Linear Equations

New Zealand

Level 6 - NCEA Level 1

Lesson

We have already learnt that a rate is a ratio between two measurements with different units.

When we graph these rates, the rate of change can be understood as the gradient, steepness or slope of a line. Further, we look at the equations in gradient-intercept form (that is, $y=mx+b$`y`=`m``x`+`b`, where $m$`m` is the gradient), the larger the absolute value of $m$`m`, the steeper the slope of the line.

For example, a line with a slope of of $4$4 is steeper than a line with a slope of $\frac{2}{3}$23. Similarly, a line with a slope of $-2$−2 is steeper than a line with a slope of $1$1, even though one is positive and one is negative.

The rate of change in a graph can be positive or negative.

The lines below have positive rates of change. Notice how as the values on the $x$`x`-axis increase, the values on the $y$`y`-axis also increase.

These next graphs have negative rates of change. Unlike graphs with a positive gradient, as the values on the $x$`x`-axis increase, the values on the $y$`y`-axis decrease.

The rate of change of a line is a measure of how steep it is. In mathematics we also call this the gradient.

The rate of change is a single value that describes:

- if a line is increasing (has positive gradient)
- if a line is decreasing (has negative gradient)
- how far up or down the line moves (change in the $y$
`y`value) with every unit step to the right (for every 1 unit increase in $x$`x`)

Take a look at this line for example. I've highlighted the horizontal and vertical steps.

We call the horizontal measurement the run** **and the vertical measurement the rise**.**

Here, for every $1$1 step across (run of $1$1), the line goes up $2$2 (rise of $2$2). This line has a rate of change of $2$2.

Sometimes it is difficult to measure how far the line goes up or down (how much the $y$`y` value changes) in 1 horizontal unit. In this case we calculate the gradient by using a formula.

$\text{Gradient }=\frac{\text{rise }}{\text{run }}$Gradient =rise run

Where you take any two points on the line whose coordinates are known or can be easily found, and look for the rise and run between them.

If you have a pair of coordinates, such as $\left(3,6\right)$(3,6) and $\left(7,-2\right)$(7,−2) we can find the gradient of the line between these points also using the rule.

$m=\frac{y_2-y_1}{x_2-x_1}$`m`=`y`2−`y`1`x`2−`x`1

Let's substitute our coordinates into this formula:

$m$m |
$=$= | $\frac{y_2-y_1}{x_2-x_1}$y2−y1x2−x1 |

$=$= | $\frac{6-\left(-2\right)}{3-7}$6−(−2)3−7 | |

$=$= | $\frac{6+2}{3-7}$6+23−7 | |

$=$= | $\frac{8}{-4}$8−4 | |

$=$= | $-2$−2 |

So the rate of change between these coordinates is $-2$−2.

**horizontal lines have a rate of change of 0**

Why?

Horizontal lines have NO rise value. The $rise=0$`r``i``s``e`=0. So:

$\text{Rate of change }$Rate of change | $=$= | $\frac{\text{Rise }}{\text{Run }}$Rise Run |

$=$= | $\frac{0}{\text{Run }}$0Run | |

$=$= | $0$0 |

It doesn't matter what the run is, the gradient will be $0$0.

Why?

Vertical lines have NO run value. The $run=0$`r``u``n`=0. So:

$\text{Rate of Change }$Rate of Change | $=$= | $\frac{\text{Rise }}{\text{Run }}$Rise Run |

$=$= | $\frac{\text{Rise }}{0}$Rise 0 |

It doesn't matter what the rise is, any division by $0$0 results in the value being undefined.

Remember

Description of rate of change: $\text{Rate of change }=\frac{\text{rise }}{\text{run }}$Rate of change =rise run

Gradient of Vertical Line = undefined

Gradient of Horizontal Line = $0$0

Rate of change can also be seen in everyday situations. Let's work through some worked examples.

What kind of slope does the following line have?

Positive

ANegative

BUndefined

CZero

D

After Mae starts running, her heartbeat increases at a constant rate.

Complete the table.

Number of minutes passed ($x$ `x`)$0$0 $2$2 $4$4 $6$6 $8$8 $10$10 $12$12 Heart rate ($y$ `y`)$49$49 $55$55 $61$61 $67$67 $73$73 $79$79 $\editable{}$ What is the constant rate the heart beat is increasing at?

Which one of the following equations describes the relationship between the number of minutes passed ($x$

`x`) and Mae’s heartbeat ($y$`y`)?$y=49x-3$

`y`=49`x`−3A$y=49x+3$

`y`=49`x`+3B$y=3x-49$

`y`=3`x`−49C$y=3x+49$

`y`=3`x`+49DIn the equation, $y=3x+49$

`y`=3`x`+49, what does $3$3 represent?The change in one minute of Mae’s heartbeat.

AThe total time Mae has run.

BThe total distance Mae has run.

C

Petrol costs a certain amount per litre. The table shows the cost of various amounts of petrol.

Number of litres ($x$x) |
$0$0 | $10$10 | $20$20 | $30$30 | $40$40 |
---|---|---|---|---|---|

Cost of petrol ($y$y) |
$0$0 | $16.40$16.40 | $32.80$32.80 | $49.20$49.20 | $65.60$65.60 |

Write an equation linking the number of litres of petrol pumped ($x$

`x`) and the cost of the petrol ($y$`y`).How much does petrol cost per litre?

How much would $47$47 litres of petrol cost at this unit price?

In the equation, $y=1.64x$

`y`=1.64`x`, what does $1.64$1.64 represent?The unit rate of cost of petrol per litre.

AThe number of litres of petrol pumped.

BThe total cost of petrol pumped.

C