Functions and algebra: Solve algebraic equations and inequalities

# Unit 1: Solve linear and quadratic equations

Dylan Busa

### Unit outcomes

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

• Solve equations with a single variable which are called linear equations
• Solve equations with a single variable that is squared (quadratic equations) by factorisation.

## 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.

1. Simplify $\scriptsize \left( {\displaystyle \frac{x}{3}-\displaystyle \frac{3}{x}} \right)\left( {\displaystyle \frac{x}{4}+\displaystyle \frac{4}{x}} \right)$
2. Factorise $\scriptsize 6{{x}^{2}}+xy-12{{y}^{2}}$
3. Simplify $\scriptsize \displaystyle \frac{a}{{2b}}-\displaystyle \frac{{2b}}{{9a}}$

Solutions

1. .
\scriptsize \begin{align*} \left( {\displaystyle \frac{x}{3}-\displaystyle \frac{3}{x}} \right)\left( {\displaystyle \frac{x}{4}+\displaystyle \frac{4}{x}} \right) &=\displaystyle \frac{x}{3}\times \displaystyle \frac{x}{4}+\displaystyle \frac{x}{3}\times \displaystyle \frac{4}{x}-\displaystyle \frac{3}{x}\times \displaystyle \frac{x}{4}-\displaystyle \frac{3}{x}\times \displaystyle \frac{4}{x}\\ &=\displaystyle \frac{{{{x}^{2}}}}{{12}}+\displaystyle \frac{{4x}}{{3x}}-\displaystyle \frac{{3x}}{{4x}}-\displaystyle \frac{{12}}{{{{x}^{2}}}}\\ &=\displaystyle \frac{{{{x}^{2}}}}{{12}}+\displaystyle \frac{4}{3}-\displaystyle \frac{3}{4}-\displaystyle \frac{{12}}{{{{x}^{2}}}}\\ &=\displaystyle \frac{{{{x}^{2}}}}{{12}}+\displaystyle \frac{7}{{12}}-\displaystyle \frac{{12}}{{{{x}^{2}}}}\end{align*}
2. $\scriptsize 6{{x}^{2}}+xy-12{{y}^{2}}=(2x+3y)(3x-4y)$
3. .
\scriptsize \begin{align*}\displaystyle \frac{a}{2b}-\displaystyle \frac{2b}{9a} &=\displaystyle \frac{9a\times a}{9a \times 2b}-\displaystyle \frac{2b\times 2b}{2b\times 9a}\\ &=\displaystyle \frac{9a^2}{18ab}-\displaystyle \frac{4b^2}{18ab}\\ &=\displaystyle \frac{9a^2-4b^2}{18ab}\\ &=\displaystyle \frac{(3a+2b)(3a-2b)}{18ab} \end{align*}

## Introduction

Equations are at the heart of almost all Mathematics. You might even say that Mathematics is all about solving equations. We use equations to find the value (or values) of some unknown quantity or quantities.

Some equations are simple to solve like $\scriptsize x+1=3$. Some equations are famously complicated and difficult to solve like this one from the theory of General Relativity $\scriptsize {{G}_{{\mu v}}}=\displaystyle \frac{{8\pi G}}{{{{c}^{4}}}}{{T}_{{\mu v}}}$. Some equations seem to have taken on a life of their own like $\scriptsize E=m{{c}^{2}}$.

You may already be familiar with how to solve equations, but just in case you are not, this unit is going to start from the very beginning by asking, ‘what is an equation?’.

### Solving simple linear equations

Before we start solving equations, let’s pause and think for a moment about what an equation is. What does the word ‘equal’ mean? What does it mean if we put two expressions on either side of an equal sign?

In an equation, two mathematical expressions are equal to each other. In this way, we are told that the value of whatever is on the left-hand side of the equal sign is equal in value to whatever is on the right-hand side of the equal sign.

We can think of the left-hand side and the right-hand side of the equation as being balanced (see Figure 1).

Now, if we have the expression $\scriptsize 2x+3$ on the left-hand side and the expression $\scriptsize 5$ on the right-hand side, the equation would be $\scriptsize 2x+3=5$ and the equation scale would look like Figure 2. This is called a linear equation because the exponent on the $\scriptsize x$ is $\scriptsize 1$.

There is one basic rule that always applies to all equations. We must keep the scales balanced. We can add, subtract, divide and multiply whatever we like to one side of the equation so long as we perform the exact same operation to the other side of the equation.

Now, how would we solve for $\scriptsize x$ in $\scriptsize 2x+3=5$?

In this case, we need to get the variable, $\scriptsize x$, all alone on the left-hand side (LHS) of the equation. The first thing we can do is to subtract $\scriptsize 3$ from the LHS. But whatever we do to the LHS we have to do to the right-hand side (RHS).

So, our equation becomes $\scriptsize 2x+3-3=5-3$. We can simplify both sides of the equation to get $\scriptsize 2x=2$.

Now, we have $\scriptsize x$ multiplied by $\scriptsize 2$ on the LHS so we can divide the LHS by $\scriptsize 2$ to get $\scriptsize x$ on its own. But whatever we do to the LHS, we have to do to the RHS, so our equation becomes $\scriptsize \displaystyle \frac{{2x}}{2}=\displaystyle \frac{2}{2}$ which we can simplify to get $\scriptsize x=1$. And we have solved for $\scriptsize x$.

Generally, this is how you should set all your equations out. You don’t always have to get the variables onto the left-hand side though. You can also get them onto the right-hand side.
\scriptsize \begin{align*}2x+3 & =5\\\therefore 2x+3-3 & =5-3\\\therefore 2x & =2\\\therefore \displaystyle \frac{{2x}}{2} & =\displaystyle \frac{2}{2}\\\therefore x & =1\end{align*}

The $\scriptsize \therefore$ symbol means ‘therefore’. Can you see how we have just constructed a logical Mathematical argument? As you get better at solving equations, you will be able to leave out some of these steps.

### Example 1.1

Solve for $\scriptsize x$:

1. $\scriptsize 4x+5=17$
2. $\scriptsize 3(x+2)=-4(x+9)$

Solutions

1. .
\scriptsize \begin{align*}4x+5 & =17\\\therefore 4x+5-5 & =17-5\quad & \text{We subtract 5 from both sides}\\\therefore 4x & =12\\\therefore \displaystyle \frac{{4x}}{4} & =\displaystyle \frac{{12}}{4}\quad & \text{We divide both sides by 4}\\\therefore x & =3\end{align*}
.
You should always check that you have solved the equation correctly by substituting your answer into the original equation to see if the LHS is equal to the RHS.

\scriptsize \begin{align*}\text{LHS: }4(3)+5=12+5=17\\\text{RHS: }17\end{align*}

The LHS = the RHS. Therefore, the solution is correct.

2. .
\scriptsize \begin{align*}3(x+2) & =-4(x+9)\quad \text{First multiply out the brackets}\\\therefore 3x+6 & =-4x-36\\\therefore 3x+4x+6 & =-4x+4x-36\quad \text{Now we add }4x\text{ to both sides}\\\therefore 7x+6 & =-36\\\therefore 7x+6-6 & =-36-6\quad \text{Now we subtract 6 from both sides}\\\therefore 7x & =-42\\\therefore \displaystyle \frac{{7x}}{7} & =\displaystyle \frac{{-42}}{7}\quad \text{Divide both sides by 7}\\\therefore x & =-6\end{align*}
.
Check that the solution is correct.

\scriptsize \begin{align*}\text{LHS: }3(-6+2)=3(-4)=-12\\\text{RHS: }-4(-6+9)=-4(3)=-12\end{align*}

The LHS = the RHS, so the solution is correct.

Remember, both of these equations we have just solved are called linear equations because the unknown, $\scriptsize x$, has an exponent of $\scriptsize 1$.

### Note

When solving equations, whatever you do to the one side of the equation, you must also do to the other side of the equation. This keeps the equation balanced and the equality between the LHS and the RHS.

### Exercise 1.1

Solve for the unknown in the following equations.

1. $\scriptsize 2x-6=8$
2. $\scriptsize 4-2y=6$
3. $\scriptsize -5x-4=11+2x+3$
4. $\scriptsize 3(x+7)=-2(x-7)$
5. $\scriptsize 7a-(3a+10)=6-2(12-9a)$

The full solutions are at the end of the unit.

## Solving linear equations with fractions

All the equations we have solved so far have had no fractions in the original equation. When there is a fraction, we need to ‘undo’ it by multiplying both sides of the equation by the lowest common denominator so that we can ‘cancel the denominator’ or create a denominator of $\scriptsize 1$. Look at the next example to see what we mean.

### Example 1.2

Solve for $\scriptsize x$:

1. $\scriptsize \displaystyle \frac{{x+2}}{3}=5$
2. $\scriptsize \displaystyle \frac{{4x-2}}{3}=\displaystyle \frac{{3x-1}}{2}$
3. $\scriptsize \displaystyle \frac{{5x}}{3}-2=\displaystyle \frac{{3x}}{5}$

Solutions

1. We need to multiply both sides of the equation by $\scriptsize 3$ so that we can create a denominator of $\scriptsize 1$ on the LHS.
.
\scriptsize \begin{align*}\displaystyle \frac{{x+2}}{3} & =5\\\therefore 3\times \left( {\displaystyle \frac{{x+2}}{3}} \right) & =3\times 5\quad \text{Multiply both sides by 3}\\\therefore x+2 & =15\\\therefore x & =15-2\quad \text{Subtract 2 from both sides}\\\therefore x & =13\end{align*}
.
Check the solution:

\scriptsize \begin{align*}\text{LHS: }\displaystyle \frac{{13+2}}{3}=\displaystyle \frac{{15}}{3}=5\\\text{RHS: }5\end{align*}

The LHS = the RHS, so the solution is correct.

2. In this equation, we have two denominators. The best way to deal with both denominators at the same time, is to multiply both sides of the equation by the lowest common denominator (LCD). In this case, the lowest common denominator is $\scriptsize 6$.
.
\scriptsize \begin{align*}\displaystyle \frac{{4x-2}}{3} & =\displaystyle \frac{{3x-1}}{2}\\\therefore 6\times \left( {\displaystyle \frac{{4x-2}}{3}} \right) & =6\times \left( {\displaystyle \frac{{3x-1}}{2}} \right)\quad \text{Multiply both sides by the LCD}\\\therefore 2(4x-2) & =3(3x-1)\quad \text{Simplify both sides: }\displaystyle \frac{6}{3}=2\text{ and }\displaystyle \frac{6}{2}=3\\\therefore 8x-4 & =9x-3\quad \text{Multiply out the brackets}\\\therefore 8x-9x-4 & =-3\quad \text{Subtract }9x\text{ from both sides}\\\therefore -x & =-3+4\quad \text{Add 4 to both sides}\\\therefore -x & =1\quad \text{Multiply both sides by }-1\\\therefore x & =-1\end{align*}
.
Check the solution:

\scriptsize \begin{align*}\text{LHS: }\displaystyle \frac{{4\times (-1)-2}}{3}=\displaystyle \frac{{-6}}{3}=-2\\\text{RHS: }\displaystyle \frac{{3\times (-1)-1}}{2}=\displaystyle \frac{{-4}}{2}=-2\end{align*}

The LHS = the RHS, so the solution is correct.

3. In this question, the LCD is $\scriptsize 15$ so we need to multiply both sides of the equation by $\scriptsize 15$.
.
\scriptsize \begin{align*}\displaystyle \frac{{5x}}{3}-2 & =\displaystyle \frac{{3x}}{5}\\\therefore 15\times \left( {\displaystyle \frac{{5x}}{3}-2} \right) & =15\times \left( {\displaystyle \frac{{3x}}{5}} \right)\quad \text{Multiply both sides by the LCD of 15}\\\therefore 15\times \left( {\displaystyle \frac{{5x}}{3}} \right)-30 & =15\times \left( {\displaystyle \frac{{3x}}{5}} \right)\quad \text{Simplify}\\\therefore 5(5x)-30 & =3(3x)\\\therefore 25x-30 & =9x\\\therefore 25x-9x-30 & =0\quad \text{Subtract }9x\text{ from both sides}\\\therefore 16x & =30\quad \text{Add 30 to both sides}\\\therefore \displaystyle \frac{{16x}}{{16}} & =\displaystyle \frac{{30}}{{16}}\quad \text{Divide both sides by 16}\\\therefore x & =\displaystyle \frac{{30}}{{16}}=\displaystyle \frac{{15}}{8}\end{align*}
.
Check the solution:

\scriptsize \begin{align*}\text{LHS: }\displaystyle \frac{{5\times \displaystyle \frac{{15}}{8}}}{3}-2=\displaystyle \frac{{\displaystyle \frac{{75}}{8}}}{3}-2=\left( {\displaystyle \frac{{75}}{8}\times \displaystyle \frac{1}{3}} \right)-2=\displaystyle \frac{{75}}{{24}}-\displaystyle \frac{{48}}{{24}}=\displaystyle \frac{{27}}{{24}}=\displaystyle \frac{9}{8}\\\text{RHS: }\displaystyle \frac{{3\times \displaystyle \frac{{15}}{8}}}{5}=\displaystyle \frac{{\displaystyle \frac{{45}}{8}}}{5}=\displaystyle \frac{{45}}{8}\times \displaystyle \frac{1}{5}=\displaystyle \frac{{45}}{{40}}=\displaystyle \frac{9}{8}\end{align*}

The LHS = the RHS, so the solution is correct.

### Exercise 1.2

Solve for the unknown in each of the following equations.

1. $\scriptsize \displaystyle \frac{{x+2}}{4}-\displaystyle \frac{{x-6}}{3}=12$
2. $\scriptsize \displaystyle \frac{{2-5a}}{3}-6=\displaystyle \frac{{4a}}{3}+2-a$
3. $\scriptsize 3-\displaystyle \frac{{y-2}}{4}=4$
4. $\scriptsize \displaystyle \frac{1}{5}(x-1)=\displaystyle \frac{1}{3}(x-2)+3$

The full solutions are at the end of the unit.

You should feel confident about solving linear equations with fractions in them. However, in all the equations we have solved thus far, the denominators had only a single term in them, and they were all constants. How would we solve an equation like $\scriptsize \displaystyle \frac{4}{x}=\displaystyle \frac{2}{{x-2}}$ where we have a denominator with more than a single term and a variable? Have a look at the next example to see the solution.

### Example 1.3

Solve: $\scriptsize \displaystyle \frac{4}{x}=\displaystyle \frac{2}{{x-2}}$

Solution

As before, we need to multiply both sides of the equation by the LCD. In this case, the LCD is $\scriptsize x(x-2)$. But before we do, there is something important to note. If we multiply both sides by $\scriptsize x(x-2)$, we need to make sure that we are not multiplying both sides by zero. This means that $\scriptsize x\ne 0$ and $\scriptsize x\ne 2$. We call these restrictions. If the solution to the equation turns out to be $\scriptsize 0$ or $\scriptsize 2$, we have to disregard it and say that there is no solution to the equation. This is just another way of saying that neither of the denominators in the original equation are allowed to be zero because we know that we cannot divide anything by zero. This is undefined.

\scriptsize \begin{align*}\displaystyle \frac{4}{x} & =\displaystyle \frac{2}{{x-2}}\quad x\ne 0;x\ne 2\\\therefore x(x-2)\times \left( {\displaystyle \frac{4}{x}} \right) & =x(x-2)\times \left( {\displaystyle \frac{2}{{x-2}}} \right)\quad \text{Multiply by the LCD of }x(x-2)\\\therefore (x-2)\times 4 & =x\times 2\quad \text{Simplify}\\\therefore 4x-8 & =2x\quad \text{Subtract }2x\text{ from both sides and add 8 to both sides}\\\therefore 2x & =8\quad \text{Divide both sides by 2}\\\therefore x & =4\end{align*}

The solution does not violate either of our restrictions so we can accept it and check that it is correct.

Check the solution:

\scriptsize \begin{align*}\text{LHS: }\displaystyle \frac{4}{4}&=1\\\text{RHS: }\displaystyle \frac{2}{{4-2}}&=\displaystyle \frac{2}{2}=1\end{align*}

The LHS = the RHS, so the solution is correct.

### Exercise 1.3

Solve the following equations, noting any restrictions.

1. $\scriptsize 5-\displaystyle \frac{7}{q}=\displaystyle \frac{{2(q+4)}}{q}$
2. $\scriptsize \displaystyle \frac{5}{{2x}}+\displaystyle \frac{1}{{6x}}=\displaystyle \frac{3}{x}+2$
3. $\scriptsize \displaystyle \frac{2}{{(b-5)}}-\displaystyle \frac{4}{{(b+5)}}=\displaystyle \frac{3}{{(b+5)}}$

The full solutions are at the end of the unit.

## Solving quadratic equations by factorisation

Besides linear equations, the other most common type of equation that you will have to solve is a quadratic equation. In quadratic equations, the biggest exponent on the unknown is $\scriptsize 2$.

$\scriptsize {{x}^{2}}-9=0$ and $\scriptsize {{x}^{2}}+4x=-3$ are more examples of quadratic equations. The highest power on the unknown in each case is $\scriptsize 2$. The standard form of a quadratic equation is $\scriptsize a{{x}^{2}}+bx+c=0$. As you will see in Example 1.4, quadratic equations must be written in this form before you solve.

All linear equations will have one solution at most. As we will see, quadratic equations have two solutions at most. We sometimes refer to the solutions of an equation as its roots. Linear equations have one root. Quadratic equations have two roots.

There are several methods we can use to solve quadratic equations but, in this unit, we will only look at solving quadratic equations using factorisation. When we use factorisation to solve any quadratic function, we have to rely on a law called the zero product property.

Think about this equation: $\scriptsize x(x-3)=0$. In order for the LHS to be equal to the RHS one or both of the factors on the LHS have to be equal to zero. In other words, $\scriptsize x=0$ or $\scriptsize (x-3)=0$ or they are both equal to zero. We can use this fact to solve this equation. Either:

$\scriptsize x=0$

or

\scriptsize \begin{align*}x-3 & =0\\\therefore x & =3\end{align*}

In other words, $\scriptsize x=0$ or $\scriptsize x=3$ are both solutions to the equation.

The zero product property states that if $\scriptsize A\times B=0$ then $\scriptsize A=0$ or $\scriptsize B=0$ or both $\scriptsize A=0$ and $\scriptsize B=0$

### Example 1.4

Solve for $\scriptsize x$, noting any restrictions:

1. $\scriptsize {{x}^{2}}-9=0$
2. $\scriptsize {{x}^{2}}-2x=8$
3. $\scriptsize -4{{x}^{2}}=-3x$
4. $\scriptsize 2{{x}^{2}}=15-7x$
5. $\scriptsize 3x=\displaystyle \frac{{54}}{{2x}}$
6. $\scriptsize \displaystyle \frac{{3({{x}^{2}}+1)+10x}}{{3x+1}}=1$

Solutions

1. The first thing we always need to do when solving a quadratic equation by factorising (in other words using the zero product property) is to get the one side of the equation equal to zero. The equation is already in standard form, so we can start by factorising the other side of the equation.
.
\scriptsize \begin{align*}{{x}^{2}}-9 & =0\quad \text{Factorise the difference of 2 squares}\\\therefore (x+3)(x-3) & =0\quad \text{Apply the zero product law}\\\therefore x+3=0\text{ } & \text{or }x-3=0\\\therefore x=-3\text{ } & \text{or }x=3\end{align*}
.
It is a good idea to check both solutions:

\scriptsize \begin{align*}x=-3\\\text{LHS: }{{\left( {-3} \right)}^{2}}-9=9-9=0\\\text{RHS: 0}\end{align*}

The LHS = the RHS, so the solution is correct.

\scriptsize \begin{align*}x=3\\\text{LHS: }{{\left( 3 \right)}^{2}}-9=9-9=0\\\text{RHS: 0}\end{align*}

The LHS = the RHS, so the solution is correct.

2. In this case, we first need to arrange the equation so that we have zero on the one side. We do this by subtracting $\scriptsize 8$ from both sides.
.
\scriptsize \begin{align*}{{x}^{2}}-2x & =8\quad \text{Subtract 8 from both sides to get the RHS = 0}\\\therefore {{x}^{2}}-2x-8 & =0\quad \text{Factorise the trinomial on the LHS}\\\therefore (x-4)(x+2) & =0\quad \text{Apply the zero product law}\\\therefore x-4=0\text{ } & \text{or }x+2=0\\\therefore x=4\text{ } & \text{or }x=-2\end{align*}
.
Check both solutions:

\scriptsize \begin{align*}x=4\\\text{LHS: }{{\text{4}}^{2}}-2\times 4=8\\\text{RHS: 8}\end{align*}The LHS = the RHS, so the solution is correct.

\scriptsize \begin{align*}x=-2\\\text{LHS: }{{\left( {-2} \right)}^{2}}-2(-2)=4+4=8\\\text{RHS: 8}\end{align*}

The LHS = the RHS, so the solution is correct.

3. Start by re-arranging the equation so that one side is equal to zero.
.
\scriptsize \begin{align*}-4{{x}^{2}} & =-3x\quad \text{Add }3x\text{ to both sides to get 0 on one side}\\\therefore -4{{x}^{2}}+3x & =0\quad \text{Multiply both sides by -1}\\\therefore 4{{x}^{2}}-3x & =0\quad \text{Factorise the LHS }\text{by taking out a common factor}\\\therefore x(4x-3) & =0\quad \text{Apply the zero product law}\\\therefore x=0\text{ } & \text{or }4x-3=0\\\therefore x=0\text{ } & \text{or }4x=3\\\therefore x=0\text{ } & \text{or }x=\displaystyle \frac{3}{4}\end{align*}
.
Don’t forget to check both solutions.
4. Once again, we need to start by re-arranging the equation so that one side is equal to zero.
.
\scriptsize \begin{align*}2{{x}^{2}} & =15-7x\quad \text{Add }7x\text{ to and subtract 15 from both sides}\\\therefore 2{{x}^{2}}+7x-15 & =0\quad \text{Factorise the trinomial}\\\therefore (2x-3)(x+5) & =0\quad \text{Apply the zero product law}\\\therefore 2x-3=0\text{ } & \text{or }x+5=0\\\therefore 2x=3\text{ } & \text{or }x=-5\\\therefore x=\displaystyle \frac{3}{2} & \text{or }x=-5\end{align*}
.
Don’t forget to check both solutions.
5. $\scriptsize 3x=\displaystyle \frac{{54}}{{2x}},x\ne 0$
We know that we may never divide by zero. Therefore, we cannot allow the unknown in the denominator to assume any value that makes the denominator zero. This means that $\scriptsize 2x\ne 0$ and therefore $\scriptsize x\ne 0$.
.
\scriptsize \begin{align*}\therefore 6{{x}^{2}} & =54\\\therefore {{x}^{2}} & =9\\\therefore x & =\pm \sqrt{9}\\\therefore x & =\pm 3\end{align*}
.
Or
.
\scriptsize \begin{align*}3x=\displaystyle \frac{{54}}{{2x}},x\ne 0\\\therefore 6{{x}^{2}}=54\\\therefore 6{{x}^{2}}-54=0\\\therefore x{{}^{2}}-9=0\\\therefore (x+3)(x-3)=0\\\therefore x=\pm 3\end{align*}
.
Don’t forget to check the solution.
6. .
\scriptsize \begin{align*}\displaystyle \frac{{3({{x}^{2}}+1)+10x}}{{3x+1}} & =1,\quad x\ne -\displaystyle \frac{1}{3}\\\therefore 3({{x}^{2}}+1)+10x & =3x+1\\\therefore 3{{x}^{2}}+3+10x-3x-1 & =0\\\therefore 3{{x}^{2}}+7x+2 & =0\\\therefore (3x+1)(x+2) & =0\\\therefore 3x+1=0\ & \text{or}\ x+2=0\\\therefore 3x=-1\ & \text{or}\ x=-2\\\therefore \xcancel{{x=-\displaystyle \frac{1}{3}}}\ & \text{or}\ x=-2\end{align*}
.
We have to disallow the solution $\scriptsize x=-\displaystyle \frac{1}{3}$ because of the restriction.
.
Don’t forget to check the solution.

### Note

1. Get one side of the equation equal to zero (write the quadratic in standard form).
2. Factorise the other side of the equation.
3. Apply the zero product property to set each factor equal to zero.
4. Solve for $\scriptsize x$ in each factor.

### Exercise 1.4

Solve the following quadratic equations, noting any restrictions.

1. $\scriptsize -{{x}^{2}}-7x=12$
2. $\scriptsize {{x}^{2}}-49=0$
3. $\scriptsize {{x}^{2}}+6x-14=13$
4. $\scriptsize 3{{x}^{2}}-18x-48=0$
5. $\scriptsize 2x(x+1)-(x-3)=6$
6. $\scriptsize {{(x+1)}^{2}}=(2x+3)(x+1)$
7. $\scriptsize (x+2)=\displaystyle \frac{{6x-12}}{{x-2}}$
8. $\scriptsize \displaystyle \frac{{4x}}{{4x+3}}=-\displaystyle \frac{3}{x}$

The full solutions are at the end of the unit.

## Summary

In this unit you have learnt the following:

• What an equation is, and that you need to keep the equation balanced by always performing the same operation to both sides of the equal sign.
• How to solve simple linear equations.
• How to solve linear equations with fractions by multiplying both sides of the equation by the LCD.
• How to solve quadratic equations by factorisation and applying the zero product property.

# Unit 1: Assessment

#### Suggested time to complete: 15 minutes

Solve for the unknown in each case:

1. $\scriptsize -7x=8(1-x)$
2. $\scriptsize 1=\displaystyle \frac{{3a-4}}{{2a+6}}$
3. $\scriptsize \displaystyle \frac{1}{4}(x-1)-1\displaystyle \frac{1}{2}(3x+2)=0$
4. $\scriptsize \displaystyle \frac{{-2}}{{({{x}^{2}}-9)}}+\displaystyle \frac{4}{{(x+3)}}=\displaystyle \frac{6}{{(x-3)}}$
5. $\scriptsize {{x}^{2}}-3=-2x$
6. $\scriptsize 2(9c-4)=9{{c}^{2}}$
7. $\scriptsize \displaystyle \frac{{3b}}{{b+2}}+1=\displaystyle \frac{4}{{b+1}}$

The full solutions are at the end of the unit.

# Unit 1: Solutions

### Exercise 1.1

1. .
\scriptsize \begin{align*}2x-6 & =8\\\therefore 2x-6+6&=8+6\\\therefore 2x & =14\\\therefore \displaystyle \frac{{2x}}{2} & =\displaystyle \frac{{14}}{2}\\\therefore x & =7\end{align*}
2. .
\scriptsize \begin{align*}4-2y & =6\\\therefore 4-4-2y & =6-4\\\therefore -2y & =2\\\therefore \displaystyle \frac{{-2y}}{{-2}} & =\displaystyle \frac{2}{{-2}}\\\therefore y & =-1\end{align*}
3. .
\scriptsize \begin{align*}-5x-4 & =11+2x+3\\\therefore -5x-4+4 & =11+2x+3+4\\\therefore -5x & =2x+18\\\therefore -5x-2x & =2x-2x+18\\\therefore -7x & =18\\\therefore \displaystyle \frac{{-7x}}{{-7}} & =\displaystyle \frac{{18}}{{-7}}\\\therefore x & =-\displaystyle \frac{{18}}{7}\end{align*}
4. .
\scriptsize \begin{align*}3(x+7)&=-2(x-7)\\\therefore 3x+21&=-2x+14\\\therefore 3x+21-21&=-2x+14-21\\\therefore 3x+2x&=-2x+2x-7\\\therefore 5x&=-7\\\therefore x&=-\displaystyle \frac{7}{5}\end{align*}
5. .
\scriptsize \begin{align*}7a-(3a+10)&=6-2(12-9a)\\\therefore 7a-3a-10&=6-24+18a\\\therefore 4a-10&=-18+18a\\\therefore 4a-18a&=-18+10\\\therefore -14a&=-8\\\therefore a&=\displaystyle \frac{{-8}}{{-14}}=\displaystyle \frac{4}{7}\end{align*}

Back to Exercise 1.1

### Exercise 1.2

1. .
\scriptsize \begin{align*}\displaystyle \frac{{x+2}}{4}-\displaystyle \frac{{x-6}}{3} & =12\\\therefore 12\times \left( {\displaystyle \frac{{x+2}}{4}} \right)-12\times \left( {\displaystyle \frac{{x-6}}{3}} \right) & =12\times 12\\\therefore 3(x+2)-4(x-6) & =144\\\therefore 3x+6-4x+24 & =144\\\therefore -x & =114\\\therefore x & =-114\end{align*}
2. .
\scriptsize \begin{align*}\displaystyle \frac{{2-5a}}{3}-6 & =\displaystyle \frac{{4a}}{3}+2-a\\\therefore 3\left( {\displaystyle \frac{{2-5a}}{3}} \right)-3(6) & =3\left( {\displaystyle \frac{{4a}}{3}} \right)+3(2-a)\\\therefore 2-5a-18 & =4a+6-3a\\\therefore -6a & =22\\\therefore a & =-\displaystyle \frac{{22}}{6}=-\displaystyle \frac{{11}}{3}\end{align*}
3. .
\scriptsize \begin{align*}3-\displaystyle \frac{{y-2}}{4} & =4\\\therefore 4(3)-4\left( {\displaystyle \frac{{y-2}}{4}} \right) & =4(4)\\\therefore 12-y+2 & =16\\\therefore -y & =2\\\therefore y & =-2\end{align*}
4. .
\scriptsize \begin{align*}\displaystyle \frac{1}{5}(x-1) & =\displaystyle \frac{1}{3}(x-2)+3\\\therefore 15\left( {\displaystyle \frac{1}{5}(x-1)} \right) & =15\left( {\displaystyle \frac{1}{3}(x-2)} \right)+15(3)\\\therefore 3(x-1) & =5(x-2)+45\\\therefore 3x-3 & =5x-10+45\\\therefore -2x & =38\\\therefore x & =-19\end{align*}

Back to Exercise 1.2

### Exercise 1.3

1. .
\scriptsize \begin{align*}5-\displaystyle \frac{7}{q} & =\displaystyle \frac{{2(q+4)}}{q}\quad q\ne 0\\\therefore 5q-7 & =2(q+4)\\\therefore 5q-7 & =2q+8\\\therefore 3q & =15\\\therefore q & =5\end{align*}
2. .
\scriptsize \begin{align*}\displaystyle \frac{5}{{2x}}+\displaystyle \frac{1}{{6x}} & =\displaystyle \frac{3}{x}+2\quad x\ne 0\\\therefore 3(5)+1 & =6(3)+12x\quad \text{Multiply both sides by the LCD of }6x\\\therefore 16 & =18+12x\\\therefore 12x & =-2\\\therefore x & =-\displaystyle \frac{1}{6}\end{align*}
3. .
\scriptsize \begin{align*}\displaystyle \frac{2}{{(b-5)}}-\displaystyle \frac{4}{{(b+5)}} & =\displaystyle \frac{3}{{(b+5)}}\quad b\ne 5;b\ne -5\\\therefore 2(b+5)-4(b-5) & =3(b-5)\\\therefore 2b+10-4b+20 & =3b-15\\\therefore -5b & =-45\\\therefore b & =9\end{align*}

Back to Exercise 1.3

### Exercise 1.4

1. .
\scriptsize \begin{align*}-{{x}^{2}}-7x & =12\\\therefore -{{x}^{2}}-7x-12 & =0\\\therefore {{x}^{2}}+7x+12 & =0\\\therefore (x+3)(x+4) & =0\\\therefore x+3=0\text{ } & \text{or }x+4=0\\\therefore x=-3\text{ } & \text{or }x=-4\end{align*}
2. .
\scriptsize \begin{align*}{{x}^{2}}-49 & =0\\\therefore (x+7)(x-7) & =0\\\therefore x+7=0\text{ } & \text{or }x-7=0\\\therefore x=-7\text{ } & \text{or }x=7\end{align*}
3. .
\scriptsize \begin{align*}{{x}^{2}}+6x-14 & =13\\\therefore {{x}^{2}}+6x-27 & =0\\\therefore (x+9)(x-3) & =0\\\therefore x=-9\text{ } & \text{or }x=3\end{align*}
4. .
\scriptsize \begin{align*}3{{x}^{2}}-18x-48 & =0\\\therefore 3({{x}^{2}}-6x-16) & =0\\\therefore {{x}^{2}}-6x-16 & =0\\\therefore (x-8)(x+2) & =0\\\therefore x=8\text{ } & \text{or }x=-2\end{align*}
5. .
\scriptsize \begin{align*}{l}2x(x+1)-(x-3) & =6\\\therefore 2{{x}^{2}}+2x-x+3 & =6\\\therefore 2{{x}^{2}}+x-3 & =0\\\therefore (2x+3)(x-1) & =0\\\therefore x=-\displaystyle \frac{3}{2}\text{ } & \text{or }x=1\end{align*}
6. .
\scriptsize \begin{align*}{{(x+1)}^{2}} & =(2x+3)(x+1)\\\therefore {{x}^{2}}+2x+1 & =2{{x}^{2}}+5x+3\\\therefore -{{x}^{2}}-3x-2 & =0\\\therefore x{}^{2}+3x+2 & =0\\\therefore (x+2)(x+1) & =0\\\therefore x=-2\text{ } & \text{or }x=-1\end{align*}
Note: You cannot divide both sides of the equation by the $\scriptsize (x+1)$ factor as this has the effect of changing the equation into a linear equation and, therefore, removing one of the solutions. Which solution (or root) does this remove?
7. .
\scriptsize \begin{align*}(x+2) & =\displaystyle \frac{{6x-12}}{{x-2}},\ x\ne 2\\\therefore (x+2)(x-2) & =6x-12\\\therefore {{x}^{2}}-4 & =6x-12\\\therefore {{x}^{2}}-6x+8 & =0\\\therefore (x-2)(x-4) & =0\\\therefore \xcancel{{x=2}}\ & \text{or}\ x=4\end{align*}
8. .
\scriptsize \begin{align*}\displaystyle \frac{{4x}}{{4x+3}} & =-\displaystyle \frac{3}{x},\ x\ne 0,\ x\ne -1\\\therefore 4{{x}^{2}} & =-3(4x+3)\\\therefore 4{{x}^{2}} & =-12x-9\\\therefore 4{{x}^{2}}+12x+9 & =0\\\therefore (2x+3)(2x+3) & =0\\\therefore x=-\displaystyle \frac{3}{2}\ & \text{or}\ x=-\displaystyle \frac{3}{2}\end{align*}
Both solutions are the same. Therefore, we say that there is only one root (or answer) for the equation and it is $\scriptsize x=-\displaystyle \frac{3}{2}$.

Back to Exercise 1.4

### Unit 1: Assessment

1. .
\scriptsize \begin{align*}-7x & =8(1-x)\\\therefore -7x & =8-8x\\\therefore x & =8\end{align*}
2. .
\scriptsize \begin{align*}1&=\displaystyle \frac{{3a-4}}{{2a+6}}\quad a\ne -3\\\therefore 2a+6 & =3a-4\\\therefore -a & =-10\\\therefore a & =10\end{align*}
3. .
\scriptsize \begin{align*}\displaystyle \frac{1}{4}(x-1)-1\displaystyle \frac{1}{2}(3x+2) & =0\\\therefore \displaystyle \frac{1}{4}(x-1)-\displaystyle \frac{3}{2}(3x+2) & =0\\\therefore (x-1)-2\times 3(3x+2) & =0\\\therefore x-1-18x-12 & =0\\\therefore -17x & =13\\\therefore x & =-\displaystyle \frac{{13}}{{17}}\end{align*}
4. .
\scriptsize \begin{align*}\displaystyle \frac{{-2}}{{({{x}^{2}}-9)}}+\displaystyle \frac{4}{{(x+3)}} & =\displaystyle \frac{6}{{(x-3)}}\quad x\ne \pm 3\\\therefore \displaystyle \frac{{-2}}{{(x+3)(x-3)}}+\displaystyle \frac{4}{{(x+3)}} & =\displaystyle \frac{6}{{(x-3)}}\\\therefore -2+4(x-3) & =6(x+3)\\\therefore -2+4x-12 & =6x+18\\\therefore -2x & =32\\\therefore x & =-16\end{align*}
5. .
\scriptsize \begin{align*}{{x}^{2}}-3 & =-2x\\\therefore {{x}^{2}}+2x-3 & =0\\\therefore (x+3)(x-1) & =0\\\therefore x=-3\text{ } & \text{or }x=1\end{align*}
6. .
\scriptsize \begin{align*}2(9c-4) & =9{{c}^{2}}\\\therefore 18c-8 & =9{{c}^{2}}\\\therefore 0 & =9{{c}^{2}}-18c+8\\\therefore 0 & =(3c-4)(3c-2)\\\therefore 3c-4=0\text{ } & \text{or }3c-2=0\\\therefore c=\displaystyle \frac{4}{3}\text{ } & \text{or }c=\displaystyle \frac{2}{3}\end{align*}
7. .
\scriptsize \begin{align*}\displaystyle \frac{{3b}}{{b+2}}+1 & =\displaystyle \frac{4}{{b+1}}\quad b\ne -2;b\ne -1\\\therefore 3b(b+1)+(b+2)(b+1) & =4(b+2)\\\therefore 3{{b}^{2}}+3b+{{b}^{2}}+3b+2 & =4b+8\\\therefore 4{{b}^{2}}+2b-6 & =0\\\therefore 2(2{{b}^{2}}+b-3) & =0\\\therefore 2{{b}^{2}}+b-3 & =0\\\therefore (2b+3)(b-1) & =0\\\therefore b=-\displaystyle \frac{3}{2}\text{ } & \text{or }b=1\end{align*}

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