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Algebraic notation

Simplify algebraic expressions SubstitutionThis topic is relevant for:

Here we will learn about equations, including solving equations, linear equations, quadratic equations, simultaneous equations and rearranging equations.

There are also equations worksheets based on Edexcel, AQA and OCR exam questions, along with further guidance on where to go next if you’re still stuck.

Equations are mathematical expressions which contain a variable and an equals sign.

We can solve an equation to find the value of the variable.

E.g.

\begin{aligned} &3x-5=7 \\\\\\ &\frac{4(x-2)}{5}=8 \\\\\\ &x^2=9 \\\\\\ &2x^2-3x-5=0 \end{aligned}**Simultaneous equations** are a pair of equations with two variables. They can be solved to find a pair of values which make both equations true at the same time.

**Step-by-step guide: **Simultaneous equations

E.g. Linear simultaneous equations

\begin{aligned} x+y&=10\\\\ x-y&=4 \end{aligned}E.g. Quadratic simultaneous equations

y=x^2-6x+8 y=2x+1**Rearranging equations** means we change the subject of the equation to display it in a different way.

**Step-by-step guide: **Rearranging equations

E.g.

The subject of the following equation is currently y .

y=3x+2We can rearrange the equation to make x the subject:

x=\frac{y-2}{3}Sometimes an equation is a formula. This is when it is used to solve a specific problem.

E.g.

Here is the formula to work out the area of a circle:

A=\pi r^2It could be rearranged to find the radius if we are given the area:

r=\sqrt{\frac{A}{\pi }}Get your free equations worksheet of 20+ questions and answers. Includes reasoning and applied questions.

DOWNLOAD FREEGet your free equations worksheet of 20+ questions and answers. Includes reasoning and applied questions.

DOWNLOAD FREEEquations and identities look very similar, but are two different things.

An **identity **is **true **for **all values **of the variable, whereas the **equation **is **true **for only **certain values**.

For example, the equation:

3x+5=14can be solved, so it is true only when x=3.

However the identity:

3(x+2)\equiv3x+6It can not be solved. The left-hand side always equals the right-hand side for all values of x.

For identities there is a special symbol that can be used ‘≡’. It is like an equals sign =, but it means ‘identical to’.

Iteration is a process in which the approximate solution is determined using an iterative formula and a known starting number. When using iteration to calculate a root of an equation, the values converge towards a fixed point.

Simple linear equations are solved by using a “balancing method” – doing the same operations to both sides of the equation. They can be checked by substituting the answer back into the original equation.

**Step-by-step guide**: Linear equations

**Linear equations – one unknown**

E.g.

Solve: 3x+4=16

**Linear equations – unknowns on both sides**

E.g.

Solve: 4x+7=2x+12

**Linear equations – with brackets**

E.g.

Solve: 3(x-5)=18

OR

**Equations with fractions**

E.g.

Solve: \frac{5}{x}=10

**Simple equations – with powers and roots**

Whilst these are **NOT **linear equations they are included here as the method used to solve them is very similar.

E.g.

Solve: 4x^2+1=35

**Step-by-step guide: **Equations with fractions

Quadratic equations are solved by using different methods. They can be checked by substituting the answer back into the original equation.

**Step-by-step guide **Quadratic equations

**Quadratic equations – by factorising**

We factorise the quadratic and then solve each factor being equal to 0 , one at a time.

E.g.

Solve: x^2-5x+6=0

\begin{aligned} x^2 -5x+6 &= 0\\\\ (x-2)(x-3) &= 0 \\\\ \end{aligned}

\begin{aligned} x-2&=0 \quad \text{or} \quad x-3=0\\\\ x&=2 \quad \text{or} \quad x=3 \\\\ \end{aligned}

**Step-by-step guide:** Solving quadratic equations by factorising

**Quadratic equations – quadratic formula**

We can use the quadratic formula to solve any quadratic equation of the form

ax^2+bx+c

The quadratic formula is: x=\frac{-b \pm \sqrt{b^{2}-4 a c}}{2 a}

E.g.

Solve: 3x^2-4x-5=0

Give your answers to 2 decimal places

\begin{aligned}
&3x^{2}-4x-5=0\\\\
x&=\frac{-b\pm\sqrt{b^2-4ac}}{2a}\\\\
a&=3, \quad b=-4, \quad c=-5\\\\
x&=\frac{-(-4)\pm\sqrt{(-4)^2-4(3)(-5)}}{2(3)}\\\\
x&=\frac{4\pm\sqrt{76}}{6}
\\\\
x&=2.1196... \quad \text{or} \quad x=-0.78629...\\\\
x&=2.12 \quad \text{or} \quad x=-0.79
\end{aligned}

**Step-by-step guide:** Quadratic formula

**Quadratic equations example – completing the square**

E.g.

Solve: x^2-4x-5=0

\begin{aligned}
x^2 -4x-5 &= 0\\\\
x^2-4x&=5\\\\
(x-2)^2 -4&=5\\\\
(x-2)^2 &= 9\\\\
x-2 &= \pm 3\\\\
x&=2\pm 3\\\\
x=5 \quad &\textrm{or} \quad x=-1
\end{aligned}

**Step-by-step guide:** Completing the square

**Quadratic equations example – graphically**

E.g.

Using the graph of y=x^2-6x+8 , solve: x^2-6x+8=0

The solutions are to be found where y=0

x=2 \quad \textrm{or} \quad x=4

**Step-by-step guide: **Solving quadratic equations graphically

**Simultaneous linear equations**

E.g.

Solve these simultaneous equations:

\begin{aligned} 2x+3y&=14\\\\ 6x-y&=12 \end{aligned}

**Step-by-step guide**: Simultaneous equations

**Quadratic simultaneous equations**

E.g.

Solve these simultaneous equations:

\begin{aligned} y&=x^2-6x+8\\\\ y&=2x+1 \end{aligned}

\[\begin{aligned}
x^2-6x+8&=2x+1\\
x^2-8x+7&=0\\
(x-7)(x-1)&=0\\
\\
x=7 \quad &\text{or} \quad x=1\\
\\
\text{substitue} \quad x=7 \quad \text{into} \quad y&=2x+1\\
y&=2(7)+1=15\\
\\
\text{substitue} \quad x=1 \quad \text{into} \quad y&=2x+1\\
y&=2(1)+1=3\\
\\
\text{solutions are} \quad x=7,\quad y=15& \quad \text{or} \quad x=1, \quad y=3\\
\end{aligned}\]

**Step-by-step guide:**Quadratic simultaneous equations

**Step-by-step guide: **Quadratic simultaneous equations

**Solve simultaneous equations graphically**

E.g.

Draw graphs to solve the simultaneous equations:

\begin{aligned} 2x+y=5 \\\\ y=2x-1 \end{aligned}

Draw the graphs on the same axes and see where they intersect.

The solution is x=1.5 and y=2

**Step-by-step guide**: Solving simultaneous equations graphically

Example:

Make t the subject of: v=u+at

\begin{aligned} &v=u+at \\\\ &v-u=at\\\\ &\frac{v-u}{a}=t\\\\ &\text{Therefore} \\\\ &t= \frac{v-u}{a} \end{aligned}**Step-by-step guide**: Make x the subject

**Solutions to equations are not always positive whole numbers**

Solutions to equations can be positive integers (whole numbers), but they can also be negative. They can also be decimals or fractions.

**Don’t try to just write down the answer**

When working with equations, take each step one at a time. Keep your workings neat and tidy so you are more likely to be correct.

**If you multiply both sides (or divide) make sure that you do this to each and every term**

When you are working with an equation and you need to multiply (or divide), it is easy to make a mistake. Make sure you apply the multiplication to every term.

For example:

Each term on both sides is multiplied by 3

1. Solve:

4x+10=30

x=5

x=10

x=6

x=9

\begin{aligned}
4x+10&=30\\\\
4x&=20\\\\
x&=5
\end{aligned}

2. Solve:

2(x+5)=3x+4

x=4

x=5

x=6

x=3

\begin{aligned}
2(x+5)&=3x+4\\\\
2x+10&=3x+4\\\\
10&=x+4\\\\
6&=x\\\\
\text{therefore} \quad x&=6
\end{aligned}

3. Solve:

\frac{x-4}{3}=7

x=31

x=25

x=27

x=29

\begin{aligned}
\frac{x-4}{3}&=7\\\\
x-4&=21\\\\
x&=25
\end{aligned}

4. Solve:

x^2+6x+8=0

x=2 \quad \text{or} \quad x=4

x=8 \quad \text{or} \quad x=6

x=-8 \quad \text{or} \quad x=-6

x=-2 \quad \text{or} \quad x=-4

\begin{aligned}
x^2+6x+8&=0\\\\
(x+2)(x+4)&=0\\\\
x=-2 \quad \text{or} \quad x&=-4
\end{aligned}

5. Solve:

\begin{aligned} 2x+5y=13 \\\\ x+2y=6 \end{aligned}

x=4 \quad \text{and} \quad y=2

x=4 \quad \text{and} \quad y=1

x=3 \quad \text{and} \quad y=1

x=3 \quad \text{and} \quad y=2

\begin{aligned}
2x+5y&=13 \quad [1]\\
x+2y&=6 \quad [2]\\
\\
[1]\times2 \quad 4x+10y&=26 \quad [3]\\
[2]\times5 \quad 5x+10y&=30 \quad[4]\\
\hline
[4]-[3] \qquad \qquad x&=4\\
\\
\text{substitute} \quad x=4 \quad \text{into} \quad[2]\\
4+2y&=6\\
2y&=2\\
y&=1\\
\\
\text{solution is} \quad x=4 \quad \text{and} \quad y=1
\end{aligned}

6. Rearrange correctly to make h the subject:

V=\pi r^2h

h=V-\pi r^2

h=\frac{\pi V}{r^2}

h=\frac{V}{\pi r^2}

h=\sqrt{\frac{V}{\pi r}}

\begin{aligned}
V &= \pi r^2 h\\\\
\frac{V}{\pi} &= r^2h\\\\
\frac{V}{\pi r^2} &=h\\\\
\text{therefore} \quad h&= \frac{V}{\pi r^2}
\end{aligned}

1. Solve:

3x-4=12

**(2 marks)**

Show answer

3x=16

**(1)**

x=5\frac{1}{3}

**(1)**

2. Solve the simultaneous equation.

\begin{aligned} 2x+y&= 15\\\\ x-y &= 6 \end{aligned}

**(3 marks)**

Show answer

Add the 2 equations

2x+x=15+6

**(1)**

\begin{aligned} 3x&= 21\\\\ x &= 7 \end{aligned}

**(1)**

**(1)**

3. Solve:

x^2-x-20=0

**(3 marks)**

Show answer

\begin{aligned}
x^2-x-20&=0\\\\
(x\pm a)(x\pm b) &= 0
\end{aligned}

**(1)**

(x-5)(x+4)=0

**(1)**

x=5 \quad \text{or} \quad x=-4

**(1)**

4. Rearrange the equation to make x the subject:

y=\frac{x}{4}-5

**(2 marks)**

Show answer

\begin{aligned}
y&=\frac{x}{4}-5\\\\
y+5&=\frac{x}{4}
\end{aligned}

**(1)**

\begin{aligned} 4(y+5)&=x\\\\ x&=4(y+5) \end{aligned}

**(1)**

You have now learned how to:

- Use algebraic methods to solve linear equations in 1 variable
- Rearrange formulae to change the subject
- To find approximate solutions of simultaneous linear equations
- Solve quadratic equations {including those that require rearrangement} algebraically by factorising
- Solve quadratic equations by completing the square
- Solve quadratic equations by using the quadratic formula
- Solve quadratic equations – find approximate solutions using a graph
- Solve 2 simultaneous linear equations in 2 variables algebraically
- Solve 2 simultaneous equations in 2 variables – one linear and one quadratic algebraically
- Solve 2 simultaneous equations in 2 variables – find approximate solutions using a graph

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8 sets of free exam practice papers written by maths teachers and examiners for Edexcel, AQA and OCR.

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