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## Algebra 2

### Unit 10: Lesson 1

Rational equations# Equations with rational expressions

CCSS.Math:

Sal solves (x²-10x+21)/(3x-12)=(x-5)/(x-4), which has one real solution and one extraneous solution.

## Video transcript

- [Voiceover] So we have a
nice little equation here dealing with rational expressions. And I encourage you to pause the video and see if you can figure
out what values of x satisfy this equation. Alright, let's work through this together. So the first thing I'd like to do is just see if I can simplify this at all, and maybe by finding some common factors between numerators and denominators, or common factors on either
side of the equal sign. So let's factor all of the
numerators and the denominators. All the ones on the right-hand
side are already done. So this thing up here, I could rewrite this as, let's see. What's product is 21? What two numbers, when I
take their product, is 21? Positive 21, so they're
going to have the same sign. And when I add them, I get negative 10. Well, negative seven and negative three, so this could be rewritten as x minus seven times x minus three. This over here, both
are divisible by three. I could rewrite this as
three times x minus four. And these are already factored. So the one thing that
jumps out at me is I have x minus four in the denominator
on the left-hand side and on the right-hand side. And so, if I were to multiply
both sides by x minus four. So actually, let me just... Let me formally replace this with that. And up here, it's not so obvious that it's going to be valuable for me to keep this factored form. So I'm just going to keep
it in this yellow form. In the expanded out form. So let me just scratch that out for now, because once I, well, let me multiply by x minus four. So, if we multiply both
sides by x minus four, and once again, why am I doing this? Just so I get rid of the x
minus four in the denominators. x minus four, and then x minus four. That and that cancels. That and that cancels, and then we're left
with, in the numerator, we're left with our x
squared minus 10 x plus 21, and, let's see, divided by three. Divided by three is equal to x minus five. Let's see, now, what we can do, and actually I could have
done it in the last step, is I can multiply both sides by three. Multiply both sides. Do that in another color just so it sticks out a little more. So I can multiply both sides by three. So multiply both sides by three. On the left-hand side,
that and that cancels, and I'll just be left with x
squared minus 10 x plus 21, and on, and I don't have
a denominator any more. My denominator is one, so
I don't need to write it. Is going to be equal to three times, let's just distribute the three. Three times x is three x. Three times negative five is negative 15. And now I can get this into
standard quadratic form by getting all of these terms
onto the left-hand side. The best way to do that, let's subtract three x from the right, but I can't just do it from the right otherwise the equality won't held. I have to do it from both sides
if I want equality to hold. And, I want to get rid
of this negative 15, so I can add 15 to both sides. So, let's do that. And what we are left with,
scroll down a little bit, so let's have a little more space. What we are going to be left with is x squared minus 13 x and then plus, what is this, plus 36. Plus 36. Did I do that right? Yep, plus 36 is equal to, is equal to zero. Alright, now let's see,
we have it as a quadratic in the standard form. How can we solve this? So first thing, can we factor this. Product of two numbers, 36. If I add them, I get negative 13. They're both going to be negative, since they have to have the same sign to get their product to be positive. And, let's see, nine and
four seem to do the trick. So, or negative nine and negative four. So, x minus four times x minus nine is equal to zero. Well, that's going to happen if either x minus four is equal to zero or x minus nine is equal to zero. We'll add four to both sides of this. This happens when x is equal to four. Add nine to both side of this. This happens when x is equal to nine. So we could say that the solutions are x equals four or x equals nine. So, x is equal to four, or x equals nine. But we need to be careful. Because we have to remember, in our original expression, x minus four was a factor
of both denominators. And so if we actually try to test x minus four in the original equation, not one of these intermediary steps, the original equation, I would end up dividing by zero right over here, and actually, I'd end up by
zero right over there as well. So the original equations, if
I tried to substitute four, they don't make sense. So this is actually an
extraneous solution. It's not going to be a solution
to the original equation. The only solution is x is equal to 9.