Video Transcript
Simplify 𝑛 of 𝑥 equals three 𝑥 over 𝑥 minus one minus 𝑥 over 𝑥 plus three.
Here, we have a function 𝑛 of 𝑥 which is found by subtracting one algebraic fraction from another. In order to simplify 𝑛 of 𝑥, we need to try and combine these two fractions into a single fraction. We know that in order to subtract numeric fractions such as the sum two-fifths minus one-third, we need to find a common denominator. And this common denominator will be the lowest common multiple of the two individual denominators. So, in the case of the sum two-fifths minus one-third, the denominator will be 15 because this is the lowest common multiple of five and three. We then find equivalent fractions for each of the original ones with the denominator of 15. And then we can perform the subtraction.
We’re going to follow a similar process for this algebraic question. First, we need to find the lowest common multiple of the two algebraic expressions 𝑥 minus one and 𝑥 plus three, which are the denominators of the two fractions. Well, these algebraic expressions have no common factors other than one. And so, their lowest common multiple is in fact their product, 𝑥 minus one multiplied by 𝑥 plus three, in the same way that the lowest common multiple of five and three, which are coprime, was their product of 15.
We’re now going to take our expression for 𝑛 of 𝑥 and rewrite it so that the denominator for each part is 𝑥 minus one multiplied by 𝑥 plus three. For the first fraction, we’d need to multiply both the numerator and denominator by 𝑥 plus three. And for the second, we’d need to multiply both the numerator and denominator by 𝑥 minus one. As we now have a common denominator for the two fractions, we can subtract them by subtracting the numerators. This gives three 𝑥 multiplied by 𝑥 plus three minus 𝑥 multiplied by 𝑥 minus one all over 𝑥 minus one multiplied by 𝑥 plus three. So, we have successfully combined to a single fraction.
The next step is to distribute each set of parentheses in the numerator. Distributing the first set gives three 𝑥 multiplied by 𝑥, which is three 𝑥 squared, plus three 𝑥 multiplied by three, which is nine 𝑥. To distribute the second set, we need to be really careful because 𝑥 minus one isn’t just multiplied by 𝑥; it is multiplied by negative 𝑥. So, we need to ensure we distribute that negative over the parentheses as well. Negative 𝑥 multiplied by 𝑥 is negative 𝑥 squared and negative 𝑥 multiplied by negative one is positive 𝑥. So, we have three 𝑥 squared plus nine 𝑥 minus 𝑥 squared plus 𝑥 all over 𝑥 minus one multiplied by 𝑥 plus three.
Simplifying by collecting the like terms in the numerator, we have three 𝑥 squared minus 𝑥 squared, which is two 𝑥 squared, and nine 𝑥 plus 𝑥, which is 10𝑥. So, our expression has simplified to two 𝑥 squared plus 10𝑥 over 𝑥 minus one multiplied by 𝑥 plus three.
The final step is to observe that the two terms in the numerator share a common factor of two and also a common factor of 𝑥. So, we can factor by two 𝑥. This gives two 𝑥 multiplied by 𝑥 plus five in the numerator and the denominator is unchanged. This fraction can’t be simplified any further as there are no common factors between the numerator and denominator other than, of course, one.
We have fully simplified 𝑛 of 𝑥 then. 𝑛 of 𝑥 is equal to two 𝑥 multiplied by 𝑥 plus five over 𝑥 minus one multiplied by 𝑥 plus three.
