There are practical or real life problems that need to be formulated as simultaneous linear equations that can be solved by the methods discussed above. Here are some examples.

Example

A farmer collected a total of L$11,000 by selling 3 cows and 5 sheep. Another farmer collected L$7,000 by selling one cow and 10 sheep. What is the price for a cow and a sheep? (Assume all cows have the same price and also the price of every sheep is the same).

Solution

Let x be the price of a cow and y be the price of a sheep.

Farmer I sold 3 cows for 3x and 5 sheep for 5y collecting a total of 11,000.
Which means, 3x + 5y = 11,000

Farmer II sold 1 cow for x and 10 sheep for 10y collecting a total of 7,000.
Which means, x + 10y = 7,000

When we consider these equations simultaneously, we get the following system of equations.

Multiplying the first equation by −2 to make the coefficients of y opposite, we get

Adding the equations we get:

Substituting x = 3,000 in one of the equations, say the second equation, we get,

3,000 + 10y = 7,000
10y = 4,000
y = 400.

Therefore, the solution is (3000, 400) showing that the price for a cow is L$3,000 and the price for a sheep is L$400 in the given unit of currency.