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Copyright © 2010 Pearson Education, Inc. All rights reserved. 4.2 – Slide 1 Chapter 4 Systems of Linear Equations and Inequalities Copyright © 2010 Pearson Education, Inc. All rights reserved. 4.2 – Slide 2 4.2 Solving Systems of Linear Equations by Substitution Copyright © 2010 Pearson Education, Inc. All rights reserved. 4.2 – Slide 3 4.2 Solving Systems of Linear Equations by Substitution Objectives 1. 2. 3. Solve linear systems by substitution. Solve special systems by substitution. Solve linear systems with fractions and decimals by substitution. Copyright © 2010 Pearson Education, Inc. All rights reserved. 4.2 – Slide 4 4.2 Solving Systems of Linear Equations by Substitution Solving Linear Systems by Substitution Example 1 Solve the system by the substitution method. 5x + 2y = 2 y = – 3x The second equation is already solved for y. This equation says that y = – 3x. Substituting – 3x for y in the first equation gives 5x + 2y = 2 5x + 2(– 3x) = 2 5x + – 6x = 2 – 1x = 2 x = –2 Copyright © 2010 Pearson Education, Inc. All rights reserved. Let y = – 3x. Multiply. Combine like terms. Multiply by – 1. 4.2 – Slide 5 4.2 Solving Systems of Linear Equations by Substitution Solving Linear Systems by Substitution Example 1 (cont.) Because x = – 2, we find y from the equation y = – 3x by substituting – 2 for x. y = – 3x = – 3(– 2) = 6 Let x = – 2. Check that the solution of the given system is (– 2, 6) by substituting – 2 for x and 6 for y in both equations. y = – 3x 5x + 2y = 2 5(– 2) + 2(6) = 2 2 = 2 6 = – 3(– 2) ? True 6 = 6 ? True The solution set of the system is {(– 2, 6)}. Copyright © 2010 Pearson Education, Inc. All rights reserved. 4.2 – Slide 6 4.2 Solving Systems of Linear Equations by Substitution Solving Linear Systems by Substitution Example 2 Solve the system 3x + 4y = 4 by the substitution method. x = 2y + 18 The second equation gives x in terms of y. Substitute 2y + 18 for x in the first equation. 3x + 4y = 4 3(2y + 18) + 4y = 4 6y + 54 + 4y = 4 10y + 54 = 4 10y = – 50 y = –5 Copyright © 2010 Pearson Education, Inc. All rights reserved. Let x = 2y + 18. Distributive property Combine like terms. Subtract 54. Divide by 10. 4.2 – Slide 7 4.2 Solving Systems of Linear Equations by Substitution Solving Linear Systems by Substitution Example 2 (continued) Because y = – 5, we find x from the equation x = 2y + 18 by substituting – 5 for y. x = 2y + 18 = 2(– 5) + 18 = 8 Let y = – 5. Check that the solution of the given system is (8, – 5) by substituting 8 for x and – 5 for y in both equations.. 3x + 4y = 4 3(8) + 4(– 5) = 4 4 = 4 x = 2y + 18 8 = 2(– 5) + 18 ? True 8 = 8 ? True The solution set of the system is {(8, – 5)}. Copyright © 2010 Pearson Education, Inc. All rights reserved. 4.2 – Slide 8 4.2 Solving Systems of Linear Equations by Substitution Solving Linear Systems by Substitution Solving a Linear System by Substitution Step 1 Solve one equation for either variable. If one of the variables has a coefficient of 1 or – 1, choose it, since it usually makes the substitution easier. Step 2 Substitute for that variable in the other equation. The result should be an equation with just one variable. Step 3 Solve the equation from Step 2. Step 4 Substitute the result from Step 3 into the equation from Step 1 to find the value of the other variable. Step 5 Check the solution in both of the original equations. Then write the solution set. Copyright © 2010 Pearson Education, Inc. All rights reserved. 4.2 – Slide 9 4.2 Solving Systems of Linear Equations by Substitution Solving Linear Systems by Substitution Example 3 Use substitution to solve the system. 3x + 2y = 26 (1) 5x – y = 13 (2) Step 1 For the substitution method, we must solve one of the equations for either x or y. Because the coefficient of y in equation (2) is – 1, we choose equation (2) and solve for y. 5x – y = 13 5x – y – 5x = 13 – 5x – y = 13 – 5x y = – 13 + 5x Copyright © 2010 Pearson Education, Inc. All rights reserved. (2) Subtract 5x. Combine like terms. Multiply by – 1. 4.2 – Slide 10 4.2 Solving Systems of Linear Equations by Substitution Solving Linear Systems by Substitution Example 3 (continued) Use substitution to solve the system. 3x + 2y = 26 (1) 5x – y = 13 (2) Step 2 Now substitute – 13 + 5x for y in equation (1). 3x + 2y = 26 3x + 2(– 13 + 5x) = 26 Copyright © 2010 Pearson Education, Inc. All rights reserved. (1) Let y = – 13 + 5x. 4.2 – Slide 11 4.2 Solving Systems of Linear Equations by Substitution Solving Linear Systems by Substitution Example 3 (continued) Use substitution to solve the system. 3x + 2y = 26 (1) 5x – y = 13 (2) Step 3 Now solve the equation from Step 2. 3x + 2(– 13 + 5x) = 26 3x – 26 + 10x = 26 13x – 26 = 26 13x – 26 + 26 = 26 + 26 13x = 52 x = 4 Copyright © 2010 Pearson Education, Inc. All rights reserved. From Step 2. Distributive property Combine like terms. Add 26. Combine like terms. Divide by 13. 4.2 – Slide 12 4.2 Solving Systems of Linear Equations by Substitution Solving Linear Systems by Substitution Example 3 (continued) Use substitution to solve the system. 3x + 2y = 26 (1) 5x – y = 13 (2) Step 4 Since y = – 13 + 5x and x = 4, y = – 13 + 5(4) = 7. Step 5 Check that (4, 7) is the solution. 5x – y = 13 5(4) – 7 = 13 3x + 2y = 26 3(4) + 2(7) = 26 12 + 14 = 26 26 = 26 ? ? True 20 – 7 = 13 13 = 13 ? ? True Since both results are true, the solution set of the system is {(4, 7)}. Copyright © 2010 Pearson Education, Inc. All rights reserved. 4.2 – Slide 13 4.2 Solving Systems of Linear Equations by Substitution Solving Linear Systems by Substitution Example 4 Use substitution to solve the system. 3x – 2y = 0 (1) 5x – 4y = 6 (2) Step 1 To use the substitution method, we must solve one of the equations for one of the variables. We choose equation (2) and solve for x. 5x – 4y = 6 5x – 4y + 4y = 6 + 4y 5x = 6 + 4y 6 4 x = + y 5 5 Copyright © 2010 Pearson Education, Inc. All rights reserved. (2) Add 4y. Combine like terms. Divide by 5. 4.2 – Slide 14 4.2 Solving Systems of Linear Equations by Substitution Solving Linear Systems by Substitution Example 4 (continued) Use substitution to solve the system. 3x – 2y = 0 (1) 5x – 4y = 6 (2) Step 2 Now substitute 6 4 for x in equation (1). + y 5 5 3x – 2y = 0 3 6 + 4 y 5 5 – 2y = 0 Copyright © 2010 Pearson Education, Inc. All rights reserved. (1) Let x = 6 4 . + y 5 5 4.2 – Slide 15 4.2 Solving Systems of Linear Equations by Substitution Solving Linear Systems by Substitution Example 4 (continued) Use substitution to solve the system. 3x – 2y = 0 (1) 5x – 4y = 6 (2) Step 3 Now solve the equation from Step 2. 3 6 + 4 y 5 5 – 2y = 0 From Step 2. 18 12 – 2y = 0 + y 5 5 Distributive property 18 2 + y = 0 5 5 Combine like terms. Copyright © 2010 Pearson Education, Inc. All rights reserved. 4.2 – Slide 16 4.2 Solving Systems of Linear Equations by Substitution Solving Linear Systems by Substitution Example 4 (continued) Use substitution to solve the system. 3x – 2y = 0 (1) 5x – 4y = 6 (2) Step 3 (continued) 18 2 + y 5 5 18 + 2 y – 18 5 5 5 2 y 5 = 0 From previous slide. = 0 – 18 5 Subtract 18 . 5 = – 18 5 Combine like terms. Copyright © 2010 Pearson Education, Inc. All rights reserved. 4.2 – Slide 17 4.2 Solving Systems of Linear Equations by Substitution Solving Linear Systems by Substitution Example 4 (continued) Use substitution to solve the system. 3x – 2y = 0 (1) 5x – 4y = 6 (2) Step 3 (continued) 2 y 5 = – 18 5 y = –9 Copyright © 2010 Pearson Education, Inc. All rights reserved. From previous slide. Multiply by 5 . 2 4.2 – Slide 18 4.2 Solving Systems of Linear Equations by Substitution Solving Linear Systems by Substitution Example 4 (continued) Use substitution to solve the system. 3x – 2y = 0 (1) 5x – 4y = 6 (2) Step 4 Find x by substituting – 9 for y in x = 6 + 5 Copyright © 2010 Pearson Education, Inc. All rights reserved. 6 4 + y. 5 5 4 (– 9) = – 6 5 4.2 – Slide 19 4.2 Solving Systems of Linear Equations by Substitution Solving Linear Systems by Substitution Example 4 (continued) Use substitution to solve the system. 3x – 2y = 0 (1) 5x – 4y = 6 (2) Step 5 Check that (– 6, – 9) is the solution. 3x – 2y 3(– 6) – 2(– 9) (– 18) – (– 18) – 18 + 18 0 = = = = = 0 0 0 0 0 5x – 4y 5(– 6) – 4(– 9) (– 30) – (– 36) (– 30) + 36 6 = = = = = 6 6 6 6 6 Both results are true, so the solution set of the system is {(– 6, – 9)}. Copyright © 2010 Pearson Education, Inc. All rights reserved. 4.2 – Slide 20 4.2 Solving Systems of Linear Equations by Substitution Solving Special Systems by Substitution Example 5 Use substitution to solve the system. y = 2x – 7 3y – 6x = – 6 (1) (2) Substitute 2x – 7 for y in equation (2). 3y – 6x = 6 (2) 3(2x – 7) – 6x = – 6 Let y = 2x – 7. 6x – 21 – 6x = – 6 Distributive property – 21 = – 6 False. This false result means that the equations in the system have graphs that are parallel lines. The system is inconsistent and the solution set is 0 . Copyright © 2010 Pearson Education, Inc. All rights reserved. 4.2 – Slide 21 4.2 Solving Systems of Linear Equations by Substitution Solving Special Systems by Substitution y Example 5 (continued) y = 2x – 7 3y – 6x = – 6. (1) (2) Copyright © 2010 Pearson Education, Inc. All rights reserved. (2) (1) x 4.2 – Slide 22 4.2 Solving Systems of Linear Equations by Substitution Solving Special Systems by Substitution Example 6 Solve the system by the substitution method. 3 = 5x – y (1) – 4y – 12 = – 20x (2) Begin by solving equation (1) for y to get y = 5x – 3. Substitute 5x – 3 for y in equation (2) and solve the resulting equation. – 4y – 12 = – 20x (2) – 4(5x – 3) – 12 = – 20x Let y = 5x – 3. – 20x + 12 – 12 = – 20x Distributive property 0 = 0 Add 20x; combine terms. This true result means that every solution of one equation is also a solution of the other, so the system has an infinite number of solutions. Copyright © 2010 Pearson Education, Inc. All rights reserved. 4.2 – Slide 23 4.2 Solving Systems of Linear Equations by Substitution Solving Special Systems by Substitution y Example 6 (continued) 3 = 5x – y – 4y – 12 = – 20x (1) (2) x The system has an infinite number of solutions – all the ordered pairs corresponding to points that lie on the common graph. Copyright © 2010 Pearson Education, Inc. All rights reserved. 4.2 – Slide 24 4.2 Solving Systems of Linear Equations by Substitution Solving Linear Systems with Fractions and Decimals Example 7 Solve the system by the substitution method. 2 x + 5y = 1 (1) 3 1 x – 4 1 y = – 13 8 8 (2) Clear equation (1) of fractions by multiplying each side by 3. 3 3 2 x 3 2 x + 5y 3 = 3 1 Multiply by 4. + 3 5y = 3 1 Distributive property 2x + 15y = 3 Copyright © 2010 Pearson Education, Inc. All rights reserved. (3) 4.2 – Slide 25 4.2 Solving Systems of Linear Equations by Substitution Solving Linear Systems with Fractions and Decimals Example 7 (cont.) Solve the system by the substitution method. 2 x + 5y = 1 (1) 3 1 x – 1 y = – 13 (2) 4 8 8 Now, clear equation (2) of fractions by multiplying each side by 8. 8 1 x – 4 8 1 x 4 – 1 y = 8 – 13 8 8 8 1y 8 = 8 – 13 8 2x – y = – 13 Copyright © 2010 Pearson Education, Inc. All rights reserved. Multiply by 8. Distributive property (4) 4.2 – Slide 26 4.2 Solving Systems of Linear Equations by Substitution Solving Linear Systems with Fractions and Decimals Example 7 (cont.) The given system of equations has been simplified to the equivalent system. 2 x + 5y = 1 (1) 2x + 15y = 3 3 1 x – 1 y = – 13 (2) 2x – y = – 13 4 8 8 (3) (4) To solve this system by substitution, equation (4) can be solved for y. 2x – y = – 13 2x – y – 2x = – 13 – 2x – y = – 13 – 2x y = 13 + 2x Copyright © 2010 Pearson Education, Inc. All rights reserved. (4) Subtract 2x. Combine like terms. Multiply by – 1. 4.2 – Slide 27 4.2 Solving Systems of Linear Equations by Substitution Solving Linear Systems with Fractions and Decimals Example 7 (cont.) The given system of equations has been simplified to the equivalent system. 2 x + 5y = 1 (1) 2x + 15y = 3 3 1 x – 1 y = – 13 (2) 2x – y = – 13 4 8 8 (3) (4) Now substitute 13 + 2x for y in equation (3). 2x + 15y 2x + 15(13 + 2x) 2x + 195 + 30x 32x + 195 = = = = Copyright © 2010 Pearson Education, Inc. All rights reserved. 3 3 3 3 (3) Let y = 13 + 2x. Distributive property Combine like terms. 4.2 – Slide 28 4.2 Solving Systems of Linear Equations by Substitution Solving Linear Systems with Fractions and Decimals Example 7 (cont.) The given system of equations has been simplified to the equivalent system. 2 x + 5y = 1 (1) 2x + 15y = 3 3 1 x – 1 y = – 13 (2) 2x – y = – 13 4 8 8 (3) (4) Now substitute 13 + 2x for y in equation (3). 32x + 195 = 3 32x + 195 – 195 = 3 – 195 32x = – 192 x = –6 Copyright © 2010 Pearson Education, Inc. All rights reserved. Combine like terms. Subtract 195. Combine like terms. Divide by 32. 4.2 – Slide 29 4.2 Solving Systems of Linear Equations by Substitution Solving Linear Systems with Fractions and Decimals Example 7 (cont.) The given system of equations has been simplified to the equivalent system. 2 x + 5y = 1 (1) 2x + 15y = 3 3 1 x – 1 y = – 13 (2) 2x – y = – 13 4 8 8 (3) (4) Substitute – 6 for x in y = 13 + 2x to get y = 13 + 2(– 6) = 1. Check by substituting – 6 for x and 1 for y in both of the original equations. The solution set is {(– 6, 1)}. Copyright © 2010 Pearson Education, Inc. All rights reserved. 4.2 – Slide 30