equation equal and valid. So how can it be extraneous if it makes the equation equal and valid? Is it just an equation that happens to work out that way, or is it something else? I hope that makes sense.

Thanks!(1 vote)- -1/6 does make the equation
`30y+5=0`

valid. It is a perfectly correct solution to that equation. The equation we are looking at, however, is not that, but`36y^2 / 30y+5=7`

. If one attempts to put -1/6 in as the solution, you would be dividing by 0 (since 30y+5=0), and since dividing by 0 is against the rules (it just doesn't work), -1/6 is an extraneous solution.

I hope this helps and answers the question you were asking!(2 votes)

- This is more of a suggestion. Sal speaks so fast during this video that I had trouble staying up. Is there a way on here to slow down the playback? Thanks.(1 vote)
- In the lower right of the video window is a "gear" symbol. Click on it. You can change video speed setting.(2 votes)

- why did he not add 2 to both sides at1:20?(1 vote)
- At0:35He Lost Me. Can Someone Explain What He Is Talking About?(1 vote)
- I tried to use another way to solve the equation, but I can just get one of the two answers:

(4-x^2)/(x-2)=4-x^2

(4-x^2)/(x-2)=(4-x^2)/1.....(1)

(x-2)(4-x^2)=(4-x^2)*1......(2)left up expression multiply right down, and left down multiply right up.

x-2=1.......................(3)left side and right side divided by (4-x^2) at the same time.

x=3

I could just get one result and I don't know where I went wrong.

Can somebody help me? thanks(1 vote)- when you divide the (4-x^2) out of your equation on both sides it kills the possibility of finding the second solution.(2 votes)

## Video transcript

Solve the equation: x squared
minus x squared minus 4 over x minus 2 is equal to 4. And they tell us that x won't
or cannot be 2 because if it was 2, then this would
be undefined. It would be dividing by 0. So let's see what x is. Let's see the x that actually
satisfies this equation. So you might be tempted to try
to express this with the common denominator of x minus
2 and then add these two expressions. But the thing that jumps out
at me initially is that we have x squared minus 4 on the
numerator, which is a difference of squares. Or if you factor it out, it's
x plus 2 times x minus 2. So we should be able to factor
this x minus 2 out. So let's do that. So if we were to rewrite it,
this is equivalent to x squared minus-- instead of
writing x squared minus 4, we know that's a difference
of squares. That is x plus 2 times
x minus 2. All of that over x minus 2. And that is equal to 4. And this whole time we're
assuming that x won't be equal to 2. And because x does not equal to
2, x minus 2 divided by x minus 2 is going
to be defined. And it will be 1. So those two will cancel out. And so we're left with x squared
minus x plus 2 is equal to 4. We can distribute the
negative sign. And I'll just arbitrarily
switch colors here. We can distribute the negative
sign, so we get x squared minus x minus 2 is equal to 4. And what we want to do is put
this in the form ax squared plus bx plus c is equal to 0. That allows us to either
factor it or apply the quadratic equation or complete
the square, or any of the ways that we know how to
solve quadratics. So let's do that. Let's get a 0 on the
right-hand side. The best way to do that is to
subtract 4 from both sides of this equation. Subtract 4 and we are left
with x squared minus x. Negative 2 minus 4
is negative 6. And then 4 minus 4 is 0. That was the whole point. So we have x squared minus
x minus 6 is equal to 0. Let me write it up here. x squared minus x minus
6 is equal to 0. And this looks factorable. We just have to think of two
numbers that when we multiply them give us negative 6. So they're going to have
different signs. When I add them I'll
get negative 1. So it looks like negative
3 and positive 2 work. So if we do x minus
3 times x plus 2. It's a little bit of trial and
error, but 6 doesn't have that many factors to deal with and
3 and 2 are one apart. They have different signs, so
that's how you can think of how we get to that conclusion. Negative 3 times 2
is negative 6. Negative 3 plus 2
is negative 1. So that is equal to 0. So we have two possible
ways to get 0. Either x minus 3 is equal to 0
or x plus 2 is equal to 0. And then if we take x minus 3 is
equal to 0, if we add 3 to both sides of that equation, we
get x is equal to 3 or, if we subtract 2 from both sides of
this equation, we get x is equal to negative 2. So both of these
are solutions. And let's apply them into
this equation to make sure that they work. Because these are solutions to
essentially, the situation where we got rid of
the x minus 2. Maybe it had some type
of side effects. So let's just make sure that
both of these actually work in the original equation. So let's try x is equal
to 3 first. So you get 3 squared minus
3 squared over 4. Sorry, 3 squared minus
4 over 3 minus 2. So this is equal to 9 minus--
3 squared is 9 --minus 4, which is 5, over 1. So 9 minus 5, which
is equal to 4. Which is exactly what we
needed it to equal. And let's try it with
negative 2. So if I take negative 2 squared
and I have a minus negative 2 squared minus
4, all of that over negative 2 minus 2. So negative 2 squared is 4. Minus negative 2 squared,
which is 4. Minus 4. All of that over negative
2 minus negative 2 over negative 4. Well 4 minus 4 is 0, so this
whole thing is just going to become 0. So this whole thing is
going to equal 4. So both of these
solutions work. And that makes sense, because
when we actually canceled this out, we actually didn't
fundamentally change anything about the equation. Only if you had the situation
where the x would have been equal to 2. That's the only thing that
you're really changing. So that's why it makes
sense that both of these solutions work.