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Current time:0:00Total duration:7:12

CCSS.Math:

- [Voiceover] So we have a
nice, little equation here that has some rational expressions in it and like always, pause the video and see if you can figure out which x's satisfy this equation. All right. Let's work through it together. Now, when I see things at
the denominator like this, my instinct is to try to not
have denominators like this. And so what we could do is to
get rid of this x minus one in the denominator on the left-hand side. We can multiply both sides of the equation times x minus one, x minus one. We're gonna multiply both
sides by x minus one. And once again, the
whole point of doing that is so that we get rid of this x minus one in the denominator right over here. And then to get rid of this x plus one in the denominator over here, we can multiply both sides of
the equation times x plus one. So, x plus one. Multiply both sides times x plus one. And so what is that going to give us? Well, on the left-hand
side, that is going to, x minus one divided by x
minus one is just gonna be one for the x's where that's defined, for x not being equal to one. And so, we're gonna have x plus one times negative two x plus four. So let me write that down. So we have x plus, I'm gonna need some space, so let me make sure I don't write too big. x plus one times negative two x, negative two x plus four is going to be equal to. Now, if we multiply both of these times three over x plus one, the x plus one is going to
cancel with the x plus one and we're gonna be left with
three times x minus one. So that is going to be
three x minus three, three x minus three, and then minus, minus one times both of these. So one times x minus one times x plus one. So, minus one times x minus one times x plus one. All I did is I multiplied, took the x minus one times x plus one, multiplied times each of these terms. When I multiply it times this first term, the x plus one and the
x plus one canceled, so I just have to multiply
three times x minus one. And then for the second term, I just multiply times both of these. And now you might recognize, if you have something x
plus one times x minus one, that's going to be x squared minus one. So I could rewrite all
of this right over here as being equal to... as being equal to x squared minus one, and once again, that's
because this is the same thing as x squared minus one. And since I'm subtracting
an x squared minus one, actually, let me just, I don't
wanna do too much on one step so let's go to the next step. So, I could multiply this out. So I could multiply x times negative two x which would give us
negative two x squared. x times four, which is going to give us plus four x and I could multiply one
times negative two x. I'm gonna subtract two x. And then one times four,
which is gonna be plus four, and then that is going to be equal to, that is going to be equal to, we have three x minus three and then we can distribute
this negative sign. So we can say minus x squared plus one. And over here, we can
simplify it a little bit. This is going to be, that
is four x minus two x is going to be, I'm gonna try to write over, yep, four x minus two x, so
that would be two x. And so this is simplified to, let's see, well this is, we have a
negative three and a one, so those two together
are going to be equal to, subtracting at two. So we can rewrite everything as, do the neutral color now, negative two x squared plus two x plus four is equal to negative x squared plus three x minus two. Now we can try to get all of this business onto the right-hand side, so let's subtract it from both sides. So we're gonna subtract, or we'll say we'll add
x squared to both sides, add x squared, that gets rid of this
white negative x squared. We subtract three x from both sides, subtract three x from both sides, add two to both sides, add two, and we will be left with, we are going to end up with. See, negative two x squared plus x squared is negative x squared. Two x minus three x is negative x. And then four plus two is six, is going to be equal to, well that's going to
cancel that, that, that, is equal to zero. I don't like having this
negative on the x squared, so let's multiply both
sides times negative one. And so if I do that, so if I just take the
negative on both sides, so I just multiply that
times negative one. Same thing, it's taking
the negative on both sides. I'm gonna get positive x squared plus x minus six is equal to zero. And we're making some good progress here. So we can factor this. And actually, let me just
do it right over here so that we can see the original problem. So, if I were to factor
this, what two numbers, their product is negative six, they're gonna have different signs since their product is negative. And they add up to one, the coefficient on the first degree term. Well, positive three
and negative two work, so I can rewrite this as x plus three times x minus two is equal to zero. Did I do that write? Yeah. Three times negative two is negative six. Three x minus two x is positive x. All right. So, I just wrote this in this quadratic in factored form. And so the way that you
get this equaling to zero is if either one of those equals zero. x plus three equals zero or x minus two is equal to zero. Well, this is going to happen if you subtract three from both sides. That's gonna happen if x
is equal to negative three or over here, if you
add two to both sides, x is equal to two. So either one of these will satisfy, but we wanna be careful. We wanna make sure that
our original equation isn't going to be undefined
for either one of these. And negative three does not make either of the denominators
equal to zero, so that's cool. And positive two does not make either of the denominators equal to zero. So it looks like we're in good shape. There's two solutions to that equation. If one of them made any of the
denominators equal to zero, then they would have been
extraneous solutions. It would have been solutions for some of our intermediate steps, but not for the actual original equations with the expressions as they're written. But this, we can feel good
about because neither of these make any of these
denominators equal to zero.