6.7 Solving Systems of Linear Equations 359 Applications of Systems of Linear Equations Our first application involves a business application known as break-even analysis. Businesses use break-even analysis to determine how many units of an item must be sold for the business to “break even,” that is, for its total revenue to equal its total cost. Suppose we let the horizontal axis of a graph represent the number of units sold and the vertical axis represent dollars. Then linear equations for cost, C, and revenue, R, can both be graphed on the same axes (Fig. 6.28). Both C and R are expressed in dollars, and both depend on the number of units. Profit, P, is the difference between revenue, R, and cost, C. Thus, P R C. = − If revenue is greater than cost, the company makes a profit. If cost is greater than revenue, the company has a loss. Initially, the cost graph is higher than the revenue graph because of fixed overhead costs such as rent and utilities. During low levels of sales, the business suffers a loss (the cost graph is greater). During higher levels of sales, the business realizes a profit (the revenue graph is greater). The point at which the two graphs intersect is called the break-even point. At that number of units sold, revenue equals cost and the business breaks even. In Section 6.2, we introduced modeling. Recall that a mathematical model is an equation or system of equations that represents a real-life situation. In Examples 9 and 10, we develop equations that model real-life situations. We could now determine x by substituting 11 17 for y in either of the original equations. Instead, let’s solve for x by eliminating the variable y from the two original equations. To do so, we multiply the first equation by 3 and the second equation by 4. − = − = + = + = x y x y x y x y 3[3 4 8] gives 9 12 24 4[2 3 9] gives 812 36 − = + = = = x y x y x x 9 12 24 8 12 36 17 60 60 17 The solution to the system is , . 60 17 11 17 ( ) 7 Now try Exercise 47 Break-even point Total revenue Profit Loss Total cost Number of Units Sold Total Revenue and Total Cost ($) Figure 6.28 Example 9 Profit and Loss in Business At a collectibles show, Richard can sell model trains for $35. The costs for making the trains are a fixed cost of $200 and a production cost of $15 apiece. a) Write an equation that represents Richard’s revenue. Write an equation that represents Richard’s cost. b) How many model trains must Richard sell to break even? c) Write an equation for the profit formula. Use the formula to determine Richard’s profit if he sells 15 model trains. d) How many model trains must Richard sell to make a profit of $600? Solution a) Let x denote the number of model trains sold. The revenue is given by the equation R x 35 ($35 times the number of trains) = and the cost is given by the equation C x 200 15 ($200 plus $15 times the number of trains) = + Steve Mann/Shutterstock
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