1. 1 Copyright © Cengage Learning. All rights reserved. 6 Systems of Equations and Inequalities

2. 6.1 LINEAR AND NONLINEAR SYSTEMS OF EQUATIONS Copyright © Cengage Learning. All rights reserved.

3. 3 Use the method of substitution to solve systems of linear equations in two variables. Use the method of substitution to solve systems of nonlinear equations in two variables. Use a graphical approach to solve systems of equations in two variables. Use systems of equations to model and solve real-life problems. What You Should Learn

4. 4 The Method of Substitution

5. 5 Most problems have involved either a function of one variable or a single equation in two variables. However, many problems in science, business, and engineering involve two or more equations in two or more variables. To solve such problems, you need to find solutions of a system of equations.

6. 6 The Method of Substitution Here is an example of a system of two equations in two unknowns. 2x + y = 5 3x – 2y = 4 A solution of this system is an ordered pair that satisfies each equation in the system. Finding the set of all solutions is called solving the system of equations. Equation 1 Equation 2

7. 7 The Method of Substitution For instance, the ordered pair (2, 1) is a solution of this system. To check this, you can substitute 2 for x and 1 for y in each equation. Check (2, 1) in Equation 1 and Equation 2: 2x + y = 5 2(2) + 1 ≟ 5 4 + 1 = 5 3x – 2y = 4 3(2) – 2(1) ≟ 4 6 – 2 = 4 Write Equation 1. Substitute 2 for x and 1 for y. Solution checks in Equation 1. Write Equation 2. Substitute 2 for x and 1 for y. Solution checks in Equation 2.

8. 8 The Method of Substitution We will study four ways to solve systems of equations, beginning with the method of substitution. Method Type of System 1. Substitution Linear or nonlinear, two variables 2. Graphical method Linear or nonlinear, two variables 3. Elimination Linear, two variables 4. Gaussian elimination Linear, three or more variables

9. 9 The Method of Substitution

10. 10 Example 1 – Solving a System of Equations by Substitution Solve the system of equations. x + y = 4 x – y = 2 Solution: Begin by solving for y in Equation 1. y = 4 – x Next, substitute this expression for y into Equation 2 and solve the resulting single variable equation for x. x – y = 2 Equation 1 Equation 2 Solve for y in Equation 1. Write Equation 2.

11. 11 Example 1 – Solution x – (4 – x) = 2 x – 4 + x = 2 2x = 6 x = 3 Finally, you can solve for y by back-substituting x = 3 into the equation y = 4 – x, to obtain y = 4 – x y = 4 – 3 cont’d Substitute 4 – x for y. Distributive Property Combine like terms. Divide each side by 2. Write revised Equation 1. Substitute 3 for x.

12. 12 Example 1 – Solution y = 1. The solution is the ordered pair (3, 1). You can check this solution as follows. Check: Substitute (3, 1) into Equation 1: x + y = 4 3 + 1 ≟ 4 4 = 4 cont’d Write Equation 1. Solve for y. Substitute for x and y. Solution checks in Equation 1.

13. 13 Example 1 – Solution Substitute (3, 1) into Equation 2: x – y = 2 3 – 1 ≟ 2 2 = 2 Because (3, 1) satisfies both equations in the system, it is a solution of the system of equations. cont’d Write Equation 2. Substitute for x and y. Solution checks in Equation 2.

14. 14 The Method of Substitution The term back-substitution implies that you work backwards. First you solve for one of the variables, and then you substitute that value back into one of the equations in the system to find the value of the other variable.

15. 15 Nonlinear Systems of Equations

16. 16 Nonlinear Systems of Equations The equations in Example 1 is linear. The method of substitution can also be used to solve systems in which one or both of the equations are nonlinear.

17. 17 Example 3 – Substitution: Two-Solution Case Solve the system of equations. 3x 2 + 4x – y = 7 2x – y = –1 Solution: Begin by solving for y in Equation 2 to obtain y = 2x + 1. Next, substitute this expression for y into Equation 1 and solve for x. 3x 2 + 4x – (2x + 1) = 7 3x 2 + 2x – 1 = 7 Equation 1 Equation 2 Substitute 2x + 1 for y in Equation 1. Simplify.

18. 18 Example 3 – Solution 3x 2 + 2x – 8 = 0 (3x – 4)(x + 2) = 0 Back-substituting these values of x to solve for the corresponding values of y produces the solutions and cont’d Write in general form. Factor. Solve for x.

19. 19 Graphical Approach to Finding Solutions

20. 20 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.

21. 21 Graphical Approach to Finding Solutions For instance, the two equations in Figure 6.1 graph as two lines with a single point of intersection; the two equations in Figure 6.2 graph as a parabola and a line with two points of intersection; and the two equations in Figure 6.3 graph as a line and a parabola that have no points of intersection. One intersection pointTwo intersection pointsNo intersection points Figure 6.1 Figure 6.2Figure 6.3

22. 22 Example 5 – Solving a System of Equations Graphically Solve the system of equations. y = ln x x + y = 1 Solution: Sketch the graphs of the two equations. From the graphs of these equations, it is clear that there is only one point of intersection and that (1, 0) is the solution point (see Figure 6.4). Equation 1 Equation 2 Figure 6.4

23. 23 Example 5 – Solution You can 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. cont’d

24. 24 Graphical Approach to Finding Solutions Example 5 shows the value of a graphical approach to solving systems of equations in two variables. Notice what would happen if you tried 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.

25. 25 Applications

26. 26 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.

27. 27 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 is sold for $60. How many pairs of shoes must be sold before the business breaks even? Solution: The total cost of producing x units is C = 5x + 300,000. Equation 1

28. 28 Example 6 – Solution 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 cont’d Equation 2.

29. 29 Example 6 – Solution 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. cont’d Substitute 60x for C in Equation 1. Subtract 5x from each side. Divide each side by 55.

30. 30 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.