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# Two more examples of solving rational equations

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- Let's solve some more rational equations.
- So let's say I have x plus 1/x is equal to 2.
- Once again, we have an x in a denominator over here, it
- looks kind of bizarre.
- But you just have to remember, if we multiply every term in
- this equation by this denominator, right here, by
- this x, right there, then essentially, you'll have x
- times 1/x and the x will disappear out of the
- denominator.
- So let's do that.
- So we have to multiply every term by that, you can't just
- multiply one term.
- So multiply this term times x, this term times x,
- and the 2 times x.
- We're just multiplying both sides of this equation by this
- denominator.
- So what do we get?
- We get x times x is x squared plus x times 1/x is just 1.
- x's cancel out.
- And then that's going to be equal to 2x.
- And this is starting to look like our
- standard quadratic equations.
- We just have to subtract 2x from both sides.
- And if we do that-- so let me subtract 2x from there,
- subtract 2x from there-- the left-hand side of our equation
- becomes x squared minus 2x plus 1.
- And the right-hand side is equal to 0.
- And we know how to solve an equation like this.
- We could either factor or use the quadratic formula.
- This actually seems pretty straightforward to factor.
- Negative 1 times negative 1 is 1.
- Negative 1 plus negative 1 is negative 2.
- So this is x minus 1 times x minus 1 is equal to 0.
- And we know, if two numbers, when you multiply them, they
- equal 0, that means one or both of those numbers have to
- be equal to 0.
- In this situation, the number is the same, right?
- This is x minus 1 squared is equal to 0, right?
- So we know that x minus 1 has to be equal to 0.
- Add 1 to both sides of this equation and we get
- x is equal to 1.
- And you can verify it.
- Verify that this works.
- 1 plus 1/1 does, indeed, equal 2.
- So it definitely does work.
- Let's do a more involved one now, now that we're warmed up.
- Let's say I have 2 over x squared plus 4x plus 3 is
- equal to 2 plus x minus 2 over x minus 3.
- So just like we've done in the last problem and in all of the
- other ones, we want to get all of these x's out of the
- denominator.
- And the best way to do it is to multiply every term in this
- equation by-- you can view it as the least common multiple
- of the denominator.
- So let me show you what I'm talking about.
- The first thing we might want to do is factor this, to see
- if this guy and this guy have anything in common.
- So let's see if we can factor it.
- What two numbers when you multiply them equal 3 and when
- you add them equal 4?
- Well, 3 and 1, right?
- So this right here, I can rewrite it as x plus 3 and--
- and really this is a plus over there, too.
- I wrote the original problem wrong.
- That is a [? product. ?]
- So this over here on the left is x plus 3 times x plus 1.
- And so, what can we multiply both sides of this equation by
- to get rid of the x plus 3 and an x plus 1 here, and an x
- plus 3 over there?
- Well, we multiply everything by x plus 3 and
- then by x plus 1.
- So this expression right here is actually divisible by this.
- So the least common multiple of x plus 3 and x squared plus
- 4x plus 3 is actually x squared plus 4x plus 3,
- because this is a multiple of that.
- It's divisible by it.
- So let's multiply both sides of the equation by this
- expression, right here.
- So we're going to get-- let me just rewrite it-- x plus 3
- times x plus 1-- and then we have this term over here--
- times 2 over x plus 3 times x plus 1 is going to be equal to
- x plus 3 times x plus 1 times this 2 times that 2, over
- there, plus x plus 3 times x plus 1 times this term-- times
- x minus 2 over x plus 3.
- Now, the whole reason why we did that is because that will
- cancel with that.
- The left-hand side of our equation will just be a 2.
- And 2 will be equal to-- well, 2 times this-- we know what
- this is equal to-- x plus 3 times x plus 1 is x squared
- plus 4x plus 3.
- We know that.
- We actually went the other way.
- We factored it.
- So it's going to be 2 times x squared plus 4x plus 3, which
- is 2 times x squared is-- I'm going to do it in that green
- color so you know we're dealing with
- this 2 right here.
- So this term right here is going to be 2 times x squared.
- So it's 2x squared plus 2 times 4x plus 8x plus 2
- times 3 plus 6.
- And now this term.
- We have an x plus 3 in the numerator.
- Or it would be, if I multiplied it times this
- fraction or this rational expression.
- That x plus 3 will cancel with that x plus 3.
- So we're just left with an x plus 1 times an x minus 2.
- An x plus 1 times x minus 2-- let me just focus in on this
- expression right here.
- I'll just box the whole thing.
- x plus 1 times x minus 2 is, let's just say it's x times x,
- which is x squared.
- x times negative 2, which is negative 2x.
- And then you have 1 times x, so plus x, and then you have 1
- times negative 2, so you have minus 2.
- And now we can simplify it.
- So once again, the left-hand side of this
- equation is still 2.
- Now on the right-hand side, let's look at the x squared
- terms. We have a 2x squared, and we have an x squared.
- Add them together, you get 3x squared.
- Then we have an 8x, and we have a minus 2x plus x.
- So what's that going to be?
- This is an 8x, and then when you add these together, you
- just get a minus-- you just get a negative x.
- So when you add everything together, you're going
- to get a plus 7x.
- Those are the x terms. And then finally, you have a-- no,
- I already used the orange-- you have a 6 and you
- have a minus 2.
- So that's going to be plus 4.
- And now let's subtract 2 from both sides of this equation.
- Let's subtract 2 from both sides.
- And we are left with 0 is equal to 3x squared plus 7x.
- We are subtracting 2 from both sides, so 4 minus 2 is plus 2.
- So we've stumbled, then we've turned this bizarre-looking
- thing into kind of a standard, plain
- vanilla quadratic equation.
- We have a non-zero coefficient out here.
- We might-- well, we just want to solve for it.
- We actually don't even have to factor it.
- So we might as well just use the quadratic formula.
- That's the easiest way to solve, especially a quadratic
- equation that has a non-one coefficient on
- the x squared term.
- So our solutions, using the quadratic formula. x is going
- to be equal to negative b-- which is negative 7-- plus or
- minus the square root of b squared-- that's 7 squared--
- which is 49 minus 4 times a-- which is 3--
- times c, which is 2.
- All of that over 2 times a, over 2 times 3 is 6.
- So what does this equal to?
- 4 times 3 is 12 times 2 is 24.
- So that, right there, is 24.
- What's 49 minus 24?
- It's 25.
- So I don't want to make that too messy, so
- let me rewrite it.
- So x is equal to negative 7 plus or minus the square root
- of 49 minus 24 is 25 over 6, which is equal to negative 7
- plus or minus 5-- the square root of 25-- over 6.
- And let's see, if we were to add 5, that would be negative
- 7 plus 5 is negative 2 over 6, which is
- equal to negative 1/3.
- That is one solution.
- And then if I do negative 7 minus 5, that is negative 12
- over 6, which is equal to negative 2.
- So those are our two solutions to this original rational
- equation right up here.
- Hopefully you enjoyed that.

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