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2 8.2 Solving Systems of Equations by Elimination

3 What You Will Learn  Solve systems of linear equations algebraically using the method of elimination  Solve systems with no solution or infinitely many solution  Use the method of elimination to solve application problems

4 The Method of Elimination

5 In this section, you will study another way to solve a system of linear equations algebraically—the method of elimination. The key step in this method is to obtain opposite coefficients for one of the variables so that adding the two equations eliminates this variable. For instance, by adding the equations 3x + 5y = 7 –3x – 2y = –1 3y = 6 you eliminate the variable x and obtain a single equation in one variable, y. Equation 1 Equation 2 Add equations.

6 The Method of Elimination

7 Example 1 – The Method of Elimination Solve the system of linear equations. 4x + 3y = 1 2x – 3y = 5 Solution: Begin by noting that the coefficients of y are opposites. So, by adding the two equations, you can eliminate y. 4x + 3y = 1 2x – 3y = 5 6x = 6 Equation 1 Equation 2 Equation 1 Equation 2 Add equations.

8 cont’d So, x = 1. By back-substituting this value into the first equation, you can solve for y, as follows. 4(1) + 3y = 1 3y = –3 y = –1 The solution is (1, –1). Check this in both of the original equations. Substitute 1 for x in Equation 1. Subtract 4 from each side. Divide each side by 3. Example 1 – The Method of Elimination

9 The Method of Elimination To obtain opposite coefficients for one of the variables, you often need to multiply one or both of the equations by a suitable constant.

10 Example 2 – The Method of Elimination Solve the system of linear equations. 2x – 3y = –7 3x + y = –5 Solution: For this system, you can obtain opposite coefficients of y by multiplying the second equation by 3. 2x – 3y = –7 3x + y = –5 Equation 1 Equation 2 Equation 1 Multiply Equation 2 by 3. Add equations. 2x – 3y = –7 9x + 3y = –15 11x = –22

11 cont’d Example 2 – The Method of Elimination So, x = –2. By back-substituting this value of x into the second equation, you can solve for y. 3x + y = –5 3(–2) + y = –5 –6 + y = –5 y = 1 The solution is (–2, 1). Substitute –2 for x. Equation 2 Simplify. Add 6 to each side.

12 cont’d Example 2 – The Method of Elimination Check this in the original equations, as follows. Substitute into Equation 1 Substitute into Equation 2 2x – 3y = –7 3x + y = –5 2(–2) – 3(1) –7 3(–2) + (1) –5 –4 – 3 = –7 –6 + 1 = –5

13 The No-Solution and Many-Solutions Cases

14 Example 5 – The Method of Elimination: No-Solution Case Solve the system of linear equations. 2x – 6y = 5 3x – 9y = 2 Solution: To obtain coefficients that differ only in sign, multiply the first equation by 3 and multiply the second equation by –2. 2x – 6y = 5 3x – 9y = 2 Equation 1 Equation 2 Multiply Equation 1 by 3. Multiply Equation 2 by –2. Add equations. 6x – 18y = 15 –6x + 18y = –4 0 = 11

15 cont’d Example 5 – The Method of Elimination: No-Solution Case Because 0 = 11 is a false statement, you can conclude that the system is inconsistent and has no solution. The lines corresponding to the two equations of this system. Note that the two lines are parallel and so have no point of intersection.

16 Example 6 – The Method of Elimination: Many-Solutions Case Solve the system of linear equations. 2x – 6y = –5 –4x + 12y = 10 Solution: To obtain coefficients that differ only in sign, multiply the first equation by 3 and multiply the second equation by –2. 2x – 6y = –5 –4x + 12y = 10 Equation 1 Equation 2 Multiply Equation 1 by 2. Equation 2. Add equations. 4x – 12y = –10 –4x + 12y = 10 0 = 0

17 cont’d Because 0 = 0 is a true statement, you can conclude that the system is dependent and has infinitely many solutions. The solutions consist of all ordered pairs (x, y) lying on the line 2x – 6y = –5. Example 6 – The Method of Elimination: Many-Solutions Case

18 Applications

19 Example 5 – Solving a Mixture Problem A company with two stores buys six large delivery vans and five small delivery vans. The first store receives 4 of the large vans and 2 of the small vans for a total cost of $200,000. The second store receives 2 of the large vans and 3 of the small vans for a total cost of $160,000. What is the cost of each type of van?

20 cont’d Example 5 – Solving a Mixture Problem Solution The two unknowns in this problem are the costs of the two types of vans. Labels: Cost of large van = x (dollars) Cost of small van = y (dollars) System: 4x + 2y = 200,000 Equation 1 2x + 3y = 160,000 Equation 2

21 cont’d Example 5 – Solving a Mixture Problem To solve this system of linear equations, use the method of elimination. To obtain coefficients of x that are opposites, multiply Equation 2 by –2. 4x + 2y = 200,000 4x + 2y = 200,000 Equation 1 2x + 3y = 160,000 –4x – 6y = –320,000 Multiply Equation 2 by –2. –4y = –120,000 Add equations. y = 30,000 Divide each side by –4. So, the cost of each small van is y = $30,000. Back-substitute this value into Equation 1 to find the cost of each large van.

22 cont’d Example 5 – Solving a Mixture Problem 4x + 2y = 200,000 Equation 1 4x + 2(30,000) = 200,000 Substitute 30,000 for y. 4x = 140,000 Simplify. x = 35,000 Divide each side by 4. The cost of each large van is x = $35,000. Check this solution in the original statement of the problem.