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Linear and Nonlinear Systems of Equations (Day 2) 7.1
Copyright © Cengage Learning. All rights reserved.
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Objectives Use a graphical approach to solve systems of equations in two variables. Use systems of equations to model and solve real-life problems.
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Graphical Approach to Finding Solutions
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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.
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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
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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
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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.
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Applications
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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.
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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?
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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.
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7.1 Example – Worked Solutions
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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
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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 = 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.
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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
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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 .
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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.
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