Equations
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Equations describe, or model, relationships between numerical quantities. Equations compare two expressions that are equal. They are written using combinations of constants, variables, and coefficients, which you can think of as the parts of a mathematical "sentence." Constants, Variables, and Coefficients Constants have a fixed value. They fall into two main categories:
Variables are quantities that can change. In mathematical expressions, variables are often represented by the letters x, y, and z, or by a letter that reflects what the variable stands for, such as r for rate, P for principal, c for cost, or t for time. A coefficient is a number that is being multiplied by a variable. For example, a wage of $15 an hour could be represented as 15h, where 15 is the coefficient and h is the variable. |
When does a marketing brand manager need to use equations?
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Mathematical expressions are built from constants, variables, and coefficients, as well as mathematical operations. In the business world, these expressions can be used to model, or describe, quantitative relationships. For example, if you rent a car for a basic charge of $40 plus $59 per day, you can model this by choosing the variable d to represent the number of days the car is rented. The cost of renting the car is described by the following expression: $40 + $59d. In this expression, 40 is a constant, the addition symbol is a mathematical operation, 59 is a coefficient of d, and d is a variable. |
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Equations with One Unknown An equation is a mathematical statement that says two expressions are equal. An equation has a "left-hand side" and a "right-hand side"; each side represents an expression that is equal to the other. The form of an equation is likely very familiar to you; it is represented below.
Consider the example of renting a car for a basic charge of $40 plus $59 per day. This could be described by the expression $40 + $59d, which serves as the left-hand side of the equation. If your total cost of renting the car is $335, the right-hand side of the equation would be $335. Because these costs are equal, you can describe their relationship with the following equation:
To determine how many days the car was rented, you would need to find the value of the unknown variable, d, by isolating it on one side of the equation. Since the two sides of an equation are equal, you can isolate a variable by adding, subtracting, multiplying, or dividing both sides by the same value in order to cancel constants and coefficients. To solve the car rental equation, you would need to isolate the variable d. To begin, subtract the constant, $40, from each side. This will cancel out the $40 in the first expression. Next, divide each side by the coefficient, $59. This will cancel out the $59 in the first expression. The result shows that the car was rented for five days. The calculation is illustrated below.
Mathematical equations abound in the business world; indeed, businesses could hardly operate without employing them to solve a wide range of problems. For example, imagine you are a retailer who wishes to make a 30 percent margin on a computer mouse that costs you $20. What price should you charge for each mouse? Before setting up this equation, identify the variables and constants.
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The retail price (R) should be 30 percent above your cost of $20. To determine your retail price, set up the equation in the following manner and solve:
To make a 30 percent margin on a computer mouse that costs you $20, you must sell the mouse at a retail price of $26. Another way to set up this equation is to describe the fact that the retail price is 130 percent of your cost. You could describe this through the following equation: As you can see, you can construct an equation in more than one way to reach the correct solution. |
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Equations with Multiple Unknowns To solve for more than one unknown, you need to use more than one equation. These multiple equations are called systems of equations. In a system of equations, there must be the same number of equations as unknowns. Many business applications, such as determining supply and demand, often require systems of equations. If you remember that it's simple algebra at the base of these potentially complex problems, you can work through them and solve them easily. The substitution method is a common way to solve systems of equations. This method has five basic steps:
To illustrate these steps, consider how you would find the values of x and y in the following system of equations. |
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The solution to this system of equations is x equals 1 and y equals 3. Check your work by substituting these values into the original system of equations. |
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View the animation below for another illustration of solving equations with multiple unknowns.
Now, consider another example of solving equations with multiple unknowns from the business world. You work for a snack food company that produces two kinds of nuts in bulk quantities (batches). There are two processing steps, roasting and packing. The time available on the roasting and packing machines is limited by the need to perform maintenance and adjustments. The following chart shows how much time is required for processing each type of product and the time available on the two machines. In an effort to run the facility as efficiently as possible, your boss asks you to determine how much of each product should be produced to maximize the available machine time. To begin determining the right product mix, choose meaningful variable names for the products. p = batches of peanuts produced Next, set up equations for the roasting and packing machines. Each batch of peanuts takes 2 hours of roast time, each batch of cashews takes 3 hours, and there are 250 hours of roast time altogether. Each batch of peanuts takes 3 hours of packing time, each batch of cashews takes 1 hour, and there are 190 hours of pack time altogether. Therefore, these are the two equations. This business situation represents two equations with two unknowns; you can solve this system of equations using substitution. The first step is to choose a variable in the first equation and isolate it—which variable you choose does not matter. For this example, isolate c in the second equation. |
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The second step is to substitute this value of c, 190 – 3p, into the other equation so that there is only one variable. Then you can solve for p. The value for p is 45.714. Substitute this value into the first equation and solve for c. You maximize the time available on the machines by making 45 batches of peanuts and 52 batches of cashews. (Although you might assume that these numbers should be rounded up to 46 and 53, doing so would overutilize the equipment. Don't round up in this instance.) 1. Identify the constants, variables, and coefficients in the following equation: 425x – (300x + 2,000) = 5,750.
2. Solve for x. 3x = 5x – 16
3. Solve for x. 10x = 100a
4. Solve for x and y. 2x + 3y = 6 4x + 5y = 2
5. Solve for x and y. 2x + 5y = 4 3x + 4y = 5
6. A snowboard manufacturer sells boards for $500 each. The cost to produce snowboards includes $150 per board and $3,000 in overhead. Identify the constants, variables, and coefficients in this problem.
7. A woman has $5,000 invested in an account with a simple annual interest rate of 6.5 percent. How long should she keep this account in order to accumulate at least $500 in interest? (Note: Interest is equal to the product of principal, interest rate, and time; therefore, the interest equation is: I = Prt.)
8. A company makes overhead projectors for use in computer-aided presentations and sells them to retailers for $1,000 each. Suppose that during a six-month period, the company's costs are as follows: Fixed costs = $540,000; variable costs = $600 per projector. How many projectors must the company make to break even? In other words, how many projectors must the company make so that the cost of producing projectors equals the revenue obtained from selling all of the projectors?
9. A man has $7,000 to deposit. He purchases a certificate of deposit earning 8.5 percent simple interest, but it carries the stipulation that he must keep his money in this investment for six months. He puts the remainder of his $7,000 into an account that is more liquid and which earns 6 percent simple interest. If the total interest earned each year is $506.25, how much was initially invested in each account? (Note: Remember from question 7 that I = Prt.)
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Solution 1