Linear and Nonlinear Systems of Equations (Day 2) 7.1 Copyright © Cengage Learning. All rights reserved.

Objectives Use a graphical approach to solve systems of equations in two variables. Use systems of equations to model and solve real-life problems.

Graphical Approach to Finding Solutions

Graphical Approach to Finding Solutions A system of two equations in two unknowns can have exactly one solution, more than one solution, or no solution. By using a graphical method, you can gain insight about the number of solutions and the location(s) of the solution(s) of a system of equations by graphing each of the equations in the same coordinate plane. The solutions of the system correspond to the points of intersection of the graphs.

Graphical Approach to Finding Solutions For instance, the two equations in Figure 7.1 graph as two lines with a single point of intersection; the two equations in Figure 7.2 graph as a parabola and a line with two points of intersection; and the two equations in Figure 7.3 graph as a parabola and a line with no points of intersection. One intersection point Two intersection points No intersection points Figure 7.1 Figure 7.2 Figure 7.3

Example 5 – Solving a System of Equations Graphically Solve the system of equations. y = ln x x + y = 1 Equation 1 Equation 2 Figure 7.4

Graphical Approach to Finding Solutions Example 5 shows the benefit of a graphical approach to solving systems of equations in two variables. Notice that by trying only the substitution method in Example 5, you would obtain the equation x + ln x = 1. It would be difficult to solve this equation for x using standard algebraic techniques.

Applications

Applications The total cost C of producing x units of a product typically has two components—the initial cost and the cost per unit. When enough units have been sold so that the total revenue R equals the total cost C, the sales are said to have reached the break-even point. You will find that the break-even point corresponds to the point of intersection of the cost and revenue curves.

Example 6 – Break-Even Analysis A shoe company invests $300,000 in equipment to produce a new line of athletic footwear. Each pair of shoes costs $5 to produce and sells for $60. How many pairs of shoes must the company sell to break even?

Applications Another way to view the solution in Example 6 is to consider the profit function P = R – C. The break-even point occurs when the profit is 0, which is the same as saying that R = C.

7.1 Example – Worked Solutions

Example 5 – Solving a System of Equations Graphically Solve the system of equations. y = ln x x + y = 1 Solution: There is only one point of intersection of the graphs of the two equations, and (1, 0) is the solution point (see Figure 7.4). Equation 1 Equation 2 Figure 7.4

Example 5 – Solution cont’d Check this solution as follows. Check (1, 0) in Equation 1: y = ln x 0 = ln 1 0 = 0 Check (1, 0) in Equation 2: x + y = 1 1 + 0 = 1 1 = 1 Write Equation 1. Substitute for x and y. Solution checks in Equation 1. Write Equation 2. Substitute for x and y. Solution checks in Equation 2.

Example 6 – Break-Even Analysis A shoe company invests $300,000 in equipment to produce a new line of athletic footwear. Each pair of shoes costs $5 to produce and sells for $60. How many pairs of shoes must the company sell to break even? Solution: The total cost of producing x units is C = 5x + 300,000. Equation 1

Example 6 – Solution cont’d The revenue obtained by selling x units is R = 60x. Because the break-even point occurs when R = C, you have C = 60x, and the system of equations to solve is C = 5x + 300,000 C = 60x Equation 2 .

Example 6 – Solution cont’d Solve by substitution. 60x = 5x + 300,000 55x = 300,000 x ≈ 5455 So, the company must sell about 5455 pairs of shoes to break even. Substitute 60x for C in Equation 1. Subtract 5x from each side. Divide each side by 55.