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Adding And Subtracting Rational Expressions

Part 1

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

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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}

Adding And Subtracting Rational Expressions

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.

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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.

And our answer is

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

This format is also acceptable:

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

Adding And Subtracting Rational Expressions 2

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.

Our answer is

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

This can’t be factored and reduced further. So, it’s the final answer.