Functions and algebra: Manipulate and simplify algebraic expressions

# Unit 2: Simplifying algebraic fractions

Dylan Busa

### Unit outcomes

By the end of this unit you will be able to:

• Add algebraic fractions
• Subtract algebraic fractions with monomial, binomial and trinomial denominators.

## What you should know

Before you start this unit, make sure you can:

Here is a short self-assessment to make sure you have the skills you need to proceed with this unit.

Add or subtract the following fractions:

1. $\scriptsize \displaystyle \frac{x}{y}+\displaystyle \frac{2}{x}$
2. $\scriptsize \displaystyle \frac{a}{{2b}}-\displaystyle \frac{{2b}}{{9a}}$

Solutions

1. .
\scriptsize \begin{align*}\displaystyle \frac{x}{y}+\displaystyle \frac{2}{x}&=\displaystyle \frac{{x\cdot x+2\cdot y}}{{xy}}\\&=\displaystyle \frac{{{{x}^{2}}+2y}}{{xy}}\end{align*}
2. .
\scriptsize \begin{align*}\displaystyle \frac{a}{{2b}}-\displaystyle \frac{{2b}}{{9a}}&=\displaystyle \frac{{9\cdot a\cdot a}}{{18ab}}-\displaystyle \frac{{2b\cdot (2b)}}{{18ab}}\\&=\displaystyle \frac{{9{{a}^{2}}-4{{b}^{2}}}}{{18ab}}\end{align*}

## Introduction

We saw in the previous unit that we can make the multiplication (and division) of algebraic fractions easier if we first factorise the numerators and denominators and then ‘cancel’ like factors in the numerator and denominator. Remember that we are not really cancelling anything. We are simply recognising that having the same factor in the numerator and denominator is equivalent to $\scriptsize 1$.

In this unit, we are going to learn how first factorising, especially the denominators, of algebraic fractions allows us to identify the lowest common denominator so that we can add and subtract more complicated algebraic fractions.

## Adding and subtracting algebraic fractions

You know that the only way to add (or subtract) fractions is to make the fractions’ denominators ‘look the same’.

$\scriptsize \displaystyle \frac{1}{2}+\displaystyle \frac{1}{3}-\displaystyle \frac{1}{5}$

The lowest common denominator of all three of these denominators is $\scriptsize 30$. Therefore, we have to change all the denominators to $\scriptsize 30$. This means multiplying the denominator of $\scriptsize 2$ by $\scriptsize 15$, the denominator of $\scriptsize 3$ by $\scriptsize 10$ and the denominator of $\scriptsize 5$ by $\scriptsize 6$.

But if we just changed the denominators without also changing the numerators, we would change the value of each fraction. So, we multiply the whole fraction of $\scriptsize \displaystyle \frac{1}{2}$, for example, by $\scriptsize \displaystyle \frac{{15}}{{15}}$. $\scriptsize \displaystyle \frac{{15}}{{15}}$ is really just another way to write $\scriptsize 1$. Hence, because we are really multiplying $\scriptsize \displaystyle \frac{1}{2}$ by $\scriptsize 1$, we are not changing its value.

\scriptsize \begin{align*}\displaystyle \frac{1}{2}+\displaystyle \frac{1}{3}-\displaystyle \frac{1}{5}&=\left( {\displaystyle \frac{1}{2}\times \displaystyle \frac{{15}}{{15}}} \right)+\left( {\displaystyle \frac{1}{3}\times \displaystyle \frac{{10}}{{10}}} \right)+\left( {\displaystyle \frac{1}{5}\times \displaystyle \frac{6}{6}} \right)\\&=\displaystyle \frac{{15}}{{30}}+\displaystyle \frac{{10}}{{30}}-\displaystyle \frac{6}{{30}}\end{align*}

Now that all the denominators are the same, we can add and subtract the numerators.
\scriptsize \begin{align*}\displaystyle \frac{{15}}{{30}}+\displaystyle \frac{{10}}{{30}}-\displaystyle \frac{6}{{30}}&=\displaystyle \frac{{15+10-6}}{{30}}\\&=\displaystyle \frac{{19}}{{30}}\end{align*}

Factorising the denominators of algebraic fractions is often an essential step in being able to identify the lowest common denominator (LCD).

### Example 2.1

Simplify the following:
$\scriptsize \displaystyle \frac{x}{{x+y}}+\displaystyle \frac{{{{x}^{2}}}}{{{{x}^{2}}-{{y}^{2}}}}$

Solution

The first thing we need to do is factorise the denominators so that we can identify the LCD.
\scriptsize \begin{align*}\displaystyle \frac{x}{{x+y}}+\displaystyle \frac{{{{x}^{2}}}}{{{{x}^{2}}-{{y}^{2}}}}&=\displaystyle \frac{x}{{x+y}}+\displaystyle \frac{{{{x}^{2}}}}{{(x+y)(x-y)}}\end{align*}

Now we need to identify the LCD. In simple terms, each factor in the denominators needs to be represented in the LCD. So, the LCD is $\scriptsize (x+y)(x-y)$.

Next, we need to multiply each fraction by the form of $\scriptsize 1$ that will change its denominator into the LCD. This basically means multiplying the numerator and denominator by the factors of the LCD that are missing from each fraction’s denominator.
\scriptsize \begin{align*}\displaystyle \frac{x}{{x+y}}+\displaystyle \frac{{{{x}^{2}}}}{{(x+y)(x-y)}}&=\left( {\displaystyle \frac{x}{{(x+y)}}\times \displaystyle \frac{{(x-y)}}{{(x-y)}}} \right)+\displaystyle \frac{{{{x}^{2}}}}{{(x+y)(x-y)}}\\&=\displaystyle \frac{{x(x-y)}}{{(x+y)(x-y)}}+\displaystyle \frac{{{{x}^{2}}}}{{(x+y)(x-y)}}\end{align*}

Now we can collect all the fractions over the LCD, add and subtract the numerators, and simplify.
\scriptsize \begin{align*}\displaystyle \frac{{x(x-y)}}{{(x+y)(x-y)}}+\displaystyle \frac{{{{x}^{2}}}}{{(x+y)(x-y)}}&=\displaystyle \frac{{x(x-y)+{{x}^{2}}}}{{(x+y)(x-y)}}\\&=\displaystyle \frac{{{{x}^{2}}-xy+{{x}^{2}}}}{{(x+y)(x-y)}}\\&=\displaystyle \frac{{2{{x}^{2}}-xy}}{{(x+y)(x-y)}}\\&=\displaystyle \frac{{x(2x-y)}}{{(x+y)(x-y)}}\end{align*}

None of the factors in the numerator cancel with the denominator, so this is the expression’s simplest form.

### Exercise 2.1

Simplify the following:

1. $\scriptsize \displaystyle \frac{2}{{xy}}+\displaystyle \frac{4}{{xz}}+\displaystyle \frac{3}{{yz}}$
2. $\scriptsize \displaystyle \frac{4}{{a+1}}-\displaystyle \frac{5}{{a-3}}$
3. $\scriptsize \displaystyle \frac{{x-1}}{{x+1}}-\displaystyle \frac{{2x+3}}{{2x+2}}$
4. $\scriptsize \displaystyle \frac{{4p}}{{p+1}}-\displaystyle \frac{{p+1}}{{4p}}$

The full solutions are at the end of the unit.

This next example is a bit more complicated but the same principles and procedure applies.

### Example 2.2

Simplify the following:
$\scriptsize \displaystyle \frac{x}{{xy+{{y}^{2}}}}+\displaystyle \frac{{2x}}{{{{x}^{2}}-{{y}^{2}}}}+\displaystyle \frac{y}{{{{x}^{2}}-xy}}$

Solution

The first thing we need to do is factorise the denominators so that we can identify the LCD.
\scriptsize \begin{align*}&\displaystyle \frac{x}{{xy+{{y}^{2}}}}+\displaystyle \frac{{2x}}{{{{x}^{2}}-{{y}^{2}}}}+\displaystyle \frac{y}{{{{x}^{2}}-xy}}\\&=\displaystyle \frac{x}{{y(x+y)}}+\displaystyle \frac{{2x}}{{(x+y)(x-y)}}+\displaystyle \frac{y}{{x(x-y)}}\end{align*}

Remember, each factor in the denominators needs to be represented in the LCD. So, the LCD is $\scriptsize xy(x+y)(x-y)$.

Now, we need to multiply each fraction by that form of $\scriptsize 1$ that will match each denominator to the LCD by multiplying the numerator and denominator by all the factors in the LCD that are missing in each fraction’s denominator.
\scriptsize \begin{align*}&\displaystyle \frac{x}{{y(x+y)}}-\displaystyle \frac{{2x}}{{(x+y)(x-y)}}+\displaystyle \frac{y}{{x(x-y)}}\\&=\left( {\displaystyle \frac{x}{{y(x+y)}}\times \displaystyle \frac{{x(x-y)}}{{x(x-y)}}} \right)+\left( {\displaystyle \frac{{2x}}{{(x+y)(x-y)}}\times \displaystyle \frac{{xy}}{{xy}}} \right)+\left( {\displaystyle \frac{y}{{x(x-y)}}\times \displaystyle \frac{{y(x+y)}}{{y(x+y)}}} \right)\\&=\displaystyle \frac{{{{x}^{2}}(x-y)}}{{xy(x+y)(x-y)}}+\displaystyle \frac{{2{{x}^{2}}y}}{{xy(x+y)(x-y)}}+\displaystyle \frac{{{{y}^{2}}(x+y)}}{{xy(x+y)(x-y)}}\end{align*}

Now we can collect all the fractions over the LCD, add and subtract the numerators, and simplify.
\scriptsize \begin{align*}&\displaystyle \frac{{{{x}^{2}}(x-y)}}{{xy(x+y)(x-y)}}+\displaystyle \frac{{2{{x}^{2}}y}}{{xy(x+y)(x-y)}}+\displaystyle \frac{{{{y}^{2}}(x+y)}}{{xy(x+y)(x-y)}}\\&=\displaystyle \frac{{{{x}^{2}}(x-y)+2{{x}^{2}}y+{{y}^{2}}(x+y)}}{{xy(x+y)(x-y)}}\\&=\displaystyle \frac{{{{x}^{3}}-{{x}^{2}}y+2{{x}^{2}}y+x{{y}^{2}}+{{y}^{3}}}}{{xy(x+y)(x-y)}}&& \text{Expand the numerator}\\&=\displaystyle \frac{{{{x}^{3}}+{{x}^{2}}y+x{{y}^{2}}+{{y}^{3}}}}{{xy(x+y)(x-y)}}&& \text{Collect like terms}\\&=\displaystyle \frac{{\left( {{{x}^{3}}+{{x}^{2}}y} \right)+\left( {x{{y}^{2}}+{{y}^{3}}} \right)}}{{xy(x+y)(x-y)}}&& \text{Factorise the numerator}\\ &&&\text{(in this case by grouping)}\\&=\displaystyle \frac{{{{x}^{2}}\left( {x+y} \right)+{{y}^{2}}\left( {x+y} \right)}}{{xy(x+y)(x-y)}}\\&=\displaystyle \frac{{\bcancel{{(x+y)}}({{x}^{2}}+{{y}^{2}})}}{{xy\bcancel{{(x+y)}}(x-y)}}&&\text{ Cancel like factors}\\&=\displaystyle \frac{{({{x}^{2}}+{{y}^{2}})}}{{(x-y)}}\end{align*}

### Exercise 2.2

Simplify the following:

1. $\scriptsize \displaystyle \frac{3}{{{{p}^{2}}-4}}+\displaystyle \frac{2}{{{{{(p-2)}}^{2}}}}$
2. $\scriptsize \displaystyle \frac{2}{{{{{(x-y)}}^{2}}}}-\displaystyle \frac{3}{{{{y}^{2}}-{{x}^{2}}}}$
3. $\scriptsize \displaystyle \frac{2}{{{{t}^{2}}+t-6}}+\displaystyle \frac{2}{{{{t}^{2}}-9}}+\displaystyle \frac{1}{{t-3}}$

The full solutions are at the end of the unit.

## Simplify more complex algebraic expressions

Sometimes you will need to add and subtract algebraic fractions that are themselves in the numerator or denominator of other fractions. When faced with these more complicated questions, it is important to simplify any algebraic fractions in these numerators and denominators first before tackling the overall fraction. Have a look at the next example.

### Example 2.3

Simplify the following:
$\scriptsize \displaystyle \frac{{y+\displaystyle \frac{1}{x}}}{{\displaystyle \frac{y}{x}}}$

Solution

In this case, we need to add the two fractions in the overall numerator first. So, let’s just focus on $\scriptsize y+\displaystyle \frac{1}{x}$:
\scriptsize \begin{align*}&y+\displaystyle \frac{1}{x}\\ & \text{LCD:}\ x\\&=\displaystyle \frac{{xy}}{x}+\displaystyle \frac{1}{x}\\&=\displaystyle \frac{{xy+1}}{x}\end{align*}

Now that we have simplified the numerator of the overall fraction, we can start to simplify the fraction. We know from unit 1 [insert link] that we can deal with a fraction divided by a fraction by multiplying by the reciprocal of the fraction in the denominator.
\scriptsize \begin{align*}\displaystyle \frac{{y+\displaystyle \frac{1}{x}}}{{\displaystyle \frac{y}{x}}}&=\displaystyle \frac{{\displaystyle \frac{{xy+1}}{x}}}{{\displaystyle \frac{y}{x}}}\\&=\displaystyle \frac{{xy+1}}{{\bcancel{x}}}\times \displaystyle \frac{{\bcancel{x}}}{y}&&\text{'Cancel' the factors}\\&=\displaystyle \frac{{(xy+1)}}{y}\end{align*}

### Example 2.4

Simplify the following:
$\scriptsize \displaystyle \frac{{\displaystyle \frac{2}{{x+1}}+\displaystyle \frac{2}{{x-1}}}}{{\displaystyle \frac{{x-1}}{{x+1}}}}$

Solution

We will start by adding the fractions in the numerator.
\scriptsize \begin{align*}&\displaystyle \frac{2}{{x+1}}+\displaystyle \frac{2}{{x-1}}\\ & \text{LCD:}\ (x+1)(x-1)\\&=\displaystyle \frac{{2(x-1)}}{{(x+1)(x-1)}}+\displaystyle \frac{{2(x+1)}}{{(x+1)(x-1)}}\\&=\displaystyle \frac{{2x-2+2x+2}}{{(x+1)(x-1)}}\\&=\displaystyle \frac{{4x}}{{(x+1)(x-1)}}\end{align*}

Now, we can continue simplifying the overall fraction.
\scriptsize \begin{align*}\displaystyle \frac{{\displaystyle \frac{2}{{x+1}}+\displaystyle \frac{2}{{x-1}}}}{{\displaystyle \frac{{x-1}}{{x+1}}}}&=\displaystyle \frac{{\displaystyle \frac{{4x}}{{(x+1)(x-1)}}}}{{\displaystyle \frac{{(x-1)}}{{(x+1)}}}}\\&=\displaystyle \frac{{4x}}{{\bcancel{{(x+1)}}(x-1)}}\times \displaystyle \frac{{\bcancel{{(x+1)}}}}{{(x-1)}}\\&=\displaystyle \frac{{4x}}{{{{{(x-1)}}^{2}}}}\end{align*}

### Exercise 2.3

Simplify the following:

1. $\scriptsize \displaystyle \frac{{\displaystyle \frac{6}{y}-\displaystyle \frac{4}{x}}}{y}$
2. $\scriptsize \displaystyle \frac{{\displaystyle \frac{3}{a}+\displaystyle \frac{b}{6}}}{{\displaystyle \frac{{2b}}{{3a}}}}$
3. $\scriptsize \displaystyle \frac{{\displaystyle \frac{a}{b}-\displaystyle \frac{b}{a}}}{{\displaystyle \frac{{a+b}}{{ab}}}}$
4. $\scriptsize \displaystyle \frac{{\displaystyle \frac{c}{{c+2}}+\displaystyle \frac{{c-1}}{{c+1}}}}{{\displaystyle \frac{{2c-1}}{{c+1}}}}$

The full solutions are at the end of the unit.

## Summary

In this unit you have learnt the following:

• How to add and subtract algebraic fractions by factorising the denominators in order to find the LCD.
• How to simplify more complex algebraic fractions with other algebraic fractions in the numerator and/or denominator.

# Unit 2: Assessment

#### Suggested time to complete: 40 minutes

1. Simplify the following:
1. $\scriptsize \displaystyle \frac{x}{{x+1}}-\displaystyle \frac{y}{{y+1}}$
2. $\scriptsize \displaystyle \frac{{3z}}{{z+1}}-\displaystyle \frac{{2z+5}}{{z-2}}$
3. $\scriptsize \displaystyle \frac{{4x}}{{4{{x}^{2}}-25}}+\displaystyle \frac{{3x}}{{2{{x}^{2}}+11x+15}}$
4. $\scriptsize \displaystyle \frac{8}{{{{a}^{2}}-1}}-\displaystyle \frac{6}{{{{a}^{2}}+2a+1}}+\displaystyle \frac{3}{{{{a}^{2}}-2a+1}}$
2. Simplify the following:
1. $\scriptsize \displaystyle \frac{{\displaystyle \frac{{2x}}{3}+\displaystyle \frac{{4x}}{7}}}{{\displaystyle \frac{x}{2}}}$
2. $\scriptsize \displaystyle \frac{{\displaystyle \frac{x}{y}-\displaystyle \frac{y}{x}}}{{\displaystyle \frac{x}{y}+\displaystyle \frac{y}{x}}}$
3. $\scriptsize \displaystyle \frac{{\displaystyle \frac{{4a+1}}{{2a-3}}-\displaystyle \frac{{2a-3}}{{2a+3}}}}{{\displaystyle \frac{a}{{4{{a}^{2}}-9}}}}$

The full solutions are at the end of the unit.

# Unit 2: Solutions

### Exercise 2.1

1. .
\scriptsize \begin{align*}&\displaystyle \frac{2}{{xy}}+\displaystyle \frac{4}{{xz}}+\displaystyle \frac{3}{{yz}}\\ & \text{LCD:}\ xyz\\&=\displaystyle \frac{{2z}}{{xyz}}+\displaystyle \frac{{4y}}{{xyz}}+\displaystyle \frac{{3x}}{{xyz}}\\&=\displaystyle \frac{{2z+4y+3x}}{{xyz}}\end{align*}
2. .
\scriptsize \begin{align*}&\displaystyle \frac{4}{{a+1}}-\displaystyle \frac{5}{{a-3}}\\ & \text{LCD:}\ (a+1)(a-3)\\&=\displaystyle \frac{{4(a-3)}}{{(a+1)(a-3)}}-\displaystyle \frac{{5(a+1)}}{{(a+1)(a-3)}}\\&=\displaystyle \frac{{4(a-3)-5(a+1)}}{{(a+1)(a-3)}}\\&=\displaystyle \frac{{4a-12-5a-5}}{{(a+1)(a-3)}}\\&=\displaystyle \frac{{-a-17}}{{(a+1)(a-3)}}\\&=\displaystyle \frac{{-(a+17)}}{{(a+1)(a-3)}}\end{align*}
3. .
\scriptsize \begin{align*}&\displaystyle \frac{{x-1}}{{x+1}}-\displaystyle \frac{{2x+3}}{{2x+2}}\\&=\displaystyle \frac{{x-1}}{{x+1}}-\displaystyle \frac{{2x+3}}{{2(x+1)}}\\ & \text{LCD:}\ 2(x+1)\\&=\displaystyle \frac{{2(x-1)}}{{2(x+1)}}-\displaystyle \frac{{(2x+3)}}{{2(x+1)}}\\&=\displaystyle \frac{{2(x-1)-(2x+3)}}{{2(x+1)}} & & \text{Write 2nd numerator in brackets so you don }\!\!'\!\!\text{ t}\\ &&& \text{make an error with the }-\text{sign}\\ &=\displaystyle \frac{{2x-1-2x-3}}{{2(x+1)}}\\&=\displaystyle \frac{{-4}}{{2(x+1)}}\\&=\displaystyle \frac{{-2}}{{(x+1)}}\end{align*}
4. .
\scriptsize \begin{align*}&\displaystyle \frac{{4p}}{{p+1}}-\displaystyle \frac{{p+1}}{{4p}}\\ & \text{LCD:}\ 4p(p+1)\\& =\displaystyle \frac{{4p(4p)}}{{4p(p+1)}}-\displaystyle \frac{{(p+1)(p+1)}}{{4p(p+1)}}\\&=\displaystyle \frac{{16{{p}^{2}}-({{p}^{2}}+2p+1)}}{{4p(p+1)}}\\&=\displaystyle \frac{{16{{p}^{2}}-{{p}^{2}}-2p-1}}{{4p(p+1)}}\\&=\displaystyle \frac{{15{{p}^{2}}-2p-1}}{{4p(p+1)}}\\&=\displaystyle \frac{{(5p+1)(3p-1)}}{{4p(p+1)}}\end{align*}

Back to Exercise 2.1

### Exercise 2.2

1. .
\scriptsize \begin{align*}&\displaystyle \frac{3}{{{{p}^{2}}-4}}+\displaystyle \frac{2}{{{{{(p-2)}}^{2}}}}\\&=\displaystyle \frac{3}{{(p+2)(p-2)}}+\displaystyle \frac{2}{{{{{(p-2)}}^{2}}}}\\ & \text{LCD:}\ (p+2){{(p-2)}^{2}}\\&=\displaystyle \frac{{3(p-2)}}{{(p+2){{{(p-2)}}^{2}}}}+\displaystyle \frac{{2(p+2)}}{{(p+2){{{(p-2)}}^{2}}}}\\&=\displaystyle \frac{{3(p-2)+2(p+2)}}{{(p+2){{{(p-2)}}^{2}}}}\\&=\displaystyle \frac{{3p-6+2p+4}}{{(p+2){{{(p-2)}}^{2}}}}\\&=\displaystyle \frac{{5p-2}}{{(p+2){{{(p-2)}}^{2}}}}\end{align*}
2. .
\scriptsize \begin{align*}&\displaystyle \frac{2}{{{{{(x-y)}}^{2}}}}-\displaystyle \frac{3}{{{{y}^{2}}-{{x}^{2}}}}\\&=\displaystyle \frac{2}{{{{{(x-y)}}^{2}}}}-\displaystyle \frac{3}{{(y+x)(y-x)}}\\&=\displaystyle \frac{2}{{{{{(x-y)}}^{2}}}}-\displaystyle \frac{3}{{(x+y)\left( {-(-y+x)} \right)}}\quad \text{Change }(y-x)\text{ into }(x-y)\text{ by taking out a factor of }-1\\&=\displaystyle \frac{2}{{{{{(x-y)}}^{2}}}}+\displaystyle \frac{3}{{(x+y)(x-y)}}\\ & \text{LCD:}\ (x+y){{(x-y)}^{2}}\\&=\displaystyle \frac{{2(x+y)}}{{{{{(x-y)}}^{2}}(x+y)}}+\displaystyle \frac{{3(x-y)}}{{(x+y){{{(x-y)}}^{2}}}}\\&=\displaystyle \frac{{2(x+y)+3(x-y)}}{{{{{(x-y)}}^{2}}(x+y)}}\\&=\displaystyle \frac{{2x+2y+3x-3y}}{{{{{(x-y)}}^{2}}(x+y)}}\\&=\displaystyle \frac{{5x-y}}{{{{{(x-y)}}^{2}}(x+y)}}\end{align*}
3. .
\scriptsize \begin{align*}&\displaystyle \frac{2}{{{{t}^{2}}+t-6}}+\displaystyle \frac{2}{{{{t}^{2}}-9}}+\displaystyle \frac{1}{{t-3}}\\&=\displaystyle \frac{3}{{(t-2)(t+3)}}+\displaystyle \frac{2}{{(t+3)(t-3)}}+\displaystyle \frac{1}{{(t-3)}}\\ & \text{LCD:}\ (t-2)(t+3)(t-3)\\&=\displaystyle \frac{{2(t-3)}}{{(t-2)(t+3)(t-3)}}+\displaystyle \frac{{2(t-2)}}{{(t-2)(t+3)(t-3)}}+\displaystyle \frac{{1(t-2)(t+3)}}{{(t-2)(t+3)(t-3)}}\\&=\displaystyle \frac{{2t-6+2t-4+({{t}^{2}}+t-6)}}{{(t-2)(t+3)(t-3)}}\\&=\displaystyle \frac{{2t-6+2t-4+{{t}^{2}}+t-6}}{{(t-2)(t+3)(t-3)}}\\&=\displaystyle \frac{{{{t}^{2}}+5t-16}}{{(t-2)(t+3)(t-3)}}\quad \text{Quadratic cannot be factorised with current techniques}\end{align*}

Back to Exercise 2.2

### Exercise 2.3

1. .
\scriptsize \begin{align*}&\displaystyle \frac{6}{y}-\displaystyle \frac{4}{x}\\ & \text{LCD:}\ xy\\&=\displaystyle \frac{{6x}}{{xy}}-\displaystyle \frac{{4y}}{{xy}}\\&=\displaystyle \frac{{6x-4y}}{{xy}}\\&=\displaystyle \frac{{2(3x-2y)}}{{xy}}\end{align*}
Therefore:
\scriptsize \begin{align*}\displaystyle \frac{{\displaystyle \frac{6}{y}-\displaystyle \frac{4}{x}}}{y}&=\displaystyle \frac{{\displaystyle \frac{{2(3x-2y)}}{{xy}}}}{y}\\&=\displaystyle \frac{{2(3x-2y)}}{{xy}}\times \displaystyle \frac{1}{y}\\&=\displaystyle \frac{{2(3x-2y)}}{{x{{y}^{2}}}}\end{align*}
2. .
\scriptsize \begin{align*}&\displaystyle \frac{3}{a}+\displaystyle \frac{b}{6}\\ & \text{LCD:}\ 6a\\&=\displaystyle \frac{{18+ab}}{{ab}}\end{align*}
Therefore:
\scriptsize \begin{align*}\displaystyle \frac{{\displaystyle \frac{3}{a}+\displaystyle \frac{b}{6}}}{{\displaystyle \frac{{2b}}{{3a}}}}&=\displaystyle \frac{{\displaystyle \frac{{18+ab}}{{ab}}}}{{\displaystyle \frac{{2a}}{{3b}}}}\\&=\displaystyle \frac{{18+ab}}{{a\bcancel{b}}}\times \displaystyle \frac{{3\bcancel{b}}}{{2a}}\\&=\displaystyle \frac{{3(18+ab)}}{{2{{a}^{2}}}}\end{align*}
3. .
\scriptsize \begin{align*}&\displaystyle \frac{a}{b}-\displaystyle \frac{b}{a}\\ & \text{LCD:}\ ab\\ &=\displaystyle \frac{{{{a}^{2}}}}{{ab}}-\displaystyle \frac{{{{b}^{2}}}}{{ab}}\\&=\displaystyle \frac{{(a+b)(a-b)}}{{ab}}\end{align*}
Therefore:
\scriptsize \begin{align*}\displaystyle \frac{{\displaystyle \frac{a}{b}-\displaystyle \frac{b}{a}}}{{\displaystyle \frac{{a+b}}{{ab}}}}&=\displaystyle \frac{{\displaystyle \frac{{(a+b)(a-b)}}{{ab}}}}{{\displaystyle \frac{{a+b}}{{ab}}}}\\&=\displaystyle \frac{{\bcancel{{(a+b)}}(a-b)}}{{\bcancel{{ab}}}}\times \displaystyle \frac{{\bcancel{{ab}}}}{{\bcancel{{(a+b)}}}}\\&=a-b\end{align*}
4. .
\scriptsize \begin{align*}&\displaystyle \frac{c}{{c+2}}-\displaystyle \frac{{c-1}}{{c+1}}\\ & \text{LCD:}\ (c+2)(c+1)\\&=\displaystyle \frac{{c(c+1)}}{{(c+2)(c+1)}}-\displaystyle \frac{{(c-1)(c+2)}}{{(c+2)(c+1)}}\\&=\displaystyle \frac{{{{c}^{2}}+c-{{c}^{2}}-c+2}}{{(c+2)(c+1)}}\\&=\displaystyle \frac{2}{{(c+2)(c+1)}}\end{align*}
Therefore:
\scriptsize \begin{align*}\displaystyle \frac{{\displaystyle \frac{c}{{c+2}}+\displaystyle \frac{{c-1}}{{c+1}}}}{{\displaystyle \frac{{2c-1}}{{c+1}}}}&=\displaystyle \frac{{\displaystyle \frac{2}{{(c+2)(c+1)}}}}{{\displaystyle \frac{{2c-1}}{{c+1}}}}\\&=\displaystyle \frac{2}{{(c+2)\bcancel{{(c+1)}}}}\times \displaystyle \frac{{\bcancel{{(c+1)}}}}{{(2c-1)}}\\&=\displaystyle \frac{2}{{(c+2)(2c-1)}}\end{align*}

Back to Exercise 2.3

### Unit 2: Assessment

1. .
1. .
\scriptsize \begin{align*}&\displaystyle \frac{x}{{x+1}}-\displaystyle \frac{y}{{y+1}}\\ & \text{LCD:}\ (x+1)(y+1)\\&=\displaystyle \frac{{x(y+1)}}{{(x+1)(y+1)}}-\displaystyle \frac{{y(x+1)}}{{(x+1)(y+1)}}\\&=\displaystyle \frac{{xy+x-xy-y}}{{(x+1)(y+1)}}\\&=\displaystyle \frac{{x-y}}{{(x+1)(y+1)}}\end{align*}
2. .
\scriptsize \begin{align*}&\displaystyle \frac{{3z}}{{z+1}}-\displaystyle \frac{{2z+5}}{{z-2}}\\&\text{LCD:}\ (z+1)(z-2)\\&=\displaystyle \frac{{3z(z-2)}}{{(z+1)(z-2)}}-\displaystyle \frac{{(2z+5)(z+1)}}{{(z+1)(z-2)}}\\&=\displaystyle \frac{{3{{z}^{2}}-6z-(2{{z}^{2}}+7z+5)}}{{(z+1)(z-2)}}\\&=\displaystyle \frac{{3{{z}^{2}}-6z-2{{z}^{2}}-7z-5}}{{(z+1)(z-2)}}\\&=\displaystyle \frac{{{{z}^{2}}-13z-5}}{{(z+1)(z-2)}}\end{align*}
3. .
\scriptsize \begin{align*}&\displaystyle \frac{{4x}}{{4{{x}^{2}}-25}}+\displaystyle \frac{{3x}}{{2{{x}^{2}}+11x+15}}\\&=\displaystyle \frac{{4x}}{{(2x+5)(2x-5)}}+\displaystyle \frac{{3x}}{{(2x+5)(x+3)}}\\ & \text{LCD:}\ (2x+5)(2x-5)(x+3)\\&=\displaystyle \frac{{4x(x+3)}}{{(2x+5)(2x-5)(x+3)}}+\displaystyle \frac{{3x(2x-5)}}{{(2x+5)(2x-5)(x+3)}}\\&=\displaystyle \frac{{4{{x}^{2}}+12x+6{{x}^{2}}-15x}}{{(2x+5)(2x-5)(x+3)}}\\&=\displaystyle \frac{{10{{x}^{2}}-3x}}{{(2x+5)(2x-5)(x+3)}}\\&=\displaystyle \frac{{x(10x-3)}}{{(2x+5)(2x-5)(x+3)}}\end{align*}
4. .
\scriptsize \begin{align*}&\displaystyle \frac{8}{{{{a}^{2}}-1}}-\displaystyle \frac{6}{{{{a}^{2}}+2a+1}}+\displaystyle \frac{3}{{{{a}^{2}}-2a+1}}\\&=\displaystyle \frac{8}{{(a+1)(a-1)}}-\displaystyle \frac{6}{{(a+1)(a+1)}}+\displaystyle \frac{3}{{(a-1)(a-1)}}\\ & \text{LCD:}\ {{(a+1)}^{2}}{{(a-1)}^{2}}\\&=\displaystyle \frac{{8(a+1)(a-1)}}{{{{{(a+1)}}^{2}}{{{(a-1)}}^{2}}}}-\displaystyle \frac{{6{{{(a-1)}}^{2}}}}{{{{{(a+1)}}^{2}}{{{(a-1)}}^{2}}}}+\displaystyle \frac{{3{{{(a+1)}}^{2}}}}{{{{{(a+1)}}^{2}}{{{(a-1)}}^{2}}}}\\&=\displaystyle \frac{{8({{a}^{2}}-1)-6({{a}^{2}}-2a+1)+3({{a}^{2}}+2a+1)}}{{{{{(a+1)}}^{2}}{{{(a-1)}}^{2}}}}\\&=\displaystyle \frac{{8{{a}^{2}}-8-6{{a}^{2}}+12a-6+3{{a}^{2}}+6a+3}}{{{{{(a+1)}}^{2}}{{{(a-1)}}^{2}}}}\\&=\displaystyle \frac{{5{{a}^{2}}+18a-11}}{{{{{(a+1)}}^{2}}{{{(a-1)}}^{2}}}}\end{align*}
2. .
1. .
\scriptsize \begin{align*}&\displaystyle \frac{{2x}}{3}+\displaystyle \frac{{4x}}{7}\\ &\text{LCD:}\ 21\\&=\displaystyle \frac{{14x}}{{21}}+\displaystyle \frac{{12x}}{{21}}\\&=\displaystyle \frac{{12x}}{{21}}\end{align*}
Therefore:
\scriptsize \begin{align*}\displaystyle \frac{{\displaystyle \frac{{2x}}{3}+\displaystyle \frac{{4x}}{7}}}{{\displaystyle \frac{x}{2}}}&=\displaystyle \frac{{\displaystyle \frac{{12x}}{{21}}}}{{\displaystyle \frac{x}{2}}}\\&=\displaystyle \frac{{12\bcancel{x}}}{{21}}\times \displaystyle \frac{2}{{\bcancel{x}}}\\&=\displaystyle \frac{{24}}{{21}}\\&=\displaystyle \frac{8}{7}\end{align*}
2. .
\scriptsize \begin{align*}&\displaystyle \frac{x}{y}-\displaystyle \frac{y}{x}\\ & \text{LCD:}\ xy\\&=\displaystyle \frac{{{{x}^{2}}}}{{xy}}-\displaystyle \frac{{{{y}^{2}}}}{{xy}}\\&=\displaystyle \frac{{(x+y)(x-y)}}{{xy}}\end{align*}
\scriptsize \begin{align*}&\displaystyle \frac{x}{y}+\displaystyle \frac{y}{x}\\ & \text{LCD:}\ xy\\&=\displaystyle \frac{{{{x}^{2}}}}{{xy}}+\displaystyle \frac{{{{y}^{2}}}}{{xy}}\\&=\displaystyle \frac{{{{x}^{2}}+{{y}^{2}}}}{{xy}}\end{align*}
Therefore:
\scriptsize \begin{align*}\displaystyle \frac{{\displaystyle \frac{x}{y}-\displaystyle \frac{y}{x}}}{{\displaystyle \frac{x}{y}+\displaystyle \frac{y}{x}}}&=\displaystyle \frac{{\displaystyle \frac{{(x+y)(x-y)}}{{xy}}}}{{\displaystyle \frac{{{{x}^{2}}+{{y}^{2}}}}{{xy}}}}\\&=\displaystyle \frac{{(x+y)(x-y)}}{{\bcancel{{xy}}}}\times \displaystyle \frac{{\bcancel{{xy}}}}{{{{x}^{2}}+{{y}^{2}}}}\\&=\displaystyle \frac{{(x+y)(x-y)}}{{{{x}^{2}}+{{y}^{2}}}}\end{align*}
3. .
\scriptsize \begin{align*}&\displaystyle \frac{{4a+1}}{{2a-3}}-\displaystyle \frac{{2a-3}}{{2a+3}}\\ & \text{LCD:}\ (2a-3)(2a+3)\\&=\displaystyle \frac{{(4a+1)(2a+3)}}{{(2a-3)(2a+3)}}-\displaystyle \frac{{(2a-3)(2a-3)}}{{(2a-3)(2a+3)}}\\&=\displaystyle \frac{{8{{a}^{2}}+14a+3-4{{a}^{2}}+12a-9}}{{(2a-3)(2a+3)}}\\&=\displaystyle \frac{{4{{a}^{2}}+26a-6}}{{(2a-3)(2a+3)}}\\&=\displaystyle \frac{{2(2{{a}^{2}}+13a-3)}}{{(2a-3)(2a+3)}}\end{align*}
Therefore:
\scriptsize \begin{align*}\displaystyle \frac{{\displaystyle \frac{{4a+1}}{{2a-3}}+\displaystyle \frac{{2a-3}}{{2a+3}}}}{{\displaystyle \frac{a}{{4{{a}^{2}}-9}}}}&=\displaystyle \frac{{\displaystyle \frac{{2(2{{a}^{2}}+13a-3)}}{{(2a-3)(2a+3)}}}}{{\displaystyle \frac{a}{{4{{a}^{2}}-9}}}}\\&=\displaystyle \frac{{2(2{{a}^{2}}+13a-3)}}{{(2a-3)(2a+3)}}\times \displaystyle \frac{{4{{a}^{2}}-9}}{a}\\&=\displaystyle \frac{{2(2{{a}^{2}}+13a-3)}}{{\bcancel{{(2a-3)}}\bcancel{{(2a+3)}}}}\times \displaystyle \frac{{\bcancel{{(2a+3)}}\bcancel{{(2a-3)}}}}{a}\\&=\displaystyle \frac{{2(2{{a}^{2}}+13a-3)}}{a}\end{align*}

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