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Algebraic Fractions

What is an algebraic fraction?

An algebraic fraction is a fraction whose numerator and denominator are algebraic expressions.

We will explain some techniques to simplify complex algebraic fractions. We know that we can simplify fractions by dividing the numerator and denominator by a common factor. This can also be done with algebraic fractions.

Example 1: Simplify each of the following fractions as far as possible:

For help check rules of exponents

a) $ \displaystyle \frac{{3x}}{6}$

The highest common factor of 3 and 6 is 3.

$ \displaystyle \frac{{3x}}{6}=\frac{{3x\div 3}}{{6\div 3}}=\frac{x}{2}$

b) $ \displaystyle \frac{{{{y}^{2}}}}{{{{y}^{5}}}}$

The highest common factor of $ \displaystyle {{{y}^{2}}}$ and $ \displaystyle {{{y}^{5}}}$ is $ \displaystyle {{{y}^{2}}}$

$ \displaystyle \frac{{{{y}^{2}}}}{{{{y}^{5}}}}=\frac{{{{y}^{2}}\div {{y}^{2}}}}{{{{y}^{5}}\div {{y}^{2}}}}=\frac{1}{{{{y}^{3}}}}$

c) $ \displaystyle \frac{{12{{p}^{3}}}}{{16{{p}^{7}}}}$

Consider the constant first. The HCF of 12 and 16 is 4, so you can divide both 12 and 16 by 4.

$ \displaystyle \frac{{12{{p}^{3}}}}{{16{{p}^{7}}}}=\frac{{3{{p}^{3}}}}{{4{{p}^{7}}}}$

Now note that the HCF of $ \displaystyle {{{p}^{3}}}$ and $ \displaystyle {{{p}^{7}}}$ is $ \displaystyle {{{p}^{3}}}$. You can divide both the numerator and the denominator by this HCF.

$\displaystyle \frac{{3{{p}^{3}}}}{{4{{p}^{7}}}}=\frac{{3{{p}^{3}}:{{p}^{3}}}}{{4{{p}^{7}}:{{p}^{3}}}}=\frac{1}{{4{{p}^{4}}}}$

d) $ \displaystyle \frac{{{{x}^{2}}-4x+3}}{{{{x}^{2}}-7x+12}}$

Notice that you can factorize both the numerator and the denominator.

$ \displaystyle \frac{{{{x}^{2}}-4x+3}}{{{{x}^{2}}-7x+12}}=\frac{{(x-3)(x-1)}}{{(x-3)(x-4)}}$

You can see that $ \displaystyle {(x-3)}$ is a factor of both the numerator and the denominator, so you can cancel this common factor.

$ \displaystyle \frac{{\cancel{{(x-3)}}(x-1)}}{{\cancel{{(x-3)}}(x-4)}}=\frac{{(x-1)}}{{(x-4)}}$

Multiplying and dividing algebraic fractions

You can use the ides explored in the multiplying and dividing fractions when multiplying or dividing algebraic fractions.

Consider the following multiplication: $ \displaystyle \frac{x}{{{{y}^{2}}}}\times \frac{{{{y}^{2}}}}{{{{x}^{3}}}}$

You know that the denominator and the numerator can be multiplied in the usual way: $ \displaystyle \frac{x}{{{{y}^{2}}}}\times \frac{{{{y}^{2}}}}{{{{x}^{3}}}}=\frac{{x{{y}^{4}}}}{{{{y}^{2}}{{x}^{3}}}}$

Now you can see that the HCF of the numerator and denominator will be $ \displaystyle x{{y}^{2}}$. If you divide through by $ \displaystyle x{{y}^{2}}$ you get: $ \displaystyle \frac{x}{{{{y}^{2}}}}\times \frac{{{{y}^{2}}}}{{{{x}^{3}}}}=\frac{{{{y}^{2}}}}{{{{x}^{2}}}}$

Example 2: Simplify each of the following:

a) $ \displaystyle \frac{4}{{3{{x}^{2}}}}\times \frac{{14{{x}^{3}}}}{{16{{y}^{2}}}}$

You can simply multiply numerators and denominators and then simplify using the methods you already know.

$\displaystyle \frac{4}{{3{{x}^{2}}}}\times \frac{{14{{x}^{3}}}}{{16{{y}^{2}}}}$$\displaystyle =\frac{{4\times 14{{x}^{3}}}}{{3{{x}^{2}}\times 16{{y}^{2}}}}$$\displaystyle =\frac{{56{{x}^{3}}}}{{48{{x}^{2}}{{y}^{2}}}}$$\displaystyle =\frac{{7x}}{{6{{y}^{2}}}}$

b) $ \displaystyle \frac{{3{{{(x+y)}}^{3}}}}{{16{{z}^{2}}}}\times \frac{{12z}}{{9{{{(x+y)}}^{7}}}}$

$\displaystyle \frac{{3{{{(x+y)}}^{3}}}}{{16{{z}^{2}}}}\times \frac{{12z}}{{9{{{(x+y)}}^{7}}}}=$$\displaystyle \frac{{36{{{(x+y)}}^{3}}z}}{{144{{{(x+y)}}^{7}}{{z}^{2}}}}$$ \displaystyle =\frac{1}{{4{{{(x+y)}}^{4}}z}}$

c) $ \displaystyle \frac{{14{{x}^{4}}{{y}^{3}}}}{9}\div \frac{{7{{x}^{2}}y}}{{18}}$

$\displaystyle \frac{{14{{x}^{4}}{{y}^{3}}}}{9}\div \frac{{7{{x}^{2}}y}}{{18}}=$$\displaystyle \frac{{14{{x}^{4}}{{y}^{3}}}}{9}\times \frac{{18}}{{7{{x}^{2}}y}}$$\displaystyle =\frac{{252{{x}^{4}}{{y}^{3}}}}{{63{{x}^{2}}y}}=4{{x}^{2}}{{y}^{2}}$

Adding and subtracting algebraic fractions

You can use common denominators when adding together algebraic fractions, just as you do with ordinary fractions.

Example 3: Write as a single fractions in its lowest terms,$ \displaystyle \frac{1}{x}+\frac{1}{y}$.

The lowest common multiple of x and y is xy. This will be the common denominator.

$ \displaystyle \frac{1}{x}+\frac{1}{y}=\frac{y}{{xy}}+\frac{x}{{xy}}=\frac{{x+y}}{{xy}}$

Example 4: Write as a single fraction in its lowest terms, $ \displaystyle \frac{1}{{x+1}}+\frac{1}{{x+2}}$.

The lowest common multiple of (x+1) and (x+2) is (x+1)(x+2).

$ \displaystyle \frac{1}{{x+1}}+\frac{1}{{x+2}}=$$ \displaystyle \frac{{x+2}}{{(x+1)(x+2)}}+\frac{{x+1}}{{(x+1)(x+2)}}=$$ \displaystyle \frac{{x+2+x+1}}{{(x+1)(x+2)}}=\frac{{2x+3}}{{(x+1)(x+2)}}$

Example 5: Write as a single fraction in its lowest terms,$ \displaystyle \frac{{3x+4}}{{{{x}^{2}}+x-6}}-\frac{1}{{x+3}}$

Firstly you should factorize the quadratic expression:

$ \displaystyle \frac{{3x+4}}{{{{x}^{2}}+x-6}}-\frac{1}{{x+3}}=\frac{{3x+4}}{{(x+3)(x-2)}}-\frac{1}{{x+3}}$

The two denominators have a common factor of x+3, and the lowest common multiple of these two denominators is (x+3)(x-2):

$\displaystyle \frac{{3x+4}}{{{{x}^{2}}+x-6}}-\frac{1}{{x+3}}=$$\displaystyle \frac{{3x+4}}{{(x+3)(x-2)}}-\frac{1}{{x+3}}=$$\displaystyle \frac{{3x+4-(x-2)}}{{(x+3)(x-2)}}=$$\displaystyle \frac{{3x+4-x+2}}{{(x+3)(x-2)}}=$$ \displaystyle \frac{{2x+6}}{{(x+3)(x-2)}}$

This may appear to be the final answer but if you factorize the numerator you will find that more can be done!

$ \displaystyle \frac{{3x+4}}{{{{x}^{2}}+x-6}}-\frac{1}{{x+3}}=\frac{{2x+6}}{{(x+3)(x-2)}}=$$ \displaystyle \frac{{2(x+3)}}{{(x+3)(x-2)}}=\frac{2}{{x-2}}$

Tip! Always check if your final numerator factorizes. If it does, then there may be more stages to go.

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