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CHAPTER 7 Algebra: Graphs, Functions, and Linear Systems © 2010 Pearson Prentice Hall. All rights reserved. 7.3 Systems of Linear Equations in Two Variables © 2010 Pearson Prentice Hall. All rights reserved. 2 Objectives 1. Decide whether an ordered pair is a solution of a linear system. 2. Solve linear systems by graphing. 3. Solve linear systems by substitution. 4. Solve linear systems by addition. 5. Identify systems that do not have exactly one ordered-pair solution. 6. Solve problems using systems of linear equations. © 2010 Pearson Prentice Hall. All rights reserved. 3 Systems of Linear Equations & Their Solutions • Two linear equations are called a system of linear equations or a linear system. • A solution to a system of linear equations in two variables is an ordered pair that satisfies both equations in the system. © 2010 Pearson Prentice Hall. All rights reserved. 4 Example 1: Determining Whether an Ordered Pair is a Solution of a System Determine whether (1,2) is a solution of the system: 2x – 3y = −4 2x + y = 4 Solution: Because 1 is the x-coordinate and 2 is the y-coordinate of (1,2), we replace x with 1 and y with 2. © 2010 Pearson Prentice Hall. All rights reserved. 5 Example 1 continued 2x – 3y = − 4 2(1) – 3(2) = −4 ? 2–6=−4 ? − 4 = − 4, TRUE 2x + y = 4 2(1) + 2 = 4 ? 2+2=4 ? 4 = 4, TRUE The pair (1,2) satisfies both equations; it makes each equation true. Thus, the pair is a solution of the system. © 2010 Pearson Prentice Hall. All rights reserved. 6 Graphing 2 lines • 2x – 3y = -4 2x + 4 = 3y y = (2/3)x + (4/3) © 2010 Pearson Prentice Hall. All rights reserved. • 2x + y = 4 y = -2x + 4 1 Solving Linear Systems by Graphing • For a system with one solution, the coordinates of the point of intersection of the lines is the system’s solution. © 2010 Pearson Prentice Hall. All rights reserved. 8 Example 2: Solving Linear Systems by Graphing Solve by graphing: x + 2y = 2 x – 2y = 6. Solution: We find the solution by graphing both x + 2y = 2 and x – 2y = 6 in the same rectangular coordinate system. We will use intercepts to graph each equation. © 2010 Pearson Prentice Hall. All rights reserved. 9 Example 2 Continued x + 2y = 2: x-intercept: Set y = 0. x+2·0=2 x=2 The line passes through (2,0). y-intercept: Set x = 0. 0 + 2y = 2 2y = 2 y=1 The line passes through (0,1). We will graph x + 2y = 2 as a blue line. © 2010 Pearson Prentice Hall. All rights reserved. 10 Example 2 Continued x – 2y = 6: x-intercept: Set y = 0. x–2·0=6 x=6 The line passes through (6,0). y-intercept: Set x = 0. 0 – 2y = 6 −2y = 6 y = −3 The line passes through (0,−3). We will graph x – 2y = 6 as a red line. © 2010 Pearson Prentice Hall. All rights reserved. 11 Example 2 Continued We see the two graphs intersect at (4,−1). Hence, this is the solution to the system. We can check this by substituting in (4,−1) into each equation and verifying That the solution is true for both equations. We leave this to the student. © 2010 Pearson Prentice Hall. All rights reserved. 12 Solving Linear Systems by the Substitution Method © 2010 Pearson Prentice Hall. All rights reserved. 13 Example 3: Solving a System by Substitution Solve by the substitution method: y = −x – 1 4x – 3y = 24. Solution: Step 1 Solve either of the equations for one variable in terms of the other. This step has been done for us. The first equation, y = −x – 1, is solved for y in terms of x. © 2010 Pearson Prentice Hall. All rights reserved. 14 Example 3 continued Step 2 Substitute the expression from step 1 into the other equation. This gives us an equation in one variable, namely 4x – 3(−x – 1) = 24. The variable y has been eliminated. © 2010 Pearson Prentice Hall. All rights reserved. 15 Example 3 continued Step 3 Solve the resulting equation containing one variable. 4x – 3(-x – 1) = 24 4x + 3x + 3 = 24 7x + 3 = 24 7x = 21 x=3 Step 4 Back-substitute the obtained value into the equation from step1. Since we found x = 3 in step 3, then we back-substitute the x-value into the equation from step 1 to find the y-coordinate. © 2010 Pearson Prentice Hall. All rights reserved. 16 Example 3 continued Step 4 (cont.) With x = 3 and y = −4, the proposed solution is (3,−4). Step 5 Check. Use this ordered pair to verify that this solution makes each equation true. We leave this to the student. © 2010 Pearson Prentice Hall. All rights reserved. 17 Your Turn • Solve the following system of equations by the substitution method. • y = 2x + 7 2x – y = -5 • -4x + y = -11 2x – 3y = 3 © 2010 Pearson Prentice Hall. All rights reserved. 1 Solving Linear Systems by the Addition Method © 2010 Pearson Prentice Hall. All rights reserved. 19 Example 5: Solving a System by the Addition Method Solve by the addition method: 3x + 2y = 48 9x – 8y = −24. Solution: Step 1 Rewrite both equations in the form Ax + By = C. Both equations are already in this form. Variable terms appear on the left and constants appear on the right. Step 2 If necessary, multiply either equation or both equations by appropriate numbers so that the sum of the x-coefficients or the sum of the ycoefficients is 0. © 2010 Pearson Prentice Hall. All rights reserved. 20 Example 5 continued 3x + 2y = 48 9x – 8y = −24 Multiply by −3 No Change Step 3 Add the equations. − 9x – 6y = − 144 9x – 8y = − 24 −14y = −168 Step 4 Solve the equation in one variable. We solve −14y = −168 by dividing both sides by −14. 14 y 168 14 14 y 12 © 2010 Pearson Prentice Hall. All rights reserved. 21 Example 5 continued Step 5 Back-substitute and find the value for the other variable. 3x + 2y = 48 3x + 2(12) = 48 3x + 24 = 48 3x = 24 x=8 Step 6 Check. The solution to the system is (8,12). We can check this by verifying that the solution is true for both equations. We leave this to the student. © 2010 Pearson Prentice Hall. All rights reserved. 22 Linear Systems Having No Solution or Infinitely Many Solutions The number of solutions to a system of two linear equations in two variables is given by one of the following: Number of Solutions What This Means Graphically Exactly one ordered-pair solution The two lines intersect at one point. No Solution The two lines are parallel. Infinitely many solutions The two lines are identical. © 2010 Pearson Prentice Hall. All rights reserved. 23 Example 7: A System with no Solution Solve the system: 4x + 6y = 12 6x + 9y = 12. Solution: Because no variable is isolated, we will use the addition method. 4x + 6y = 12 Multiply by 3 6x + 9y = 12 Multiply by -2 Add: The false statement 0 = 12 indicates that the system has no solution. The solution is the empty set, Ø. © 2010 Pearson Prentice Hall. All rights reserved. 24 Example 8: A System with Infinitely Many Solutions Solve the system: y = 3x – 2 15x – 5y = 10. Solution: Because the variable y is isolated in y = 3x – 2, the first equation, we will use the substitution method. © 2010 Pearson Prentice Hall. All rights reserved. 25 Example 8 continued The statement 10 = 10 is true. Hence, this indicates that the system has infinitely many solutions. © 2010 Pearson Prentice Hall. All rights reserved. 26 Modeling with Systems of Equations: Making Money (and Losing It) Revenue and Cost Functions A company produces and sells x units of a product. • Revenue Function R(x) = (price per unit sold)x • Cost Function C(x) = fixed cost + (cost per unit produced)x The point of intersection of the graphs of the revenue and cost functions is called the break-even point. © 2010 Pearson Prentice Hall. All rights reserved. 27 Example 9: Finding a Break-Even Point A company is planning to manufacture radically different wheelchairs. Fixed cost will be $500,000 and it will cost $400 to produce each wheelchair. Each wheelchair will be sold for $600. a. Write the cost function, C, of producing x wheelchairs. b. Write the revenue function, R, from the sale of x wheelchairs. c. Determine the break-even point. Describe what this means. © 2010 Pearson Prentice Hall. All rights reserved. 28 Example 9 Continued Solution: a. The cost function is the sum of the fixed cost and the variable cost. b. The revenue function is the money generated from the sale of x wheelchairs. © 2010 Pearson Prentice Hall. All rights reserved. 29 Example 9 continued c. The break even point occurs where the graphs of C and R intersect. Thus, we find this point by solving the system C(x) = 500,000 + 400x R(x) = 600x, Or y = 500,000 + 400x y = 600x. © 2010 Pearson Prentice Hall. All rights reserved. 30 Example 9 continued Using substitution, we substitute 600x in for y in the first equation: 600x = 500,000 + 400x 200x = 500,000 x = 2500 Back-substituting 2500 for x in either of the system’s equations (or functions), we obtain © 2010 Pearson Prentice Hall. All rights reserved. 31 Example 9 continued The break-even point is (2500, 1,500,000). This means that the company will break even if it produces and sells 2500 wheelchairs for $1,500,000. © 2010 Pearson Prentice Hall. All rights reserved. 32 The Profit Function • The profit, P(x), generated after producing and selling x units of a product is given by the profit function P(x) = R(x) – C(x), where R and C are the revenue and cost, respectively. The profit function, P(x), for the previous example is P(x) = R(x) – C(x) = 600x – (500,000 + 400x) = 200x – 500,000. © 2010 Pearson Prentice Hall. All rights reserved. 33