Introduction
Systems equations refers to a set of algebraic equations which can be solved together to obtain some values for one, two, or more variables depending on given information. In business management and accounting systems equations have been widely used in cost, price, and volume of sales analysis while making projections for sales and profit margin (Wallace, nd). The most common type of system equation in business accounting for cost-price-volume (CPV) are linear equations. Non-linear equations are also applied to solve demand and supply relations based on various models. Economic models are all based on systems equations hence a part and parcel of economics. Systems equations can be solved through substitution, graphing, matrix, and simultaneous method (Gundersen, 2009). Simulated case will be used to show how system equation can be developed and applied in cost-price-volume analysis for two products.
Example
In campus canteen there are two types of soft drinks sold. Express revenue in a system equation given one brand sells at $0.50 and the other cost $0.60 per cup.
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System equation for two products can be expressed as “Ax + By = C” where A&B represent the constant and x&y represent variables. For an equation to be considered systems, each constant must have a variable but the value of the constant can be zero. System equation for revenue raised would be
0.50x + 0.6y = SALES
b- Calculate the units of x and y sold if revenue raised was $283 and total number units sold were 530
| Selling Price | quantity sold | Revenue |
| 0.5 | x | 0.5x |
| 0.6 | y | 0.6y |
| 530 | 283 |
System Equations Obtained
Equation I- x + y = 530
Equation II- 0.50x + 0.6y = $283
Equation can be simplified further as by multiplying by ten
10(0.50x + 0.6y = $283)
Equation I- 10x + 10y = 5300
Equation II - 5x + 6y = $ 2830
Solve through substitution method
Equation I - x + y = 530
x = 530-y – equation iii
Replace ‘x’ in equation II with equation iii above
Equation II - 5x + 6y = $ 2830
5(530 – y) + 6y = 2830
2650 -5y + 6y = 2830
6y – 5y = 2830 – 2650
Y = 180
Use equation iii to get units of ‘x’
X = 530-y = (530 – 180) = 350
| Selling Price | quantity sold | Revenue |
| 0.5 | 350 | 175 |
| 0.6 | 180 | 108 |
| 530 | 283 |
If we want to make $350 by raising the prices of soda to sell at a better ratio of 0.6 to the 0.5 soda, the new ratio would be
0.6: 0.5 = 6x:5y
Total units sold=6x +5y= 11 units
X + y =11
New prices x1 and y1 – Eq 1
Revenue = 6x1 + 5y1= 350 – Eq2
Let x1 and y1 be represented by x and y respectively
X + y = 11
Y= 11-x –Eq 3
Replace y in Eq 2 with Eq 3
6x +5(11-x)=350
6x + 55 -5x= 350
X= 295
To get the units of y sold, equate ratios to units calculated
0.6 – 295x
0.5- y
0.6y= 295(0.5)
Y= 305.
To make $350 at new prices, the firm should sell 295 units of x and 305 units of y.
References
Gundersen, E. (2009). Cost–Volume–Profit Analysis [pdf]. Retrieved from: http://www.pearsoncanada.ca/media/highered-showcase/multi-product-showcase
Tyler Wallace. (nd). Systems of Equations - Value Problems [pdf]. Retrieved from: http://www.wallace.ccfaculty.org/book/4.5%20Value.pdf
