# Systems of Linear Equations

A Linear Equation is an **equation** for a **line**.

A **System** of Equations is when we have **two or more equations** working together.

An example will help:

### Example: You versus Horse

It's a race!

You can run **0.2 km** every minute.

The Horse can run **0.5 km** every minute. But it takes 6 minutes to saddle the horse.

**How far can you get before the horse catches you?**

We can make **two** equations (**d**=distance in km, **t**=time in minutes)

- You run at 0.2km every minute, so
**d = 0.2t** - The horse runs at 0.5 km per minute, but we take 6 off its time:
**d = 0.5(t−6)**

So we have a **system** of equations (that are **linear**):

Do you see how the horse starts at 6 minutes, but then runs faster?

It seems you get caught after 10 minutes ... you only got 2 km away.

Run faster next time.

So now you know what a System of Linear Equations is.

Let us continue to find out more about them ....

## Solving

A linear equation is not always in the form **y = 3x+2**,

It can also be something like **y − 3x = 2**

Or **−3x + y = 2**

These are all the same linear equation

**And there can be many ways to solve linear equations!**

Let us see another example:

### Example: Solve these two equations:

- x + y = 6
- −3x + y = 2

The two equations are shown on this graph:

Our task is to find where the two lines cross.

OK, we can see where they cross, but let's solve it using Algebra!

Hmmm ... how should we solve this? **There can be many ways!** In this case both equations have "y" so let's try subtracting the second equation from the first:

Which simplifies to:

So now we know that **x=1** is on both lines.

And we can find the matching value of **y** using either of the two original equations (because we know they have the same value at x=1). Let's use the first one (you can try the second one yourself):

And the solution is:

x = 1 and y = 5

And the graph shows us we are right!

## Linear Equations

Only simple variables are allowed. **No x ^{2}, y^{3}, √x, etc**:

Linear vs non-linear

## Dimensions

A Linear Equation can be in 2 dimensions ... (such as x and y) |
||

... or 3 dimensions (such as x, y and z) ... |
||

... or 4 dimensions ... | ||

... or more! |

## Common Variables

For the equations to "work together" they share one or more variables:

A System of Equations has **two or more equations** in **one or more variables**

## Many Variables

So a System of Equations could have **many **equations and **many **variables.

### Example: 3 equations in 3 variables

2x | + | y | − | 2z | = | 3 |

x | − | y | − | z | = | 0 |

x | + | y | + | 3z | = | 12 |

There can be any combination:

- 2 equations in 3 variables,
- 6 equations in 4 variables,
- 9,000 equations in 567 variables,
- etc.

## Solutions

When the number of equations is the **same** as the number of variables there is **likely** to be a solution. Not guaranteed, but likely.

In fact there are only three possible cases:

**No**solution**One**solution**Infinitely many**solutions

When there is **no solution** the equations are called **"inconsistent"**.

**One **or ** infinitely many solutions** are called

**"consistent"**

Here is a diagram for **2 equations in 2 variables**:

## Independent

**"Independent"** means that each equation gives new information.

Otherwise they are **"Dependent"**.

Also called "Linear Independence" and "Linear Dependence"

### Example:

- x + y = 3
- 2x + 2y = 6

Those equations are **"Dependent"**, because they are really the **same equation**, just multiplied by 2.

So the second equation gave **no new information**.

## Where the Equations are True

The trick is to find where **all** equations are **true at the same time**.

**True?** What does that mean?

### Example: You versus Horse

The "you" line is **true all along its length** (but nowhere else).

Anywhere on that line **d** is equal to **0.2t**

- at t=5 and d=1, the equation is
**true**(Is d = 0.2t? Yes, as**1 = 0.2×5**is true) - at t=5 and d=3, the equation is
**not**true (Is d = 0.2t? No, as**3 = 0.2×5 is not true**)

Likewise the "horse" line is also **true all along its length** (but nowhere else).

But only at the point where they **cross** (at t=10, d=2) are they **both true**.

So they have to be true ** simultaneously** ...

... that is why some people call them **"Simultaneous Linear Equations"**

## Solve Using Algebra

Instead of plotting a graph we can use Algebra:

### Example: You versus Horse

Let us solve it using Algebra.

The system of equations is:

- d = 0.2t
- d = 0.5(t−6)

**In this case** it seems easiest to set them equal to each other:

d = 0.2t = 0.5(t−6)

**0.5(t−6)**:0.2t = 0.5t − 3

**0.5t**from both sides:−0.3t = −3

**−0.3**:t = −3/−0.3 =

**10**minutes

*Now we know when you get caught!*

**t**we can calculate

**d**:d = 0.2t = 0.2×10 =

**2**km

And our solution is:

t = 10 minutes and d = 2 km

## Algebra vs Graphs

Why use Algebra when graphs are so easy? Because:

More than 2 variables can't be solved by a simple graph.

So Algebra comes to the rescue with two popular methods:

- Solving By Substitution
- Solving By Elimination

We will see each one, with examples in 2 variables, and in 3 variables. Here goes ...

## Solving By Substitution

These are the steps:

- Write one of the equations so it is in the style
**"variable = ..."** **Replace**(i.e. substitute) that variable in the other equation(s).**Solve**the other equation(s)- (Repeat as necessary)

Here is an example with **2 equations in 2 variables**:

### Example:

- 3x + 2y = 19
- x + y = 8

We can start with **any equation** and **any variable**.

Let's use the second equation and the variable "y" (it looks the simplest equation).

Write one of the equations so it is in the style "variable = ...":

We can subtract x from both sides of x + y = 8 to get **y = 8 − x**. Now our equations look like this:

- 3x + 2y = 19
**y = 8 − x**

Now replace "y" with "8 − x" in the other equation:

- 3x + 2
**(8 − x)**= 19 - y = 8 − x

Solve using the usual algebra methods:

Expand **2(8−x)**:

- 3x +
**16 − 2x**= 19 - y = 8 − x

Then **3x−2x = x**:

**x**+ 16 = 19- y = 8 − x

And lastly **19−16=3**

**x = 3**- y = 8 − x

Now we know what **x** is, we can put it in the **y = 8 − x** equation:

- x = 3
- y = 8
**− 3**= 5

And the answer is:

x = 3

y = 5

*Note: because there is a solution the equations are "consistent"*

Check: why don't you check to see if x = 3 and y = 5 works in both equations?

## Solving By Substitution: 3 equations in 3 variables

OK! Let's move to a **longer** example: **3 equations in 3 variables**.

*This is not hard to do... it just takes a long time!*

### Example:

- x + z = 6
- z − 3y = 7
- 2x + y + 3z = 15

We should line up the variables neatly, or we may lose track of what we are doing:

x | + | z | = | 6 | |||||

− | 3y | + | z | = | 7 | ||||

2x | + | y | + | 3z | = | 15 |

WeI can start with any equation and any variable. Let's use the first equation and the variable "x".

Write one of the equations so it is in the style "variable = ...":

x |
= |
6 − z |
|||||||

− | 3y | + | z | = | 7 | ||||

2x | + | y | + | 3z | = | 15 |

Now replace "x" with "6 − z" in the other equations:

(Luckily there is only one other equation with x in it)

x | = | 6 − z | ||||||||

− | 3y | + | z | = | 7 | |||||

2(6−z) |
+ | y | + | 3z | = | 15 |

Solve using the usual algebra methods:

**2(6−z) + y + 3z = 15** simplifies to **y + z = 3**:

x | = | 6 − z | |||||||

− | 3y | + | z | = | 7 | ||||

y | + | z | = | 3 |

Good. We have made some progress, but not there yet.

Now **repeat the process**, but just for the last 2 equations.

Write one of the equations so it is in the style "variable = ...":

Let's choose the last equation and the variable z:

x | = | 6 − z | |||||||

− | 3y | + | z | = | 7 | ||||

z |
= |
3 − y |

Now replace "z" with "3 − y" in the other equation:

x | = | 6 − z | |||||||

− | 3y | + | 3 − y |
= | 7 | ||||

z | = | 3 − y |

Solve using the usual algebra methods:

**−3y + (3−y) = 7** simplifies to **−4y = 4**, or in other words **y = −1**

x | = | 6 − z | |||||||

y |
= |
−1 |
|||||||

z | = | 3 − y |

Almost Done!

Knowing that **y = −1 **we can calculate that **z = 3−y = 4**:

x | = | 6 − z | |||||||

y | = | −1 | |||||||

z |
= |
4 |

And knowing that **z = 4 **we can calculate that **x = 6−z = 2**:

x |
= |
2 |
|||||||

y | = | −1 | |||||||

z | = | 4 |

And the answer is:

x = 2

y = −1

z = 4

Check: please check this yourself.

We can use this method for 4 or more equations and variables... just do the same steps again and again until it is solved.

Conclusion: Substitution works nicely, but does take a long time to do.

## Solving By Elimination

Elimination can be faster ... but needs to be kept neat.

"Eliminate" means to **remove**: this method works by removing variables until there is just one left.

The idea is that we **can safely**:

**multiply**an equation by a constant (except zero),**add**(or subtract) an equation on to another equation

Like in these examples:

### WHY can we add equations to each other?

Imagine two really simple equations:

x − 5 = 3

5 = 5

We can add the "5 = 5" to "x − 5 = 3":

x − 5 + 5 = 3 + 5

x = 8

Try that yourself but use 5 = 3+2 as the 2nd equation

It will still work just fine, because both sides are equal (that is what the = is for!)

We can also swap equations around, so the 1st could become the 2nd, etc, if that helps.

OK, time for a full example. Let's use the **2 equations in 2 variables** example from before:

### Example:

- 3x + 2y = 19
- x + y = 8

**Very** important to keep things neat:

3x | + | 2y | = | 19 | |||

x | + | y | = | 8 |

Now ... our aim is to **eliminate** a variable from an equation.

First we see there is a "2y" and a "y", so let's work on that.

**Multiply** the second equation by 2:

3x | + | 2y | = | 19 | |||

2x |
+ | 2y |
= | 16 |

**Subtract** the second equation from the first equation:

x |
= |
3 |
|||||

2x | + | 2y | = | 16 |

**Yay! Now we know what x is!**

Next we see the 2nd equation has "2x", so let's halve it, and then subtract "x":

**Multiply** the second equation by **½** (i.e. divide by 2):

x | = | 3 | |||||

x |
+ |
y |
= |
8 |

**Subtract** the first equation from the second equation:

x | = | 3 | |||||

y |
= |
5 |

**Done!**

And the answer is:

x = 3 and y = 5

And here is the graph:

The blue line is where **3x + 2y = 19** is true

The red line is where **x + y = 8** is true

At x=3, y=5 (where the lines cross) they are **both** true. **That** is the answer.

Here is another example:

### Example:

- 2x − y = 4
- 6x − 3y = 3

Lay it out neatly:

2x | − | y | = | 4 | |||

6x | − | 3y | = | 3 |

**Multiply** the first equation by 3:

6x |
− | 3y |
= |
12 |
|||

6x | − | 3y | = | 3 |

**Subtract** the second equation from the first equation:

0 |
− | 0 |
= |
9 |
|||

6x | − | 3y | = | 3 |

0 − 0 = 9 ???

**What is going on here?**

Quite simply, there is no solution.

They are actually parallel lines: |

And lastly:

### Example:

- 2x − y = 4
- 6x − 3y = 12

Neatly:

2x | − | y | = | 4 | |||

6x | − | 3y | = | 12 |

**Multiply** the first equation by 3:

6x |
− | 3y |
= |
12 |
|||

6x | − | 3y | = | 12 |

**Subtract** the second equation from the first equation:

0 |
− | 0 |
= |
0 |
|||

6x | − | 3y | = | 3 |

0 − 0 = 0

**Well, that is actually TRUE! Zero does equal zero ...**

... that is because they are really the same equation ...

... so there are an Infinite Number of Solutions

They are the same line: |

And so now we have seen an example of each of the three possible cases:

**No**solution**One**solution**Infinitely many**solutions

## Solving By Elimination: 3 equations in 3 variables

Before we start on the next example, let's look at an improved way to do things.

Follow this method and we are less likely to make a mistake.

First of all, eliminate the variables **in order**:

- Eliminate
**x**s first (from equation 2 and 3, in order) - then eliminate
**y**(from equation 3)

So this is how we eliminate them:

We then have this "triangle shape":

Now start at the bottom and **work back up** (called "Back-Substitution")

(put in** z** to find **y**, then **z **and** y** to find **x**):

And we are solved:

ALSO, we will find it is easier to do **some** of the calculations in our head, or on scratch paper, rather than always working within the set of equations:

### Example:

- x + y + z = 6
- 2y + 5z = −4
- 2x + 5y − z = 27

Written neatly:

x | + | y | + | z | = | 6 | |||

2y | + | 5z | = | −4 | |||||

2x | + | 5y | − | z | = | 27 |

First, eliminate **x** from 2nd and 3rd equation.

There is no x in the 2nd equation ... move on to the 3rd equation:

**Subtract 2 times the 1st equation from the 3rd equation** (just do this in your head or on scratch paper):

And we get:

x | + | y | + | z | = | 6 | |||

2y | + | 5z | = | −4 | |||||

3y |
− | 3z |
= |
15 |

Next, eliminate **y** from 3rd equation.

We **could** subtract 1½ times the 2nd equation from the 3rd equation (because 1½ times 2 is 3) ...

... but we can **avoid fractions** if we:

- multiply the 3rd equation by
**2**and - multiply the 2nd equation by
**3**

and *then* do the subtraction ... like this:

And we end up with:

x | + | y | + | z | = | 6 | |||

2y | + | 5z | = | −4 | |||||

z |
= |
−2 |

We now have that "triangle shape"!

Now go back up again "back-substituting":

We know **z**, so **2y+5z=−4** becomes **2y−10=−4**, then** 2y=6**, so **y=3**:

x | + | y | + | z | = | 6 | |||

y |
= |
3 |
|||||||

z | = | −2 |

Then **x+y+z=6** becomes **x+3−2=6**, so **x=6−3+2=5**

x |
= |
5 |
|||||||

y | = | 3 | |||||||

z | = | −2 |

And the answer is:

x = 5

y = 3

z = −2

Check: please check for yourself.

## General Advice

Once you get used to the Elimination Method it becomes easier than Substitution, because you just follow the steps and the answers appear.

But sometimes Substitution can give a quicker result.

- Substitution is often easier for small cases (like 2 equations, or sometimes 3 equations)
- Elimination is easier for larger cases

And it always pays to look over the equations first, to see if there is an easy shortcut ... so experience helps.