# More Advanced Equations

Continued from: Simple EquationsOur page Introducing Simple Equations explains how to solve straightforward equations, even those with basic algebra.

This page discusses more complex equations, including those involving fractions, and two particular problems that you may encounter: simultaneous equations and quadratic equations.

Most importantly, it makes clear that these equations, like others, conform to rules, and that you can still manipulate them, as long as you remember to do the same thing to both sides of the equation.

### Brackets in Algebra

In algebraic equations, you will quite often come across brackets which need to be multiplied together, for example *(x + 5)(x + 4)*.

In this case, you need to multiply each term in the first bracket by each term in the second bracket and add them all together, that is, multiply x by x, x by 4, then x by 5, then 4 by 5.

x times x is of course x^{2}, x times 4 is 4x, x times 5 is 5x, and 4 x 5 is 20.

(x + 5)(x + 4) = x^{2} + 5x + 4x + 20 = x^{2} + 9x + 20

### Equations with Fractions

Equations with fractions look a bit daunting, but there is a simple trick to make them easier to solve.

** Cross-multiplication** involves removing the fractions by multiplying both sides by each denominator, in turn. For more about working with fractions, see our page on

**Fractions**.

### Worked Example

2 + x |
= |
9 + x |

3 |
5 |

To remove the fractions, multiply both sides of the equation by first 3, and then 5.

3(2 + x) |
= |
3(9 + x) |

3 |
5 |

On the left, the two 3s cancel out, leaving only 2 + x. On the right, the brackets multiply out to make 27 + 3x

2 + x |
= |
27 + 3x |

5 |

Now multiply by 5. Again, it will cancel out on the right, and you will end up with:

**5(2 + x) = 27 + 3x**

**10 + 5x = 27 + 3x**

Take 10 from each side to get numbers on the right hand side only, and you end up with:

**5x = 17 + 3x**

Take 3x from each side to get all the x values on the left, and you end up with:

**2x = 17**

**x = 8.5**

**Note that x does not always have to be a whole number.**

## Simultaneous Equations

Sometimes you may be given two equations, both involving the same variables, both of which are true. These are called *simultaneous equations*. Having both pieces of information will enable you to solve them.

First of all, you need to rearrange one to obtain a value for one of the variables. Once you have a value for x, say, then you can substitute it into the other to find the value of y. These simultaneous equations are often indicated by a long curly bracket around their right hand side to link them together.

### Worked Example

2x = 6 |
} |

y = 4x + 5 |

If 2x = 6, then **x = 3**.

By substituting 3 for x in the second equation, you can solve it to find out what y is.

y = (4 x 3) + 5 = 12 + 5 = 17. **y = 17**

## Quadratic Equations

An equation which takes the form ax^{2} + bx + c = 0 is called a *quadratic equation*.

a, b and c are all numbers, and in any given equation may all be the same or may be different. They can also be negative or positive.

Examples of quadratic equations are:

**2x**. In this equation, a = 2, b = 5 and c = 10.^{2}+ 5x + 10 = 0**3x**. In this equation, a = 3, b = -3 and c = 9.^{2}- 3x + 9 = 0**52x**In this equation, a = 52, b = 1 and c = -45.^{2}+ x -45 = 0.

There are several different ways to solve these equations:

**1. By Factoring**

Factoring involves identifying the two brackets that have been multiplied together to make the equation.

Because you make a quadratic equation by multiplying out two expressions in brackets (x + a number)(x + another number), every one *that has a solution* can be written in this two-bracket form.

**(x + m)(x + n) = x ^{2} + (m + n)x + mn**

This means that when you have an equation in the form x^{2} + bx + c, you are looking for two numbers such that when they are multiplied you get c, and when they are added you get b. You will normally be able to see straight away if these exist as whole numbers.

### Worked Example

**x ^{2} + 9x + 20 = 0**

You know that 4 x 5 = 20, and 4 + 5 = 9.

The two brackets are therefore (x + 4)(x + 5), and the two solutions of the equation are **x = 4 and x = 5**.

**2. Using a Formula**

If the two factors are not obvious, the next step is to use a formula. All quadratic equations that can be solved will give an answer using the formula:

x = |
b ± √(b^{2}-4ac) |

2a |

In this case, a is the coefficient of x^{2}, b of x, and c is the number at the end when the equation is in the form ax^{2} + bx + c = 0.

Any equation which has **only** terms with x^{2}, x and numbers can be turned into the form ax^{2} + bx + c = 0, and then solved using the formula.

Because you can have b plus or minus the square root, quadratic equations always have two solutions.

It is important to remember that some quadratic equations do not have a ‘real’ answer. For example, if b^{2}-4ac is negative, then there will be no real answer, because you cannot have a square root of a minus number, except in the form of an imaginary number (there is more about imaginary numbers on our page on special numbers and concepts).

### Conclusion

Having read this page and followed the examples, you should now be feeling more confident about your ability to handle even quite complex equations.

Just remember the golden rule:

*Always do the same thing to each side of the equation*

If you do, then you will be fine.

Continue to:

Simple Statistical Analysis

Set Theory

More numeracy skills:

Introduction to Trigonometry

Probability | BODMAS

Averages (Mean, Median and Mode)