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2. Algebra
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Australia
Year 9

2.06 Modelling linear equations

Lesson
Worksheet
Practice
Lesson

Application of linear equations

Now that we know how:

  • to graph linear relationships
  • to find the equations of linear functions
  • to use algebra and graphs to extract information
  • to find intercepts and constant values, and
  • that the slope of a linear function represents constant change.

We can put this to use to solve a range of real life applications.

It's all the same mathematics, but this time you will have a context to apply it to.

Real-world situations often involve the collection of data, which is commonly displayed in a table of values. The independent variable is always displayed in the first row of the table, followed by the dependent variable in the second row.

We may first want to decide whether the values in the table represent a linear relationship. To do this, we could write an equation showing the linear relationship between the two data.

Once we know that the relationship between the variables is linear, we can work out the gradient and vertical intercept and express the relationship as a linear equation (or function).

Examples

Example 1

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

\text{Number of litres }(x)010203040
\text{Cost of petrol }(y)016.4032.8049.2065.60
a

Write an equation linking the number of litres of petrol pumped (x) and the cost of the petrol (y).

Worked Solution
Create a strategy

The cost of petrol is equal to the price per litre times the number of litres pumped.

Apply the idea

Find the price per litre:

\displaystyle y\displaystyle =\displaystyle \dfrac{16.40}{10}Divide 16.40 by 10
\displaystyle =\displaystyle 1.64Evaluate

Since we get the price per litre, we can write the equation as y=1.64x.

b

How much does petrol cost per litre?

Worked Solution
Create a strategy

Divide the cost of the petrol (y) by the number of litres (x).

Apply the idea
\displaystyle \text{Cost per litre}\displaystyle =\displaystyle \dfrac{16.40}{10}Divide 16.40 by 10
\displaystyle =\displaystyle \$1.64Evaluate
c

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

Worked Solution
Create a strategy

Using the equation from part (a), subtitute x=47.

Apply the idea
\displaystyle y\displaystyle =\displaystyle 1.64xWrite the equation
\displaystyle =\displaystyle 1.64 \times 47Substitute x=47
\displaystyle =\displaystyle \$77.08Evaluate
d

In the equation, y=1.64x, what does 1.64 represent?

A
The unit rate of cost of petrol per litre.
B
The number of litres of petrol pumped.
C
The total cost of petrol pumped.
Worked Solution
Create a strategy

By looking on part (a), 1.64 represents the rate at which y increases.

Apply the idea

The correct answer is option A: The unit rate of cost of petrol per litre.

Example 2

Kerry currently pays \$50 a month for her internet service. She is planning to switch to a fibre optic cable service.

a

Write an equation for the total cost T of Kerry's current internet service over a period of n months.

Worked Solution
Create a strategy

To find the total cost, we need to multiply the cost of internet a month by the number of months.

Apply the idea

T=\text{Total cost},\, n=\text{ Number of months},\, \$50=\text{Cost of internet per month}

T= 50n

b

For the fibre optic cable service, Kerry pays a one-off amount of \$1200 for the installation costs and then a monthly fee of \$25. Write an equation of the total cost T of Kerry's new internet service over n months.

Worked Solution
Create a strategy

Multiply the monthly fee by the number of months, and then add it to the installation fee.

Apply the idea

\text{Total cost}=T ,\,\text{Monthly fee} = \$25 ,\, \text{Number of months}=n,\,\text{Installation fee}=\$1200

T = 1200 + 25 n

c

Fill in the table of values for the total cost of the current internet service, given by T=50n.

n16121824
T\text{ (dollars)}
Worked Solution
Create a strategy

Substitute the value of n in T=50n.

Apply the idea
\displaystyle T\displaystyle =\displaystyle 50\times 1Substitute n=1
\displaystyle =\displaystyle 50Evaluate
\displaystyle T\displaystyle =\displaystyle 50\times 6Substitute n=6
\displaystyle =\displaystyle 300Evaluate
\displaystyle T\displaystyle =\displaystyle 50\times 12Substitute n=12
\displaystyle =\displaystyle 600Evaluate
\displaystyle T\displaystyle =\displaystyle 50\times 18Substitute n=18
\displaystyle =\displaystyle 900Evaluate
\displaystyle T\displaystyle =\displaystyle 50\times 24Substitute n=24
\displaystyle =\displaystyle 1200Evaluate

The complete table of values for the total cost of the current internet service:

n16121824
T\text{ (dollars)}503006009001200
d

Fill in the table of values for the total cost of the fibre optic cable service, given by T=25n+1200.

n16121824
T\text{ (dollars)}
Worked Solution
Create a strategy

Substitute the value of n in T=25n + 1200.

Apply the idea
\displaystyle T\displaystyle =\displaystyle 25 \times 1 + 1200Substitute n=1
\displaystyle =\displaystyle 1225Evaluate
\displaystyle T\displaystyle =\displaystyle 25 \times 6 + 1200Substitute n=6
\displaystyle =\displaystyle 1350Evaluate
\displaystyle T\displaystyle =\displaystyle 25 \times 12 + 1200Substitute n=12
\displaystyle =\displaystyle 1500Evaluate
\displaystyle T\displaystyle =\displaystyle 25 \times 18 + 1200Substitute n=18
\displaystyle =\displaystyle 1650Evaluate
\displaystyle T\displaystyle =\displaystyle 25 \times 24 + 1200Substitute n=24
\displaystyle =\displaystyle 1800Evaluate

The complete table of values for the total cost of the fibre optic internet service:

n16121824
T\text{ (dollars)}12251350150016501800
Idea summary

  • The independent variable is always displayed in the first row of the table, followed by the dependent variable, in the second row.

  • The slope is the constant rate of change. If the x-value is increasing by 1, then it will be the increase in the y-values.

  • The y-intercept is the value of y when x=0. This may not actually be given in the table, so you may have to count backwards using the rate of change.

Outcomes

AC9M9A05

use mathematical modelling to solve applied problems involving change including financial contexts; formulate problems, choosing to use either linear or quadratic functions; interpret solutions in terms of the situation; evaluate the model and report methods and findings

AC9M9M05

use mathematical modelling to solve practical problems involving direct proportion, rates, ratio and scale, including financial contexts; formulate the problems and interpret solutions in terms of the situation; evaluate the model and report methods and findings

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