System of Linear Equations, Statistics Problem Example

Bob bought 2 oranges and 4 apples and gave shillings 100 but was given a balance of shillings 64. In the same shop, Justus bought 5 oranges and 6 apples and paid 70. From the two purchases we can get the prices of each commodity.

Let an orange be represented by R and an apple by A.

Bob bought 2 oranges (2R) and 4 apples (4A). Justus bought 5 oranges (5R) and 6 apples (6A). Assuming that they only spent the money in the purchases of apples and oranges, we shall have;

2R + 4A = (100 – 64)…………………….Bob’s

5R + 6A = 70………………………………..Justus’s

Let Bob’s be equation pp and Justus’s be ppp

2R + 4A = 36…………………pp

5R + 6A = 70………………..ppp

To get the price of each commodity, we solve the two equations.

We multiply pp by 5 ands ppp by two (to eliminate R i.e. to make sure that the coefficients of R in both cases are equal). Thus we shall have;

10R + 20A = 180…………………..pp x 5

10R + 12A = 140………………….ppp x 2

We subtract the two to eliminate R and remain with A. subtraction 2ppp from 5pp we get;

(10R + 20A) – (10R + 12A) = 180 – 140

8A = 40

A = 5; to get R we substitute the value of A in any equation. When it is substituted in pp we get;

2R + 4(5) = 36

2R = 36 – 20

2R = 16

R = 8; Therefore the price of an apple is shillings 5 and Orange 8.

References

Frederic Miller, P. et al. (2009). System of Linear Equations. pg 13-14. Publishing House Ltd.

Girko, L. (1996). Theory of linear algebraic equations with random coefficients: Pg 43-45. Publisher Alperton Press, University of Michigan.