3.1 Systems of Equations
Back when studying linear equations, we found the intersection of two lines. Doing so allowed us to solve interesting problems by finding a pair of values that satisfied two different equations. While we didn’t call it this at the time, we were solving a system of equations. To start out, we’ll review an example of the type of problem we’ve solved before.
Example of Breaking Even
A small business produces soap and lotion gift baskets. Labor, utilities, and other fixed expenses cost $6,000 a month. Each basket costs $8 to produce, and sells for $20. How many baskets does the company need to sell each month to break even?
In business terms, “break even” means for revenue (money brought in) to equal costs. While this problem can be approached in several ways, we’ll approach here by first creating two linear functions, one for the costs, and another for revenue.
Let’s define n to be the number of gift baskets the company sells in a month. There are $6,000 of fixed costs each month, and costs increase by $8 for each basket, so we can write the linear function for costs, C, as:
Each sale brings in $20, so the revenue, R, after selling n baskets will be:
To find the break-even point, we are looking for the number of baskets where the revenue will equal the costs. In other words, if we were to graph the two linear functions, we are looking for the point that lies on both lines; the solution is the point that satisfies both equations.
In this case we could probably solve the problem from the graph itself, but we can also solve it algebraically by setting the equations equal:
The break even point is at 500 baskets. The company must sell 500 baskets a month, at which point their revenue of $10,000 will cover their total costs of $10,000.
The example above illustrates one type of system of equations, one where both equations are given in functional form. When the equations are written this way, it is easy to solve the system using substitution, by setting the two outputs equal, and solving for the input. However, many system of equations problems aren’t written this way.
Another Example of Solving a System of Equations
A company produces a basic and premium version of its product. The basic version requires 20 minutes of assembly and 15 minutes of painting. The premium version requires 30 minutes of assembly and 30 minutes of painting. If the company has staffing for 3,900 minutes of assembly and 3,300 minutes of painting each week. If the company wants to fully utilize all staffed hours, how many of each item should they produce?
Notice first that this problem has two variables, or two unknowns – the number of basic products to make, and the number of premium products to make. There are also two constraints – the hours of assembly and the hours of painting available. This is going to give us two equations in two unknowns, what we call a 2 by 2 system of equations.
We’ll start by defining our variables:
b: the number of basic product