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## Algebra (all content)

### Course: Algebra (all content)ย >ย 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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# Equation with two rational expressions (old example)

Sal solves the equation 4/(p-1)=5/(p+3). Created by Sal Khan and Monterey Institute for Technology and Education.

## Want to join the conversation?

- Is 0 positive or negative?(3 votes)
- 0 is neither positive nor negative it is just 0(67 votes)

- What does he mean by "undefined"?(6 votes)
- you can't divide any number by zero because you cannot multiply zero by any number to get the numerator, ever. saying undefined means that a problem would represent this kind of situation- when the problem just can't happen.(18 votes)

- At0:55, why does he try to find the excluded values when there is only one answer to the equation?(4 votes)
- In general when you see an equation like this, it's not always easy to see what the solution is, but it is relatively straight-forward to find the values for which the fractions are undefined by having a zero in the denominator. This step is necessary because if you proceeded to solve this type of equation and the solution was indeed one of the excluded values, then this equation doesn't have a solution because the equation is undefined at this specific value.(6 votes)

- At the end of solving the equation, why did you multiply both sides by -1? I don't see where it came from if we cancelled out the -1 when multiplying both sides by (p-1).(3 votes)
- Because we had a -p = -17 but we want to know what p equals, not -p.

We had that -p because earlier in the steps taken we had subtracted 5p from 4p(0 votes)

- why don't you just cross-multiply?(2 votes)
- Cross-multiplication is a shortened version of what Sal did. It's the same method, excepting skipping a few steps and accounting for that.(1 vote)

- Why is Sal not showing videos of rational expressions involving logarithms, exponents, roots and other operations? I think those rational expressions are just as important.(2 votes)
- Can anyone tell me how to solve a rational equation when there is a variable as the denominator? for example: 3/y + 4/3=-1

I understand how to solve rational equations but just not when there is a variable in the denominator.(2 votes)- If you divide both sides of the problem (each individual part) by 3 you can isolate the Y. So you'd get y + 4/3/3 ( <--same as 4/3 x 1/3) = -1/3

Once you've done this move the ( 4/3 x 1/3) to the other side of the equal sign.

So the unsimplified answer would be y= -1/3 - ( 4/3 x 1/3)(1 vote)

- can you help me plz on this problum 1/2-j+2=4/2-j(2 votes)
- The answer is j = 1/2 = 0.500,

you can check this for step to step solution:

http://www.tiger-algebra.com/drill/1/(2-j)_2=4/(2-j)/(1 vote)

- why did you subtract -5 to both sides ?(2 votes)
- don't you mean 5p? he did it so that you could have all of the p's on the left side and all constants on the right - basic linear equations(1 vote)

- Why are these called "rational" equations?(1 vote)
- Because they deal with ratios. The "rational" part comes from the word "ratios", not rational as being logical/reasonable. Ratios are essentially fractions.(3 votes)

## Video transcript

Solve the equation and
find excluded values. And what they're talking about,
about finding excluded values, is we need to think about what
values would make these either side of this equation undefined. And the reason why
we want to do that is because as we
manipulate this, we might lose things
in the denominator. And then we might
get some answer. But if it's one of
those things that made the original, the original
expressions, or either side of the original
equation undefined then that wouldn't be
a legitimate solution. So that's what they're talking
about, the excluded values. So what values do we have
to exclude here right from the get go? Well 4 over p minus
1 won't be defined if p is 1 because if p is
1, then this, then you're going to be dividing by 0. And that's undefined. So we know that p
cannot be equal to 1. And over here, if p
was at negative 3, then this denominator would be
0 and it would be undefined. And so p cannot be equal
to 1 or negative 3. So these right here are
our excluded values. So now let's try to solve. Let's try to solve
this equation. And I'm going to
rewrite it over here. So we have 4 over p minus 1
is equal to 5 over p plus 3. So the first thing we
could do, especially because we can assume now that
neither of these expressions are 0. And this is going to
be defined, since we've excluded these values of
p, is to get the p minus 1 out of the denominator. We can multiply the left
hand side by p minus 1. But remember, this
is an equation. If you want them to
continue to be equal, anything you do
left hand side, you have to do to the
right hand side. So I'm multiplying by p minus 1. Now I also want to get this p
plus 3 out of the denominator here on the right hand side. So the best way to do that is
multiply the right hand side by p plus 3. But if I do that to
the right hand side, I also have to do that
to the left hand side. p plus 3. And so what happens? We have a p minus 1 in
the numerator, p minus 1 at the denominator. These cancel out. So you have just a 1
of the denominator, or you have no
denominator anymore. And the left hand
side simplifies to 4 times p plus
3, or if you were to distribute the
4, 4 times p plus 3. So that is 4p plus 12. And then the right
hand side, you have plus 3 canceling
with a p plus 3. This is p plus 3
divided by p plus 3. And all you're left with
is 5 times p minus 1. If you distribute the
5 you get 5p minus 5. And now this is a pretty
straightforward linear equation to solve. We just want to isolate
the p's on one side and the constants on the other. So let's subtract
5p from both sides. I'll switch colors. So let's subtract
5p from both sides. And we get on the
left hand side, 4p minus 5p is
negative p plus 12. Is equal to, these cancel
out, is equal to negative 5. And then we could subtract
12 from both sides. And we get, these cancel out,
we get negative p is equal to, negative 5 minus
12 is negative 17. And we're almost done. We can multiply both
sides by negative 1 or divide both
sides by negative 1 depending on how
you want to view it. And we get, negative
one times negative p is. So let me just scroll
down a little bit so I have a little
bit of real estate. That's positive
p is equal to 17. p is equal to 17. And let's verify that
this really works. Well, it wasn't one of our
excluded values, but just in case, let's verify
that it really works. If we go, if we have p is 17. We get 4 over 17 minus 1,
needs to be equal to 5 over 17 plus 3. I'm just putting 17 in for p,
because that's our solution. So this is the same
thing as 4 over 16, needs to be the same
thing as 5 over 20. Or 4/16 is the
same thing as 1/4, and that needs to be
the same thing as 5/20, which is the same thing as 1/4. So it all checks out. So these are excluded values. And lucky for us, this
wasn't one of them.