Video: Simplifying a Quotient of Two Rational Functions

Given that 𝑛₁(π‘₯) = π‘₯ + (16/(π‘₯ βˆ’ 8)), 𝑛₂(π‘₯) = 9π‘₯ + (144/(π‘₯ βˆ’ 8)), and 𝑛(π‘₯) = 𝑛₁(π‘₯) Γ· 𝑛₂(π‘₯), determine 𝑛(π‘₯) in its simplest form.

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Video Transcript

Given that 𝑛 one of π‘₯ equals π‘₯ plus 16 over eight minus π‘₯, 𝑛 two of π‘₯ equals nine π‘₯ plus 144 over eight minus π‘₯, and 𝑛 of π‘₯ equals 𝑛 one π‘₯ divided by 𝑛 two π‘₯, determine 𝑛 of π‘₯ in its simplest form.

We’re going to be working out 𝑛 of π‘₯, which is the quotient of 𝑛 one of π‘₯ and 𝑛 two of π‘₯. So let’s divide π‘₯ plus 16 over π‘₯ minus eight β€” remember that’s 𝑛 one β€” by 𝑛 two, which is known as π‘₯ plus 144 over π‘₯ minus eight. Before we do that, we’re going to need to simplify the expressions in each of these set of parentheses. By adding a denominator of one to π‘₯ and nine π‘₯, we see that we can add the fractions in each expression. We’ll create a common denominator in this first expression by multiplying the first term by π‘₯ minus eight over π‘₯ minus eight. That gives us π‘₯ times π‘₯ minus eight over π‘₯ minus eight plus 16 over π‘₯ minus eight. We will multiply nine π‘₯ over one by π‘₯ minus eight over π‘₯ minus eight too. That gives us nine π‘₯ times π‘₯ minus eight over π‘₯ minus eight plus 144 over π‘₯ minus eight.

And then, we add the numerators. Our first expression becomes π‘₯ times π‘₯ minus eight plus 16 all over π‘₯ minus eight. And our second expression becomes nine π‘₯ times π‘₯ minus eight plus 144 over π‘₯ minus eight. Then, we recall that to divide by a fraction, we multiply by the reciprocal of that same fraction. In this case, we multiply by π‘₯ minus eight over nine π‘₯ times π‘₯ minus eight plus 144. And then if we look carefully, we see we can cross cancel. We can divide by π‘₯ minus eight. We then multiply the numerator of the first fraction by the numerator of the second and repeat that process for the denominator. So we get π‘₯ times π‘₯ minus eight plus 16 times one, which is just that our original expression, over nine π‘₯ times π‘₯ minus eight plus 144.

Next, we distribute our parentheses. We multiply π‘₯ by π‘₯ and π‘₯ by negative eight. And we get π‘₯ squared minus eight π‘₯. Similarly, we multiply nine π‘₯ by π‘₯ and nine π‘₯ by negative eight. So we end up with π‘₯ squared minus eight π‘₯ plus 16 over nine π‘₯ squared minus 72π‘₯ plus 144. And in fact, it might look like we’re finished. But we need to find 𝑛 of π‘₯ in its simplest form. So here, we notice that we have a common factor in the denominator. That common factor is nine. And if we factor nine out of the denominator, we get nine times π‘₯ squared minus eight π‘₯ plus 16. Now, notice we have a common factor. We can divide both the numerator and the denominator by π‘₯ squared minus eight π‘₯ plus 16. And so, we’re left with one-ninth.

𝑛 of π‘₯ equals one-ninth.

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