## Table of Contents

## What is a Linear Function?

- A linear function is an equation with two variables that takes the shape of a straight line
- Linear functions have 2 variables: $x$ and $y$
- They usually take the form of $y=mx+b,$ where m and b are numerical values (e.g. 4, 2.8, $\pi$, $\sqrt[3]{7},$ etc.)
- $b$ can be equal to zero, but $m$ cannot
- Sometimes, linear equations will come in the form $0=ax+by+c$ or $ax+by+c=d$. In these cases, it’s usually best to rearrange the equation into the form of $y=mx+b$

## How do I Graph a Linear Function?

- construct a table of values, where the independent variable $(x)$ is the first row, and the dependent variable $(y)$ is the second row
- For example, for $y=2x+7$:

$x$ | -2 | -1 | 0 | 1 | 2 |
---|---|---|---|---|---|

$y$ | 3 | 5 | 7 | 9 | 11 |

- Draw a number plane with the $x$ variable on the horizontal axis, and the $y$ variable on the vertical axis
- Plot the points from your table of values onto the number plane
- Connect the points together with a straight line (USE A RULER!!!)

## How do I calculate the gradient of a linear function?

- Sometimes, you don’t have to! If the equation is in the form $y=mx+b,$ $m$ is the gradient ðŸ˜„
- If you’re given a table of values, you can use the
*slope-intercept formula*to calculate the value of $m$: $$\color{orange}{m=\frac{y_{2}-y_{1}}{x_{2}-x_{1}}}$$ - $y_2$ and $y_1$ are any two y values from the table of values, while $x_2$ and $x_1$ are the x-values which correspond to $y_2$ and $y_1$
- For example, using the table of values below:
$x$ -2 -1 0 1 2 $y$ 3 5 7 9 11 - We can take $y_2 =5,y_1=7,$ which means $x_2=-1$ and $x_1=0$. Substituting these into the formula: $$\begin{gather*}m=\frac{5-7}{-1-0} \\ =\frac{-2}{-1}=\textcolor{orange}{2}\end{gather*}$$
- Which matches the original equation we had $(y=\textcolor{orange}{2}{x+7)}$

- For example, using the table of values below:

## How do I find where a linear equation crosses the y-axis?

- Like with the gradient, the general form of a linear equation gives this away
- The value of $b$ is the y-coordinate of the y-intercept (the x-coordinate is 0)
- Therefore, for all linear equations in general form, the y-intercept (also known as $y_0$) is at $(0,b)$
- If you don’t have $b,$ find $m$ first, and then substitute $x=0,y=y_0$, and rearrange for $b$

### How can I determine if 2 lines are parallel?

- If two lines have the same gradient $(m),$ but different y-intercepts $(b),$ then they are parallel.

## What are simultaneous equations?

- Simultaneous equations are equations with more than 1 unknown, that have an infinite number of solutions.
- For example, $y=2x+2$ could be (1,4), (2,6), or any valid value of $x$.

- To solve simultaneous equations, we use a second equation alongside it.
- For example, $y=2x+2$ and $y=3x+1$.
- There is only one solution for which both the equations are true.

- As a general rule, for every unknown value, you need at least that many equations to find a solution.
- For linear equations, there will usually be 2 unknowns, so you need 2 equations to find 1 solution.

### How do we solve simultaneous equations?

- There are 2 main methods: graphical and algebraic.
- Graphical is simpler, but can be less precise, while algebraic takes more practice to get right, but once you do, it’s much quicker.
- You may be asked to solve simultaneous equations using a specific method, which is why we’ll go through both.
- We’ll use the pair from earlier ($y=2x+2$ and $y=3x+1$) for the examples.

#### Graphical Method

- This method is pretty straightforward: graph both equations and see where they intersect.
- The point of intersection is your solution.

- $(1,4)$ is the point of intersection.

#### Solving Algebraically by Substitution

- So there are actually 2 methods of algebraically solving simultaneous equations, but each has a slightly different use case. Substitution is slightly easier, so we’ll start with this one.

Solve $y=2x+2 \bbox[5px, border: 2px solid grey]{1},y=3x+1 \bbox[5px, border: 2px solid grey]{2}$ using the substitution method.

Rearrange one of the equations to make $x$ the subject.

- If one of the equations has $m=1,$ it’s usually easier to rearrange that one.
- We’ll rearrange $y=2x+2$.

$\begin{gather*}y=2x+2 \\ y-2=2x \\ \frac{y-2}{2}=x \bbox[5px, border: 2px solid grey]{3}\end{gather*}$

Substitute this into the equation you didn’t rearrange.

- Basically, replace $x$ in $\bbox[5px, border: 2px solid grey]{2}$ with equation $\bbox[5px, border: 2px solid grey]{3}$

$y=3\left(\frac{y-2}{2}\right)+1$

Rearrange to make $y$ the subject, then solve.

$\begin{gather*}y=\frac{3y-6}{2}+1 \\ y=\frac{3y}{2}-3+1\\y=\frac{3y}{2}-2\\y-\frac{3y}{2}=-2 \\ -\frac{y}{2}=-2 \\ y-2\times-2 \\ y=4\end{gather*}$

Take your newly found $y$ value, and substitute it into equation $\bbox[5px, border: 2px solid grey]{3}$ to find $x$

$x=\frac{4-2}{2}=\frac{2}{2}=1$

Answer the question.

Therefore $x=1, y=4.$

#### Solving Algebraically by Elimination

- This method is essentially subtraction.

Solve $y=2x+2 \bbox[5px, border: 2px solid grey]{1},y=3x+1 \bbox[5px, border: 2px solid grey]{2}$ using the elimination method.

- Convert equations into the form $ax+by=c$

For equation 1:

$\begin{gather*}0=2x+2-y \\ -2=2x-y \\ 2=y-2x\end{gather*}$

For equation 2:

$\begin{gather*}0=3x+1-y \\ -1=3x-y \\ 1=y-3x\end{gather*}$

Subtract one from the other

- It doesn’t really matter which you subtract from which, you get the same answer either way.

$\begin{gather*} \phantom{-y}2=y-2x \\ -\phantom{y}\underline{1=y-3x} \\ \phantom{-y}1=0y+1x \end{gather*}$

$\therefore x=1$

Substitute the resulting x value into one of the original equations

$y=(2\times1)+2=2+2=4$

Answer the question.

Therefore $x=1, y=4.$