Lesson Video: Rational Equations Mathematics

In this video, we will learn how to solve rational equations.

16:07

Video Transcript

In this video, we’ll learn how to solve rational equations. These are equations that involve fractions. We’ll recap how to find common denominators of a pair of algebraic fractions and learn how this process can help us to rearrange and thus solve these kinds of equations. We begin by recalling that algebraic fractions obey the same rules for arithmetic that numerical fractions do. When multiplying algebraic fractions, we multiply the numerator of the first fraction by the numerator of the second. And then we repeat that process with the denominators.

And what about dividing algebraic fractions? Well, to divide by an algebraic fraction or any fraction, we multiply by the reciprocal of that fraction. So here, we multiply by one over π‘₯ plus one over four or four over π‘₯ plus one. And then we multiply as normal. We’ll consider two types of example for adding and subtracting algebraic fractions.

Consider one over three π‘₯ plus two over π‘₯ squared. To add these fractions, we need to find a common denominator. In this case, we see that the highest power of π‘₯ is two. But we also have this coefficient of three to consider. So the common denominator we’re going to use will be three π‘₯ squared. We multiply the numerator and denominator of our first fraction by π‘₯. And we get π‘₯ over three π‘₯ squared. And multiplying the numerator and denominator of our second fraction by three gives us six over three π‘₯ squared. Once we have a common denominator, we simply add the numerators. And so this fraction becomes π‘₯ plus six over three π‘₯ squared. So sometimes, we can find a common denominator by using observation.

But what about π‘₯ over π‘₯ plus one minus four over π‘₯ plus three? In this case, it’s not instantly obvious what the common denominator is. So we find a denominator that absolutely must be common to both. This is found by multiplying one denominator by the other, giving us a common denominator of π‘₯ plus one times π‘₯ plus three. We multiply the numerator and denominator of our first fraction by π‘₯ plus three. So the numerator becomes π‘₯ times π‘₯ plus three, which is π‘₯ squared plus three π‘₯. And then we multiply the numerator and denominator of our second fraction by π‘₯ plus one, which gives us four π‘₯ plus four.

We’re now ready to subtract the numerators, being really careful to take into account that we’re subtracting. And so we’re going to subtract both four π‘₯ and four. This gives us π‘₯ squared minus π‘₯ minus four over π‘₯ plus one times π‘₯ plus three. So let’s look at how to apply this process to help us solve algebraic equations.

Solve the equation a half equals two over three π‘₯ minus one.

When asking us to solve, this question wants us to rearrange to make π‘₯ the subject. So how are we going to do that? There are two techniques, and we’ll consider both. We have a bit of an issue here. And that is that the π‘₯ is on the denominator of our fraction. So we might decide to begin by multiplying everything by three π‘₯ to counter this. If we do that, we need to multiply every single term in this equation by three π‘₯. So instead, let’s look to deal with the negative one to begin with. We begin by adding one to both sides of our equation. That gives us a half plus one equals two over three π‘₯. We can write one as one over one. And then to add it to one-half, we create a common denominator of two by multiplying the numerator and denominator of this fraction by two.

So we get a half plus two over two equals two over three π‘₯. A half plus two over two is three over two. So our equation is three over two equals two over three π‘₯. We’re now going to deal with that awkward three π‘₯ on the denominator of our fraction. And we’re going to counter this by multiplying both sides of our equation by three π‘₯. On the left-hand side, we get three over two times three π‘₯, which we can write as three over two times three π‘₯ over one. And then we simply multiply the numerators and the denominators to get nine π‘₯ over two. Then on the right-hand side, we’re simply left with two. Note here that a common mistake would be to think that we’re left with six π‘₯. But remember, multiplying by three π‘₯ is the inverse to dividing by three π‘₯. So what happens is these three π‘₯’s cancel.

Next, we’ll multiply by two to get rid of this fraction on the left-hand side. Nine π‘₯ divided by two times two is nine π‘₯. And two times two is four. Remember, we’re trying to make π‘₯ the subject. So our last step is to divide through by nine. And that gives us π‘₯ is equal to four-ninths. Remember, at this stage, we could check our solution for π‘₯ by substituting it back into the original equation. Instead, we’re going to consider an alternative technique to help us solve this problem. It involves writing the number negative one as negative one over one. And then we’re going to create a common denominator on that right-hand side.

The common denominator of three π‘₯ and one is three π‘₯. So we’re going to multiply the numerator and denominator of our second fraction by three π‘₯. When we do, we get two over three π‘₯ minus three π‘₯ over three π‘₯. And since the denominators of these fractions are now equivalent, we can simply subtract the numerators. And our equation becomes a half equals two minus three π‘₯ over three π‘₯. Like we did in our previous method, we’re going to multiply through by three π‘₯ to get rid of this denominator. And on the left-hand side, we get three π‘₯ times a half which is three π‘₯ over two. Then on the right-hand side, this simply eliminates our denominator. So we end up with two minus three π‘₯.

Next, we’ll multiply by two to get rid of the denominator on the left-hand side. We make sure we multiply everything on the right-hand side by two. And our equation is now three π‘₯ equals four minus six π‘₯. Then, whenever we’re solving an equation with π‘₯’s on both sides, we look to deal with these smallest value of π‘₯. Here, that’s negative six π‘₯. So the inverse operation is to add six π‘₯ to both sides. That gives us nine π‘₯ equals four. And then our last step is to divide through by nine. Once again, we find π‘₯ is equal to four-ninths.

In our next example, we’ll consider how we can solve an algebraic equation when the fraction is equal to zero.

Find π‘₯ given π‘₯ minus 20 over π‘₯ plus 10 is equal to zero.

Let’s consider what’s really going on in our equation. We have some fraction, and it’s equal to zero. Now, this tells us nothing about the denominator of our fraction. In fact, we don’t really care about the denominator. For our fraction to be equal to zero, the numerator itself must be equal to zero, since zero divided by any number is zero. So to solve this equation, we simply need to solve π‘₯ minus 20 equals zero. And to solve this equation, we just add 20 to both sides, giving us π‘₯ is equal to 20. We might even wish to check our solution by substituting it back into the original expression. When π‘₯ is equal to 20, π‘₯ minus 20 over π‘₯ plus 10 is 20 minus 20 over 20 plus 10. That’s zero divided by 30, which is, of course, zero. So we know our solution π‘₯ equals 20 is correct.

We’ll now consider how to solve a rational equation that contains two algebraic fractions.

Given that seven π‘₯ over π‘₯ minus three is equal to 16π‘₯ over π‘₯ plus three minus nine, find the value of π‘₯.

To solve this equation, we’re going to begin by looking simply at the right-hand side. We have the sum of a rational expression, that’s the 16π‘₯ over π‘₯ plus three, and an integer. That’s negative nine. So we’re going to begin by subtracting nine from our fraction by creating a common denominator. Now, currently, the denominator of nine is one. Remember, we’ve got nine ones. And we’re going to multiply the numerator and denominator of this fraction by π‘₯ plus three. That will create a denominator of just π‘₯ plus three. That gives us 16π‘₯ over π‘₯ plus three minus nine π‘₯ plus three over π‘₯ plus three.

Once their denominators are equal, we can simply subtract their numerators. So our numerator becomes 16π‘₯ minus nine lots of π‘₯ plus three. We’ll now distribute these parentheses, remembering that we’re multiplying everything inside by negative nine. When we do, we get 16π‘₯ minus nine π‘₯ minus 27 over π‘₯ plus three, which simplifies to seven π‘₯ minus 27 over π‘₯ plus three. We’ll now bring down the left-hand side of our equation. And we can see that seven π‘₯ over π‘₯ minus three is equal to seven π‘₯ minus 27 over π‘₯ plus three. And we have a couple of options here. We could subtract one of our fractions so that we end up with some expression equal to zero. We then create a common denominator of π‘₯ plus three times π‘₯ minus three.

In this case, though, we’re not going to do that. We’re just going to look to get rid of the denominators of our fractions. To get rid of the denominator on the left-hand side, we’re going to multiply by π‘₯ minus three. That leaves us simply with seven π‘₯ on the left-hand side. And on the right, we’re really multiplying by π‘₯ minus three over one. So we get seven π‘₯ minus 27 over π‘₯ plus three times π‘₯ minus three over one. We multiply as normal by multiplying the numerators and then separately multiplying the denominators. And the right-hand side is now seven π‘₯ minus 27 times π‘₯ minus three over π‘₯ plus three.

Next, we’ll multiply by π‘₯ plus three to get rid of the denominator on the right-hand side. And we now have seven π‘₯ times π‘₯ plus three equals seven π‘₯ minus 27 times π‘₯ minus three. Note though instead of multiplying by π‘₯ minus three in one step and then π‘₯ plus three in another, we could have initially cross multiplied by π‘₯ minus three and π‘₯ plus three and achieved the same result. We’re now going to distribute both pairs of parentheses. On the right-hand side, that’s achieved by multiplying everything inside by seven π‘₯ to get seven π‘₯ squared plus 21π‘₯. And on the right-hand side, we multiply everything in the second set of parentheses by seven π‘₯ and then by negative 27. And that gives us seven π‘₯ squared minus 21π‘₯ minus 27π‘₯ plus 81. This on the right-hand side simplifies to seven π‘₯ squared minus 48π‘₯ plus 81.

We’re aiming to solve for π‘₯. So let’s begin by subtracting seven π‘₯ squared from both sides of our equation. This has the effect of just eliminating any π‘₯ squareds. And we’re left with 21π‘₯ equals negative 48π‘₯ plus 81. We then add 48π‘₯ to get 69π‘₯ equals 81. And finally, we divide through by 69. In doing so, we find π‘₯ is equal to 81 over 69 which simplifies by dividing through by three to 27 over 23. And so the value of π‘₯ that satisfies our equation is 27 over 23.

We’ll now look at how we can solve an algebraic equation that involves three fractions.

Solve three over 𝑛 squared minus four plus one over 𝑛 plus two equals two over 𝑛 minus two.

We have three algebraic fractions. And there are a number of ways we can solve this. The key to any of these methods, though, is spotting that 𝑛 squared minus four could be factored. We can write it using the difference of two squares. We write it as 𝑛 minus two times 𝑛 plus two. And then we notice that this is the product of our other two denominators. So we could create a common denominator of 𝑛 minus two times 𝑛 plus two and gather all our terms on the left-hand side. Alternatively, we could subtract one over 𝑛 plus two from both sides of our equation. Let’s see what that would look like.

Our equation becomes three over 𝑛 squared minus four equals two over 𝑛 minus two minus one over 𝑛 plus two. Next, we’ll multiply the numerator and denominator of our first fraction by 𝑛 plus two and of our second by 𝑛 minus two, creating a common denominator of 𝑛 minus two times 𝑛 plus two. The numerators on the right-hand side become, respectively, two times 𝑛 plus two and one times 𝑛 minus two. And in fact, we saw we could write the denominator on the left-hand side as 𝑛 minus two times 𝑛 plus two.

Now, the denominators are equal on the right-hand side. We’re going to subtract one times 𝑛 minus two from two times 𝑛 plus two. And the expression on the right-hand side becomes two times 𝑛 plus two minus one times 𝑛 minus two over 𝑛 minus two times 𝑛 plus two. Notice now that the denominators of these two fractions are equal. And we’re told the fractions themselves are equal. So this means our numerators must be equal also. So three must be equal to two times 𝑛 plus two minus one times 𝑛 minus two.

Let’s distribute these parentheses, remembering that, on our second set of parentheses, we’re multiplying everything by negative one and our equation becomes three equals two 𝑛 plus four minus 𝑛 plus two. We simplify the right-hand side to get 𝑛 plus six. And we now see we have quite a simple equation that we can solve for 𝑛. We subtract six from both sides, giving us 𝑛 equals negative three. So the solution to our equation is 𝑛 equals negative three.

Remember, we could check the solution by substituting it back into the original equation and making sure that both sides are then equal.

In our final example, we’ll look at how to solve an equation where we have rational expressions being divided.

Solve π‘₯ plus one over π‘₯ minus one divided by two π‘₯ plus 10 over π‘₯ plus one equals one-half.

There’s quite a lot going on here. But notice the left-hand side is the division of two algebraic fractions. So let’s deal with that first. We recall that to divide by a fraction, we multiply by the reciprocal of that fraction. The reciprocal of two π‘₯ plus 10 and π‘₯ plus one is π‘₯ plus one over two π‘₯ plus 10. So we’re going to work out π‘₯ plus one over π‘₯ minus one times π‘₯ plus one over two π‘₯ plus 10. Note that an alternative method we have for dividing algebraic fractions is to create a common denominator and then simply divide the numerators. In this case, that might be slightly more long winded.

Once we have it in this form, once we’re essentially multiplying two fractions, we simply multiply their numerators and then individually multiply their denominators. So the fraction is now π‘₯ plus one times π‘₯ plus one over π‘₯ minus one times two π‘₯ plus 10. Let’s distribute these parentheses. In doing so, we get π‘₯ squared plus two π‘₯ plus one over two π‘₯ squared minus [plus] eight π‘₯ minus 10. Let’s bring down the right-hand side of our equation, and so that our entire rational expression is equal to one-half.

We now want to eliminate the denominator of our fractions. We’re going to begin by multiplying both sides by two π‘₯ squared plus eight π‘₯ minus 10. Now, we could if we wanted at the same time multiply both sides by two. But we don’t need to do that here. And that’s because the right-hand side of our equation becomes a half times two π‘₯ squared plus eight π‘₯ minus 10. Notice that every term in our quadratic expression here is a multiple of two. So let’s multiply each term individually by one-half. And so the right-hand side becomes π‘₯ squared plus four π‘₯ minus five. We’ll subtract π‘₯ squared from both sides of our equation.

And since we’re solving, we want to make π‘₯ the subject. So let’s deal with the smallest number of π‘₯. Let’s subtract two π‘₯ from both sides. That gives us one is equal to two π‘₯ minus five. Then we add five to both sides to get six is equal to two π‘₯. And finally, we divide through by two, giving us a solution π‘₯ is equal to three.

In this video, we recapped the fact that algebraic fractions obey the same rules for arithmetic that numerical fractions do. We saw that we can use these rules to help us solve equations involving algebraic fractions by rearranging to make π‘₯ the subject. Finally, we saw that if we’re working with an algebraic fraction that’s equal to zero, we can simply set its numerator equal to zero and solve for π‘₯.

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