# Part 1

In the first video, we are going to add and subtract rational expressions.

• Only $19.95/mo. No commitment. Cancel anytime. For example: $\frac{3}{2x}+\frac{5}{2x}$ Simply combine the numerator and the denominator together if the denominators are the same and we have $\frac{8}{2x}$ Let’s look at another example: $\frac{2}{6n}+\frac{4}{2mn}$ Since the denominators are not the same, multiple each fraction to achieve the least common multiple $\frac{(m)2}{(m)6n}+\frac{(3)4}{(3)2mn}$ $\frac{2m}{6mn}+\frac{12}{6mn}$ Now, we can add the terms together $\frac{2m+12}{6mn}$ In order to reduce the expression, factor if necessary $\frac{2(m+6)}{6mn}$ Reduce, and our final answer is $\frac{(m+6)}{3mn}$ ## Video-Lesson Transcript Let’s go over how to add and subtract rational expressions. We have two rational expressions here $\dfrac{3}{2x} + \dfrac{5}{2x}$ Just like any fractions that need to add or subtract, we have to have a common denominator. In this case, we do. So, we have the same denominator and then just add the numerator. Our answer is $= \dfrac{8}{2x}$ Let’s look at a more difficult one. $\dfrac{5b - 2}{9b + 45} - \dfrac{b - 2}{9b + 45}$ Again, we have common denominators. So we’ll just copy it. We’ll subtract the numerator as necessary. $= \dfrac{5b - 2 - (b - 2)}{9b + 45}$ Then distribute the negative sign $= \dfrac{5b - 2 - b + 2}{9b + 45}$ Then, let’s combine the like terms $= \dfrac{4b}{9b + 45}$ Let’s look at an example where we don’t have common denominators. $\dfrac{2}{6n} + \dfrac{4}{2mn}$ What we want to do first is to write this with common denominators. So, we need the least common multiple of the two denominators. ${6n}$ and $2mn$ What can we multiply these by to give us something that will equal each other? Let’s look at the numbers first and identify the least common multiple. ${6n}$ will be $6mn$, $2mn$ will be $6mn$ We multiplied $6n$ by $m$ and $2mn$ by $3$. So here’s our solution: $\dfrac{(m) 2}{(m) 6n} + \dfrac{(3) 4}{(3) 2mn}$ $= \dfrac{2m}{6mn} + \dfrac{12}{6mn}$ Now we can add these together. $= \dfrac{2m + 12}{6mn}$ Let’s look at it further. We can still reduce this since there’s a factor $= \dfrac{2 (m + 6)}{6mn}$ $= \dfrac{m + 6}{3mn}$ # Part 2 In the second video, we are going to add and subtract rational expressions that are more complex. • Only$19.95/mo. No commitment. Cancel anytime.

For example: $\frac{6}{n-6}+\frac{3}{n-1}$

Multiply each fraction to achieve the least common multiple
$\frac{(n-1)6}{(n-1)(n-6)}+\frac{(n-6)3}{(n-6)(n-1)}$

Use the distributive property to multiply
$\frac{6n-6}{(n-1)(n-6)}+\frac{3n-8}{(n-1)(n-6)}$

Now that we have common denominators, add the numerators together
$\frac{9n-24}{(n-1)(n-6)}$

Factor any expression if necessary
$\frac{3(3n-8)}{(n-1)(n-6)}$

Since nothing can be reduced, we are going back to the original expression
$\frac{9n-24}{(n-1)(n-6)}$

Use FOIL (First, Outside, Inside, Last) to multiply
$\frac{9n-24}{n^2-6n-n+6}$

Combine like terms, and we have
$\frac{9n-24}{n^2-7n+6}$

## Video-Lesson Transcript

Let’s go over how to add and subtract rational expressions part 2.

We’re going to focus on binomials on the denominator.

For example:

$\dfrac{6}{n - 6} + \dfrac{3}{n - 1}$

First, we want to write this with common denominators.

What could be the least common multiple of these two?

$n - 6$,
$n - 1$

Let’s take a step back and take a look at adding fractions.

If we have

$\dfrac{2}{3} + \dfrac{1}{2}$

what is the least common multiple of $3$ and $2$?

It’s $6$.

We got that by simply multiplying the two numbers.

Now, we’ll have

$= \dfrac{(2) 2}{(2) 3} + \dfrac{(3) 1}{(3) 2}$
$= \dfrac{4}{6} + \dfrac{3}{6}$
$= \dfrac{7}{6}$

So, we’ll do the same thing here.

Let’s just multiply these together.

$\dfrac{6}{n - 6} + \dfrac{3}{n - 1}$
$= \dfrac{6 (n - 1)}{(n - 6) (n - 1)} + \dfrac{3 (n - 6)}{(n - 1) (n - 6)}$
$= \dfrac{6n - 6}{(n - 1) (n - 6)} + \dfrac{3n - 18}{(n - 1) (n - 6)}$

Now, let’s add the numerators together.

$= \dfrac{9n - 24}{(n - 6) (n - 1)}$

The reason why we didn’t multiply the denominators together is because it’s a lot easier to factor this way.

Now, we can factor the numerator only and keep the denominator as it is.

$= \dfrac{3 (3n - 8)}{(n - 6) (n - 1)}$

But there isn’t anything to cancel out.

So, let’s just solve it.

$= \dfrac{9n - 24}{n^2 - 7n + 6}$

This format is also acceptable:

$= \dfrac{9n - 24}{(n - 6) (n - 1)}$

Let’s look at this one

$\dfrac{3}{n - 4} + \dfrac{2n - 8}{n^2 - n + 12}$

So, we’ll just multiply the denominators together to get the common denominator.

This will be a bit more complicated because we’re going to multiply a binomial to a trinomial.

Multiplying these is going to give us a big expression. If we just multiply them upfront.

$= \dfrac{3 (n^2 - n - 12)}{n - 4 (n^2 - n - 12)} + \dfrac{2n - 8 (n - 4)}{n^2 - n + 12 (n - 4)}$

This is going to be confusing.

We’ll still get the correct answer but this is a lot of work.

So, let’s look at another option.

What we can do is to look at the factors of these denominators.

$n - 4$ stays as is

Let’s factor $n^2 - n - 12$ out

$= (n - 4) (n + 3)$

The first denominator is already $n - 4$, so we’ll just multiply it by

$n + 3$.

At this point, we already have common denominators.

$= \dfrac{3 (n + 3)}{n - 4 (n + 3)} + \dfrac{2n - 8 }{(n - 4) (n + 3)}$

The numerator for the second expression stays the same because all we did was just to factor its denominator.

The denominator didn’t change in the second expression.

$= \dfrac{3n + 9}{n - 4 (n + 3)} + \dfrac{2n - 8 }{(n - 4) (n + 3)}$

We have like terms here so let’s combine.

$= \dfrac{5n + 1}{(n - 4) (n + 3)}$