# Simplifying Algebraic Fractions

This section explores simplifying algebraic fractions; Here we focus on simplifying expressions with fractions by factorising and cancelling, adding, subtracting, multiplying and adding algebraic fractions and lastly solving equations which involve fractions. You must have prior knowledge of fraction arithmetic and factorising simple polynomials.

## Cancelling algebraic fractions

Let us look at cancelling algebraic fractions first. We shall look at algebraic fractions that involve quadratic expressions.Example Simplify the following fraction;

x + 3/x² + 5x + 6

Answer:

x + 3/(x + 3)(x + 2)

x + 3/x² + 5x + 6 = ~~(x + 3)~~/~~(x + 3)~~(x + 2)

= 1/x + 2

= 1/x + 2

Explanation: We have to factorise the denominator of the fraction to simplify it first. This will make it easier to evaluate. We do so to get the following outcome.

x + 3/(x + 3)(x + 2)

We can see a common factor in both the numerator and the denominator so we can cancel these out;
x + 3/x² + 5x + 6 = ~~(x + 3)~~/~~(x + 3)~~(x + 2)

= 1/x + 2

= 1/x + 2

We have divided (x + 3) at the top with (x + 3) at the bottom, so the result is 1.

We have managed to workout the fraction, i.e: Simplify it.Example Simplify the following fraction.

4/m + 3/x

Answer:

4/m + 3/x = 4×x + 3×m/mx

4/m + 3/x = 4x + 3m/mx

Explanation: We need to find the common denominator first. We can do this simply by multiplying both denominators in the fraction.

4/m + 3/x = 4×x + 3×m/mx

Lastly we simplify by cleaning up the fraction of the multiplication signs. We get the answer as being.
4/m + 3/x = 4x + 3m/mx

Example Simplify the following fraction.

2/x² + 5x + 6 + 5/x²-4

Answer:

2/(x + 2)(x + 3) + 5/(x + 2)(x – 2)

2(x – 2) + ?/(x + 2)(x + 3)(x – 2)

2(x – 2) + 5(x + 3)/(x + 2)(x + 3)(x – 2)

2x – 4 + 5x + 15/(x + 2)(x + 3)(x – 2)

7x + 11/(x + 2)(x + 3)(x – 2)

Explanation: First we need to find the common denominator as we’ve done above, but before we do that we need to factorise the denominators as shown below;

2/(x + 2)(x + 3) + 5/(x + 2)(x – 2)

You can see above that both fraction denominators have a factor of (x + 2) so we only include this once, because it will be cancelled out in the end anyway. You can try to include the extra (x + 2) and see what happens later. Remember what you multiply in the denominator must be multiplied with the numerator to make it fair otherwise the fractions won’t be equivalent. The lowest common denominator we find here is;
(x + 2)(x + 3)(x – 2)

Now we must multiply the numerator as well. We multiply 2 by (x – 2) because to find the common denominator we multiplied by (x – 2)
2(x – 2) + ?/(x + 2)(x + 3)(x – 2)

For the fraction which has 5 to have the common denominator we must multiply 5 with (x + 3) as shown below;
2(x – 2) + 5(x + 3)/(x + 2)(x + 3)(x – 2)

Next we expand the brackets.
2x – 4 + 5x + 15/(x + 2)(x + 3)(x – 2)

And lastly we simplify the fraction.
7x + 11/(x + 2)(x + 3)(x – 2)

## Multiplying and dividing algebraic fractions

In the following examples we’re going to look at multiplying and dividing algebraic fractions.

Example Simplify the following fraction.

The following example involves division.
Answer:
~~(x + 7)~~/2(x – 1) × 3x²(x + 2)/4(x + 7)

x + 7/2x(x – 1) × 3x³(x + 2)/4(x + 7)

x + 7/2~~x~~(x – 1) × 3x

^{3}^{2}(x + 2)/4(x + 7)1/2(x – 1) × 3x²(x + 2)/4

x × 3x²(x + 2)/2(x – 1) × 4

3x²(x + 2)/8(x – 1)

Explanation:
~~(x + 7)~~/2(x – 1) × 3x²(x + 2)/4(x + 7)
That leaves;

x + 7/2x(x – 1) × 3x³(x + 2)/4(x + 7)

First we check whether any numbers can cancel. Here 3 cannot cancel with 2 or 4 so we have to move on to the next step. Next we check the variables or letters. We see that the x in one fraction will cancel with one of the xs in the x^{3}of the other fraction. We can cancel these;x + 7/2~~x~~(x – 1) × 3x

We can also cancel out x+7 from the numerator of the first fraction with the denominator if the second fraction as shown below;
^{3}^{2}(x + 2)/4(x + 7)1/2(x – 1) × 3x²(x + 2)/4

Since we cannot cancel any more we simplify just multiply what is left.
x × 3x²(x + 2)/2(x – 1) × 4

The final answer then becomes;
3x²(x + 2)/8(x – 1)

Example Simplify the following fraction.

4X + 12/X² – 4 ÷ 6x – 12x – 90/5x – 10

Answer:
~~4~~(x + 3)/(x + 2)(x – 2) × 5(x – 2)/~~6~~(x + 3)(x – 5)

4x + 12/x² – 4 ÷ 6x² – 12x – 90/5x – 10

4(x + 3)/(x + 2)(x – 2) ÷ 6(x + 3)(x – 5)/5(x – 2)

4(X + 3)/(X + 2)(X – 2) × 5(x – 2)/6(x + 3)(x – 5)

^{2}

_{3}

2~~(x + 3)~~/(x + 2)(x – 2) × 5(x – 2)/3~~(x + 3)~~(x – 5)

2/(x + 2)~~x – 2~~ × 5~~(x – 2)~~/3(x – 5)

2/(x + 2) × 5(x – 2)/3(x – 5)

10/3(x + 2)(x – 5)

Explanation: Look at each numerator and denominator and see whether you can find their common factors. If you look carefully you will see that;
We rewrite the expression as shown below;
Now we have the fraction as;
~~4~~(x + 3)/(x + 2)(x – 2) × 5(x – 2)/~~6~~(x + 3)(x – 5)
We can also cancel out the factor of (x+3)

- 4x+12 has a common factor of 4
- 6x
^{2}-12x -90 has common factor of 6 - 5x – 10 has a common factor of 5
- x
^{2}– 4 has no common factors

4x + 12/x² – 4 ÷ 6x² – 12x – 90/5x – 10

Now we can factorise the quadratics in each fraction if any.
- x
^{2}-2x-15 factorises to (x+3)(x-5) - x
^{2}-4 factorises to (x+2)(x-2)

4(x + 3)/(x + 2)(x – 2) ÷ 6(x + 3)(x – 5)/5(x – 2)

Next we turn the second fraction upside down and change the division sign to multiplication as shown below.
4(X + 3)/(X + 2)(X – 2) × 5(x – 2)/6(x + 3)(x – 5)

Now we can cancel out the common factors. We can divide 4 and 6 by 2.
^{2}

_{3}

2~~(x + 3)~~/(x + 2)(x – 2) × 5(x – 2)/3~~(x + 3)~~(x – 5)

Also (x-2) can be cancelled out.
2/(x + 2)~~x – 2~~ × 5~~(x – 2)~~/3(x – 5)

We can’t cancel out any more which leaves;
2/(x + 2) × 5(x – 2)/3(x – 5)

Now we can multiply what is left together. This leaves;
10/3(x + 2)(x – 5)

…as the final simplified answer.
LINK TO PAGE

Copy and paste HTML code into your page.

Url:

**Oh snap!**Presentation file not Found!

**Oh snap!**Practice file Found!

why are there no expressions that show subtraction. It moves from addition to multiplication to division

Thanks for the article… It helped me a lot

Yeah Bro this is very useful to those who are new to simplifying algebraic fractions. Thanks a lot mate, id rate this an 8! GOOD JOB KEEP ON COMING WITH THE GOOD WORK BUDS! SwAgmEiSter free styling out! Peace out Bruh’s!