6 3 Solving Systems by Elimination Preview Warm

6 -3 Solving Systems by Elimination Preview Warm Up California Standards Lesson Presentation

6 -3 Solving Systems by Elimination Warm Up Simplify each expression. 1. 3 x + 2 y – 5 x – 2 y – 2 x 2. 5(x – y) + 2 x + 5 y 7 x 3. 4 y + 6 x – 3(y + 2 x) y 4. 2 y – 4 x – 2(4 y – 2 x) – 6 y Write the least common multiple. 5. 3 and 6 6 6. 4 and 10 20 7. 6 and 8 24 8. 2 and 5 10

6 -3 Solving Systems by Elimination California Standards 9. 0 Students solve a system of two linear equations in two variables algebraically and are able to interpret the answer graphically. Students are able to solve a system of two linear inequalities in two variables and to sketch the solution sets.

6 -3 Solving Systems by Elimination Another method for solving systems of equations is elimination. Like substitution, the goal of elimination is to get one equation that has only one variable. To do this by elimination, you add the two equations in the system together. Remember that an equation stays balanced if you add equal amounts to both sides. Consider the system x – 2 y = – 19 5 x + 2 y = 1 Since 5 x + 2 y = 1, you can add 5 x + 2 y to one side of the first equation and 1 to the other side and the balance is maintained

6 -3 Solving Systems by Elimination Since – 2 y and 2 y have opposite coefficients, you can eliminate the y-term by adding two equations. The result is one equation that has only one variable: 6 x = – 18.

6 -3 Solving Systems by Elimination Solving Systems of Equations by Elimination Step 1 Write the system so that like terms are aligned. Step 2 Eliminate one of the variables. Step 3 Solve for the variable not eliminated in Step 2. Step 4 Substitute the value of the variable into one of the original equations and solve for the other variable. Step 5 Write the answers from Steps 2 and 3 as an ordered pair, (x, y), and check your answer.

6 -3 Solving Systems by Elimination Later in this lesson you will learn how to multiply one or more equations by a number in order to produce opposites that can be eliminated.

6 -3 Solving Systems by Elimination Additional Example 1: Elimination Using Addition Solve 3 x – 4 y = 10 by elimination. x + 4 y = – 2 Step 1 3 x – 4 y = 10 Step 2 x + 4 y = – 2 4 x + 0 = Step 3 4 x = 8 4 4 x=2 8 Write the system so that like terms are aligned. Add the equations to eliminate the y-terms. Simplify and solve for x. Divide both sides by 4.

6 -3 Solving Systems by Elimination Additional Example 1 Continued Step 4 x + 4 y = – 2 2 + 4 y = – 2 – 2 Write one of the original equations. Substitute 2 for x. Subtract 2 from both sides. 4 y = – 4 4 y – 4 4 4 y = – 1 Step 5 (2, – 1) Divide both sides by 4. Write the solution as an ordered pair.

6 -3 Solving Systems by Elimination Check It Out! Example 1 Solve y + 3 x = – 2 by elimination. 2 y – 3 x = 14 Check your answer. Step 1 Step 2 y + 3 x = – 2 2 y – 3 x = 14 3 y + 0 Step 3 = 12 3 y = 12 Write the system so that like terms are aligned. Add the equations to eliminate the x-terms. Simplify and solve for y. Divide both sides by 3. y=4

6 -3 Solving Systems by Elimination Check It Out! Example 1 Continued Step 4 y + 3 x = – 2 4 + 3 x = – 2 – 4 3 x = – 6 3 3 x = – 2 Step 5 (– 2, 4) Write one of the original equations. Substitute 4 for y. Subtract 4 from both sides. Divide both sides by 3. Write the solution as an ordered pair.

6 -3 Solving Systems by Elimination Check It Out! Example 1 Continued Check Substitute (– 2, 4) into both equations in the system. y + 3 x = – 2 4 + 3(– 2) 4 + (– 6) – 2 – 2 2 y – 3 x = 14 2(4) – 3(– 2) 8+6 14 14

6 -3 Solving Systems by Elimination When two equations each contain the same term, you can subtract one equation from the other to solve the system. To subtract an equation add the opposite of each term.

6 -3 Solving Systems by Elimination Additional Example 2: Elimination Using Subtraction Solve 2 x + y = – 5 by elimination. 2 x – 5 y = 13 Step 1 2 x + y = – 5 Step 2 – (2 x – 5 y = 13) 2 x + y = – 5 – 2 x + 5 y = – 13 Step 3 0 + 6 y = – 18 y = – 3 Add the opposite of each term in the second equation. Eliminate the x term. Simplify and solve for y.

6 -3 Solving Systems by Elimination Additional Example 2 Continued Step 4 2 x + y = – 5 2 x + (– 3) = – 5 2 x – 3 = – 5 +3 +3 Write one of the original equations. Substitute – 3 for y. 2 x Simplify and solve for x. = – 2 x = – 1 Step 5 (– 1, – 3) Add 3 to both sides. Write the solution as an ordered pair.

6 -3 Solving Systems by Elimination Remember! Remember to check by substituting your answer into both original equations.

6 -3 Solving Systems by Elimination Check It Out! Example 2 Solve 3 x + 3 y = 15 by elimination. – 2 x + 3 y = – 5 Check your answer. Step 1 Step 2 Step 3 3 x + 3 y = 15 – (– 2 x + 3 y = – 5) 3 x + 3 y = 15 + 2 x – 3 y = +5 5 x + 0 5 x = 20 x=4 Add the opposite of each term in the second equation. Eliminate the y term. Simplify and solve for x.

6 -3 Solving Systems by Elimination Check It Out! Example 2 Continued Step 4 3 x + 3 y = 15 3(4) + 3 y = 15 12 + 3 y = 15 – 12 3 y = 3 y=1 Step 5 (4, 1) Write one of the original equations. Substitute 4 for x. Subtract 12 from both sides. Simplify and solve for y. Write the solution as an ordered pair.

6 -3 Solving Systems by Elimination Check It Out! Example 2 Continued Check Substitute (4, 1) into both equations in the system. 3 x + 3 y = 15 3(4) + 3(1) 12 + 3 15 15 – 2 x + 3 y = – 5 – 2(4) + 3(1) – 5 – 8 + 3 – 5 – 5

6 -3 Solving Systems by Elimination In some cases, you will first need to multiply one or both of the equations by a number so that one variable has opposite coefficients.

6 -3 Solving Systems by Elimination Additional Example 3 A: Elimination Using Multiplication First Solve the system by elimination. x + 2 y = 11 – 3 x + y = – 5 Step 1 Step 2 Step 3 x + 2 y = 11 – 2 (– 3 x + y = – 5) x + 2 y = 11 +(6 x – 2 y = +10) 7 x + 0 = 21 7 x = 21 x=3 Multiply each term in the second equation by – 2 to get opposite y-coefficients. Add the new equation to the first equation. Simplify and solve for x.

6 -3 Solving Systems by Elimination Additional Example 3 A Continued Step 4 x + 2 y = 11 3 + 2 y = 11 – 3 2 y = 8 y=4 Step 5 (3, 4) Write one of the original equations. Substitute 3 for x. Subtract 3 from each side. Simplify and solve for y. Write the solution as an ordered pair.

6 -3 Solving Systems by Elimination Additional Example 3 B: Elimination Using Multiplication First Solve the system by elimination. – 5 x + 2 y = 32 2 x + 3 y = 10 Step 1 2 (– 5 x + 2 y = 32) 5 (2 x + 3 y = 10) Step 2 – 10 x + 4 y = 64 +(10 x + 15 y = 50) Step 3 19 y = 114 Multiply the first equation by 2 and the second equation by 5 to get opposite x-coefficients Add the new equations. y=6 Simplify and solve for y.

6 -3 Solving Systems by Elimination Additional Example 3 B Continued Step 4 2 x + 3 y = 10 2 x + 3(6) = 10 2 x + 18 = 10 – 18 Step 5 2 x = – 8 x = – 4 (– 4, 6) Write one of the original equations. Substitute 6 for y. Subtract 18 from both sides. Simplify and solve for x. Write the solution as an ordered pair.

6 -3 Solving Systems by Elimination Helpful Hint Use the techniques for finding a common denominator when trying to find values to multiply each equation by. To review these techniques, see Skills Bank p. SB 8.

6 -3 Solving Systems by Elimination Check It Out! Example 3 a Solve the system by elimination. Check your answer. 3 x + 2 y = 6 –x + y = – 2 Step 1 3 x + 2 y = 6 3(–x + y = – 2) Multiply each term in the Step 2 second equation by 3 to get 3 x + 2 y = 6 opposite x-coefficients. +(– 3 x + 3 y = – 6) Add the new equation to 0 + 5 y = 0 the first equation. Step 3 5 y = 0 Simplify and solve for y. y=0

6 -3 Solving Systems by Elimination Check It Out! Example 3 a Continued Step 4 –x + y = – 2 –x + (0) –x + 0 –x x Step 5 = = – 2 – 2 2 (2, 0) Write one of the original equations. Substitute 0 for y. Simplify and solve for x. Write the solution as an ordered pair.

6 -3 Solving Systems by Elimination Check It Out! Example 3 a Continued Check Substitute (2, 0) into both equations in the system. 3 x + 2 y = 6 3(2) + 2(0) 6+0 6 6 –x + y = – 2 + 0 – 2 – 2

6 -3 Solving Systems by Elimination Check It Out! Example 3 b Solve the system by elimination. Check your answer. 2 x + 5 y = 26 – 3 x – 4 y = – 25 Step 1 3 (2 x + 5 y = 26) 2 (– 3 x – 4 y = – 25) Step 2 6 x + 15 y = 78 +(– 6 x – 8 y = – 50) Step 3 0 + 7 y = 28 y =4 Multiply the first equation by 3 and the second equation by 2 to get opposite x-coefficients Add the new equations. Simplify and solve for y.

6 -3 Solving Systems by Elimination Check It Out! Example 3 b Continued Step 4 2 x + 5 y = 26 2 x + 5(4) = 26 2 x + 20 = 26 – 20 2 x = 6 x=3 Step 5 (3, 4) Write one of the original equations. Substitute 4 for y. Subtract 20 from both sides. Simplify and solve for x. Write the solution as an ordered pair.

6 -3 Solving Systems by Elimination Check It Out! Example 3 b Continued Check Substitute (3, 4) into both equations in the system. 2 x + 5 y = 26 2(3) + 5(4) 6 + 20 26 26 – 3 x – 4 y = – 25 – 3(3) – 4(4) – 9 – 16 – 25

6 -3 Solving Systems by Elimination Additional Example 4: Application Paige has $7. 75 to buy 12 sheets of felt and card stock for her scrapbook. The felt costs $0. 50 per sheet, and the card stock costs $0. 75 per sheet. How many sheets of each can Paige buy? Write a system. Use f for the number of felt sheets and c for the number of card stock sheets. Step 1 0. 50 f + 0. 75 c = 7. 75 f+ c = 12 The cost of felt and card stock totals $7. 75. The total number of sheets is 12.

6 -3 Solving Systems by Elimination Additional Example 4 Continued Step 2 0. 50 f + 0. 75 c = 7. 75 + (– 0. 50)(f + c = 12) Multiply the second equation by – 0. 50 to 0. 50 f + 0. 75 c = 7. 75 get opposite f+ (– 0. 50 f – 0. 50 c = – 6) coefficients. 0. 25 c = 1. 75 Add this equation to the first equation to eliminate the f-term. Step 3 c=7 Simplify and solve for c.

6 -3 Solving Systems by Elimination Additional Example 4 Continued Step 4 f + c = 12 f + 7 = 12 – 7 f = 5 Step 5 (5, 7) Write one of the original equations. Substitute 7 for c. Subtract 7 from both sides. Write the solution as an ordered pair. Paige can buy 5 sheets of card stock and 7 sheets of felt.

6 -3 Solving Systems by Elimination Check It Out! Example 4 What if…? Sally spent $14. 85 to buy 13 flowers. She bought lilies, which cost $1. 25 each, and tulips, which cost $0. 90 each. How many of each flower did Sally buy? Write a system. Use l for the number of lilies and t for the number of tulips. Step 1 1. 25 l + 0. 90 t = 14. 85 The cost of lilies and tulips totals $14. 85. l+ t = 13 The total number of flowers is 13.

6 -3 Solving Systems by Elimination Check It Out! Example 4 Continued Step 2 1. 25 l + 0. 90 t = 14. 85 Multiply the second equation by – 0. 90 to get + (– 0. 90)(l + t) = 13 opposite t-coefficients. 1. 25 l + 0. 90 t = 14. 85 Add this equation to the + (– 0. 90 l – 0. 90 t = – 11. 70) first equation to eliminate the t-term. Step 3 0. 35 l = 3. 15 Simplify and solve for l. l=9

6 -3 Solving Systems by Elimination Check It Out! Example 4 Continued Step 4 Step 5 l + t = 13 9 + t = 13 – 9 t = 4 (9, 4) Write one of the original equations. Substitute 9 for l. Subtract 9 from both sides. Write the solution as an ordered pair. Sally bought 9 lilies and 4 tulips.

6 -3 Solving Systems by Elimination All systems can be solved in more than one way. For some systems, some methods may be more appropriate than others.

6 -3 Solving Systems by Elimination

6 -3 Solving Systems by Elimination Lesson Quiz Solve each system by elimination. 1. 2 x + y = 25 (11, 3) 3 y = 2 x – 13 2. – 3 x + 4 y = – 18 (2, – 3) x = – 2 y – 4 – 2 x + 3 y = – 15 (– 3, – 7) 3 x + 2 y = – 23 4 pairs of athletic socks and 3 pairs of dress socks 4. Harlan has $44 to buy 7 pairs of socks. Athletic socks cost $5 per pair. Dress socks cost $8 per pair. How many pairs of each can Harlan buy? 3.