*Linear models can be used in practical situations.*

Nathias is a computer salesperson. He is paid $450 a week and receives 5% commission on his sales, How much does he earn if his sales are:

a) $0

b) $2000

A linear equation should be constructed given that $P is Nathias’ earnings and s is his amount of sales.

P = 0.05s + 450

From this, *s* can be substituted to find *P* for the appropriate sales. For $0 sales, Nathias will earn $450. For $2000 of sales, Nathias will earn $550. This information can also be represented in a graph.

Simultaneous equations must be solved when there is one pair of numbers to satisfy two linear relations. If both equations are graphed, the solution to the simultaneous equations will be the point where the two lines intersect. In Further Mathematics, it is recommended that a CAS Cx calculator or alternative, is used to solve simultaneous equations.

In this method, one equation is substituted into another in order to minimise the variables to one.

*Solve the following pair of simultaneous equations:
Equation 1: y = 3x + 4
Equation 2: y = x + 8*

The *y *of Equation 2 can be substituted as the *y* for Equation 1 or vice versa. By doing this, a new equation is produced:

This can be easily solved for x.

When *x* = 2, substitute this into either Equation 1 or Equation 2.

Equation 2:

y = 2 + 8

y = 10

Therefore, the solution to this pair of simultaneous equations is: *x* = 2, *y *= 10.

In a real life situation, if *y *= the price of bananas and *x* represented the price of apples, these equations could represent the total price of the fruits in two different transactions. The objective can be to find the cost of one banana and one apple. By solving the simultaneous equations, it was found that an apple was $2 and a banana was $10.

If the coefficient of one or the unknowns is the same in both equations, subtracting one equation from the other will minimise the variables to one. This allows the equation to be solved. In some cases, it is necessary to multiple one or both of the equations so that one of the coefficients of either *x* or *y* is the same in the two equations.

*Solve the following simultaneous equations:
Equation 1: 3x + 4y = 12
Equation 2: 3x + 9y = 27*

*x* can be eliminated. In order to do this, Equation 1 should be subtracted from Equation 2. The other way around will also work out. This results in the equation:

When *y* = 3,

3x + 4(3) = 12

x = 0

Therefore the solution to this pair of simultaneous equations is *x* = 0 and *y* = 3.

It costs a certain amount of money to manufacture products. Retailers must take this into account when setting their prices as they would need to sell a certain number of their item in order to equate (or ‘break-even’) with the manufacture price in order to start profiting.

The cost (C) associated with making a certain calculator is given by:

C = 5n + 2000

where n represents the number of calculators manufactured.

Each calculator is sold for $54. Calculate the cost and revenue if 500 calculators are made. How many must be sold to break-even?

From this information it can be inferred that the equation for revenue is R = 54n.

If 500 calculators are made, the cost for this is:

The revenue would be:

In order to break even, the cost must equal the revenue. This means,

**See also:**

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