6.1/6.2 SYSTEMS OF LINEAR EQUATIONS Copyright © Cengage Learning. All rights reserved.

Use the method of substitution to solve systems of linear equations in two variables. Use the method of elimination to solve systems of linear 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

The Method of Substitution

The Method of Substitution 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.

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

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.

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

The Method of Substitution

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.

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.

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.

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.

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.

Another Example – Using Substitution Solve by substitution: x + 5y = 17 x – 6y = –16 Let’s solve the top equation for x: x = 17 – 5y Next, substitute 17 – 5y for x in the bottom equation: (17 – 5y) – 6y = –16 17 – 11y = –16 – 11y = –33 y = 3 Then, back-substitute 3 for y in one of the original equations: Let’s pick Equation 1: x + 5(3) = 17 x + 15 = 17

Example (continued) Solve for x: x = 2 So, our solution set is (2, 3). Check solutions in both equations: Equation 1: x + 5y = 17 2 + 5(3) = 17 2 + 15 = 17 17 = 17 Solution checks Equation 2: x – 6y = –16 2 – 6(3) = –16 2 – 18 = –16 –16 = –16 Solution checks

The Method of Elimination

The Method of Elimination We have studied two methods for solving a system of equations: substitution and graphing. Now we will study the method of elimination. The key step in this method is to obtain, for one of the variables, coefficients that differ only in sign so that adding the equations eliminates the variable. Equation 1 Equation 2 Add equations.

The Method of Elimination Note that by adding the two equations, you eliminate the x -terms and obtain a single equation in y. Solving this equation for y produces y = 2, which you can then back substitute into one of the original equations to solve for x.

The Method of Elimination To perform the Method of Elimination on a system of two linear equations in two variables x and y, do the following steps: Multiply either one or both equations by coefficients so that when the equations are added vertically, one of the variables cancels out. Examples: If you have 3y in the top equation and y in the bottom equation, simply multiply the bottom equation by -3. Adding vertically will eliminate the y terms. If you have 5x in the top equation and 2x in the bottom equation, you could multiply the top by 2 and the bottom by -5. Thus, when adding the equations vertically, 10x and -10x will cancel out. Add the equations vertically. One of the variables should cancel out. Solve the equation that results from Step 2. Back-substitute the solution into one of the original equations to find the solution for the other variable. Check your solutions in both of the original equations.

Example 2 – Solving a System of Equations by Elimination Solve the system of linear equations. Solution: For this system, you can obtain coefficients for the y terms that differ only in sign by multiplying Equation 2 by 4. Equation 1 Equation 2 Write Equation 1. Multiply Equation 2 by 4. Add equations. Solve for x.

By back-substituting into Equation 1, you can solve for y. 2x – 4y = –7 – 1 – 4y = –7 – 4y = –6 The solution is cont’d Example 2 – Solution Write Equation 1. Substitute -1/2 for x. Solve for y. Add 1 to both sides. Simplify.

Check this in the original system, as follows. Check: 2x – 4y = –7 –1 – 6 = –7 5x + y = –1 cont’d Example 2 – Solution Write original Equation 1. Substitute into Equation 1. Equation 1 checks. Write original Equation 2. Substitute into Equation 2. Equation 2 checks.

The Method of Elimination In Example 2, the two systems of linear equations (the original system and the system obtained by multiplying by constants) are called equivalent systems because they have precisely the same solution set.

Another Example of Using Elimination Solve the System using Elimination: 2x + 5y = 4 5x + 8y = 5 Let’s eliminate the x’s. So, we shall multiply the top equation by 5, and the bottom equation by -2. 5(2x + 5y = 4) 10x + 25y = 20 -2(5x + 8y = 5) -10x – 16y = -10 Add the equations vertically: 9y = 10

Example (continued) Solve for y: y = 10/9 Back substitute y = 10/9 into one of the original equations. Let’s substitute 10/9 for y in Equation 1: Equation 1: 2x + 5y = 4 Substitute: 2x + 5(10/9) = 4 Solve for x: 2x + 50/9 = 4 2x = -14/9 x = -14/18 or -7/9 So, the solution to the system of equations is (-7/9, 10/9). Now, let’s check the solution set in both equations…

Example (continued) Equation 1: 2x + 5y = 4 Check solution: 2(-7/9) + 5(10/9) = 4 -14/9 + 50/9 = 4 36/9 = 4 4 = 4 Solution checks. Equation 2: 5x + 8y = 5 Check solution: 5(-7/9) + 8(10/9) = 5 -35/9 + 80/9 = 5 45/9 = 5 5 = 5 Solution checks.

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 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 point Two intersection points No intersection points Figure 6.1 Figure 6.2 Figure 6.3

Example – Solving a System of Equations Graphically Solve the system of equations. x – y = 2 (1/2)x + 3y = 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 (2, 0) is the solution point (see picture). Equation 1 Equation 2 To graph these lines on the TI-84, change x – y = 2 to y = x – 2 and change (1/2)x + 3y = 1 to y = (-1/6)x + 1/3.

Example 5 – Solution You can check this solution as follows. Check (2, 0) in Equation 1: x – y = 2 2 – 0 = 2 2 = 2 Check (2, 0) in Equation 2: (1/2)x + 3y = 1 (1/2)(2) + 3(0) = 1 1 + 0 = 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

Graphical Interpretation of Solutions

Graphical Interpretation of Solutions It is possible for a general system of equations to have exactly one solution, two or more solutions, or no solution. If a system of linear equations has two different solutions, it must have an infinite number of solutions.

Graphical Interpretation of Solutions A system of linear equations is consistent if it has at least one solution. A consistent system with exactly one solution is independent, whereas a consistent system with infinitely many solutions is dependent. A system is inconsistent if it has no solution.

Example 4 – Recognizing Graphs of Linear Systems Match each system of linear equations with its graph in Figure 6.7. Describe the number of solutions and state whether the system is consistent or inconsistent. Figure 6.7

Example 4 – Solution a. The graph of system (a) is a pair of parallel lines (ii). The lines have no point of intersection, so the system has no solution. The system is inconsistent. b. The graph of system (b) is a pair of intersecting lines (iii). The lines have one point of intersection, so the system has exactly one solution. The system is consistent. c. The graph of system (c) is a pair of lines that coincide (i). The lines have infinitely many points of intersection, so the system has infinitely many solutions. The system is consistent.

Finding Inconsistent Systems Algebraically Consider the following example of an inconsistent system. Solve by using substitution: x + 4y = 8 2x + 8y = 5 By solving the top equation for x, we have x = 8 – 4y. Then, if we substitute 8 – 4y for x in the bottom equation, we get 2(8 – 4y) + 8y = 5 16 – 8y + 8y = 5 16 = 5 FALSE 16 is clearly not equal to 5, so there is no solution to this system.

Finding A Dependent System Algebraically Consider the following consistent system of equations: 4x – 6y = 5 -12x + 18y = -15 Let’s solve by elimination. Multiply the top equation by 3 so that the x terms will eventually cancel out when we add vertically. 12x – 18y = 15 -12x + 18y = -15 Add: 0 = 0 Since it’s obviously true that 0 = 0, then there are infinitely many solutions to this system of equations. Note that the two equations represent the same line.

Applications of Systems of Linear Equations

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 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

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 .

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.

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.

Applications At this point, you may be asking the question “How can I tell which application problems can be solved using a system of linear equations?” The answer comes from the following considerations. 1. Does the problem involve more than one unknown quantity? 2. Are there two (or more) equations or conditions to be satisfied? If one or both of these situations occur, the appropriate mathematical model for the problem may be a system of linear equations.

Example 8 – An Application of a Linear System An airplane flying into a headwind travels the 2000-mile flying distance between Chicopee, Massachusetts and Salt Lake City, Utah in 4 hours and 24 minutes. On the return flight, the same distance is traveled in 4 hours. Find the airspeed of the plane and the speed of the wind, assuming that both remain constant. Solution: The two unknown quantities are the speeds of the wind and the plane.

If r1 is the speed of the plane and r2 is the speed of the wind, then r1 – r2 = speed of the plane against the wind r1 + r2 = speed of the plane with the wind as shown in Figure 6.10. cont’d Example 8 – Solution Figure 6.10

Using the formula distance = (rate)(time) for these two speeds, you obtain the following equations. 2000 = (r1 – r2) 2000 = (r1 + r2)(4) These two equations simplify as follows. cont’d Example 8 – Solution Equation 1 Equation 2

To solve this system by elimination, multiply Equation 2 by 11. cont’d Example 8 – Solution Write Equation 1. Multiply Equation 2 by 11. Add equations.

So, miles per hour and miles per hour. cont’d Example 8 – Solution Speed of plane Speed of wind