Video: Solving Rational Equations in One Variable

Consider the equation 1/(π‘₯ βˆ’ 3) βˆ’ 2/(π‘₯ + 3) = (3 + π‘₯)/(9 βˆ’ π‘₯Β²). If you multiply both sides by π‘₯Β² βˆ’ 9 and rearrange so that the right-hand side is 0, what equation do you get? What can you say about the solutions of the original equation?

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Video Transcript

Consider the equation one over π‘₯ minus three minus two over π‘₯ plus three is equal to three plus π‘₯ over nine minus π‘₯ squared. If you multiply both sides by π‘₯ squared minus nine and rearrange so that the right-hand side is zero, what equation do you get? What can you say about the solutions of the original equation?

We’ll start with what we were given. And then we’re trying to multiply both sides of this equation by π‘₯ squared minus nine. For the left-hand side, that means we’ll need to distribute π‘₯ squared minus nine to both of these terms. We multiply the numerator by π‘₯ squared minus nine. And our first term becomes π‘₯ squared minus nine over π‘₯ minus three. The second term looks like this, two times π‘₯ squared minus nine over π‘₯ plus three. On the right-hand side, we’ll have three plus π‘₯ times π‘₯ squared minus nine over nine minus π‘₯ squared.

But we want to move the term on the right-hand side to the left-hand side. If we subtract three plus π‘₯ times π‘₯ squared minus nine over nine minus π‘₯ squared from both sides of the equation, we get this. We’ve accomplished one of our goals. And that’s to get the right-hand side to be equal to zero. But now we need to simplify these three terms. To do this, let’s break them apart and simplify each term by itself and then plug it back in.

Our first term is π‘₯ squared minus nine over π‘₯ minus three. Here we have to notice a pattern. π‘₯ squared minus nine is the difference of two squares. And we rewrite the difference of two squares as π‘Ž plus 𝑏 times π‘Ž minus 𝑏. In our case, π‘₯ squared minus nine is the same thing as saying π‘₯ squared minus three squared. And π‘₯ squared minus three squared is the same thing as π‘₯ plus three times π‘₯ minus three. When we rewrite it this way, we see that the π‘₯ minus three in the numerator and the denominator cancel out and that our first term is only π‘₯ plus three. We’ll substitute that back in for our first term.

Now when it comes to our second term, notice that we’re doing subtraction. We can’t leave out that negative sign. To accommodate for that, we’ll say negative two times π‘₯ squared minus nine over π‘₯ plus three. Again, we have the same difference of squares. And we can rewrite the numerator as negative two times π‘₯ plus three times π‘₯ minus three. π‘₯ plus three in the numerator and the denominator cancel out. And we’re left with negative two times π‘₯ minus three. We can distribute that negative two. Negative two times π‘₯ equals negative two π‘₯. Negative two times negative three equals positive six. In place of our second term, we’ll write negative two π‘₯ plus six.

Now we have our third term, which again has a negative in front of it. So we’ll write negative one times three plus π‘₯ times π‘₯ squared minus nine all over nine minus π‘₯ squared. Notice that we have π‘₯ squared minus nine in the numerator and nine minus π‘₯ squared in the denominator. We can think of this as having a positive π‘₯ squared minus nine and a positive nine minus π‘₯ squared. But if we multiply the numerator by negative one, we would end up with negative π‘₯ squared and positive nine. So that’s what we wanna do. We wanna multiply negative one by π‘₯ squared minus nine. When we do that, we get negative π‘₯ squared plus nine.

Remember that when it comes to multiplying, we can multiply the numerator in any order. So we’ve just multiplied negative one by π‘₯ squared minus nine. From there, we wanna switch these two addends. Instead of saying negative π‘₯ squared plus nine, we’ll say nine minus π‘₯ squared times three plus π‘₯ all over nine minus π‘₯ squared. The nine minus π‘₯ squared in the numerator and the nine minus π‘₯ squared in the denominator cancel out. And our third term is just three plus π‘₯. In place of our third term, we write three plus π‘₯. And we know that this is equal to zero.

We now have π‘₯ plus three minus two π‘₯ plus six plus three plus π‘₯. So we’ll combine the like terms, π‘₯ minus two π‘₯ plus π‘₯. Well, if we add the two π‘₯s together, we’ll get positive two π‘₯ minus two π‘₯. So they cancel completely. Then, we add three plus six plus three, which equals 12. Our simplified statement is that 12 equals zero. The first part of our question is then 12 equals zero. The second half wants to know what we can say about the solutions of the original equation. When we’re solving an equation and we end up with a simplified form that’s an untrue statement β€” 12 equals zero is untrue β€” we can say that there are no solutions to the equation. There is no value for π‘₯ that will make that statement true because zero is never equal to 12.

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