1. You can also solve systems of equations with the elimination method. With elimination, you get rid of one of the variables by adding or subtracting equations. You may have to multiply one or both equations by a number to create variable terms that can be eliminated. The elimination method is sometimes called the addition method or linear combination. Reading Math

2. Use elimination to solve the system of equations. Example 2A: Solving Linear Systems by Elimination 3x + 2y = 4 4x – 2y = –18 Step 1 Find the value of one variable. 3x + 2y = 4 + 4x – 2y = –18 The y-terms have opposite coefficients. First part of the solution 7x = –14 x = –2 Add the equations to eliminate y.

3. Example 2A Continued Step 2 Substitute the x-value into one of the original equations to solve for y. 3(–2) + 2y = 4 2y = 10 y = 5 Second part of the solution The solution to the system is (–2, 5).

4. Use elimination to solve the system of equations. Example 2B: Solving Linear Systems by Elimination 3x + 5y = –16 2x + 3y = –9 Step 1 To eliminate x, multiply both sides of the first equation by 2 and both sides of the second equation by –3. Add the equations. First part of the solution y = –5 2(3x + 5y) = 2(–16) –3(2x + 3y) = –3(–9) 6x + 10y = –32 –6x – 9y = 27

5. Example 2B Continued Second part of the solution 3x + 5(–5) = –16 3x = 9 3x – 25 = –16 x = 3 Step 2 Substitute the y-value into one of the original equations to solve for x. The solution for the system is (3, –5).

6. Example 2B: Solving Linear Systems by Elimination Check Substitute 3 for x and –5 for y in each equation. 3x + 5y = –16 2x + 3y = –9 –16 3(3) + 5(–5) –16 2(3) + 3(–5)–9

7. Use elimination to solve the system of equations. 4x + 7y = –25 –12x –7y = 19 Check It Out! Example 2a Step 1 Find the value of one variable. The y-terms have opposite coefficients. First part of the solution –8x = –6 4x + 7y = –25 – 12x – 7y = 19 Add the equations to eliminate y. x =

8. Check It Out! Example 2a Continued 3 + 7y = –25 7y = –28 Second part of the solution Step 2 Substitute the x-value into one of the original equations to solve for y. 4 ( ) + 7y = –25 y = –4 The solution to the system is (, –4).

9. Use elimination to solve the system of equations. 5x – 3y = 42 8x + 5y = 28 Step 1 To eliminate x, multiply both sides of the first equation by –8 and both sides of the second equation by 5. Add the equations. First part of the solution y = –4 Check It Out! Example 2b 49y = –196 –8(5x – 3y) = –8(42) 5(8x + 5y) = 5(28) –40x + 24y = –336 40x + 25y = 140

10. Second part of the solution 5x – 3(–4) = 42 5x = 30 5x + 12 = 42 x = 6 Step 2 Substitute the y-value into one of the original equations to solve for x. The solution for the system is (6,–4). Check It Out! Example 2b

11. Check Substitute 6 for x and –4 for y in each equation. 5x – 3y = 42 8x + 5y = 28 42 5(6) – 3(–4) 42 8(6) + 5(–4)28 Check It Out! Example 2b

12. In Lesson 3–1, you learned that systems may have infinitely many or no solutions. When you try to solve these systems algebraically, the result will be an identity or a contradiction. An identity, such as 0 = 0, is always true and indicates infinitely many solutions. A contradiction, such as 1 = 3, is never true and indicates no solution. Remember!

13. Classify the system and determine the number of solutions. Example 3: Solving Systems with Infinitely Many or No Solutions 3x + y = 1 2y + 6x = –18 Because isolating y is straightforward, use substitution. Substitute (1–3x) for y in the second equation. Solve the first equation for y. 3x + y = 1 2(1 – 3x) + 6x = –18 y = 1 –3x 2 – 6x + 6x = –18 2 = –18 Distribute. Simplify. Because 2 is never equal to –18, the equation is a contradiction. Therefore, the system is inconsistent and has no solution. x

14. Classify the system and determine the number of solutions. 56x + 8y = –32 7x + y = –4 Because isolating y is straightforward, use substitution. Substitute (–4 –7x) for y in the first equation. Solve the second equation for y. 7x + y = –4 56x + 8(–4 – 7x) = –32 y = –4 – 7x 56x – 32 – 56x = –32 Distribute. Simplify. Because –32 is equal to –32, the equation is an identity. The system is consistent, dependent and has infinite number of solutions. Check It Out! Example 3a –32 = –32

15. Classify the system and determine the number of solutions. 6x + 3y = –12 2x + y = –6 Because isolating y is straightforward, use substitution. Substitute (–6 – 2x) for y in the first equation. Solve the second equation. 2x + y = –6 6x + 3(–6 – 2x)= –12 y = –6 – 2x 6x –18 – 6x = –12 Distribute. Simplify. Because –18 is never equal to –12, the equation is a contradiction. Therefore, the system is inconsistent and has no solutions. Check It Out! Example 3b –18 = –12 x

16. A veterinarian needs 60 pounds of dog food that is 15% protein. He will combine a beef mix that is 18% protein with a bacon mix that is 9% protein. How many pounds of each does he need to make the 15% protein mixture? Example 4: Zoology Application Let x present the amount of beef mix in the mixture. Let y present the amount of bacon mix in the mixture.

17. Example 4 Continued Write one equation based on the amount of dog food: Amount of beef mix plus amount of bacon mix equals xy 60. 60+ = Write another equation based on the amount of protein: Protein of beef mix plus protein of bacon mix equals 0.18x0.09y protein in mixture. 0.15(60) + =

18. Solve the system. x + y = 60 0.18x +0.09y = 9 x + y = 60 y = 60 – x First equation 0.18x + 0.09(60 – x) = 9 0.18x + 5.4 – 0.09x = 9 0.09x = 3.6 x = 40 Solve the first equation for y. Substitute (60 – x) for y. Distribute. Simplify. Example 4 Continued

19. Substitute the value of x into one equation. Substitute x into one of the original equations to solve for y. 40 + y = 60 y = 20 Solve for y. The mixture will contain 40 lb of the beef mix and 20 lb of the bacon mix. Example 4 Continued

20. A coffee blend contains Sumatra beans which cost $5/lb, and Kona beans, which cost $13/lb. If the blend costs $10/lb, how much of each type of coffee is in 50 lb of the blend? Let x represent the amount of the Sumatra beans in the blend. Check It Out! Example 4 Let y represent the amount of the Kona beans in the blend.

21. Write one equation based on the amount of each bean: Amount of Sumatra beans plus amount of Kona beans equals xy 50. 50 + = Write another equation based on cost of the beans: Cost of Sumatra beans plus cost of Kona beans equals 5x5x 13y cost of beans. 10(50) += Check It Out! Example 4 Continued

22. Solve the system. x + y = 50 5x + 13y = 500 x + y = 50 y = 50 – x First equation 5x + 13(50 – x) = 500 5x + 650 – 13x = 500 –8x = –150 x = 18.75 Solve the first equation for y. Substitute (50 – x) for y. Distribute. Simplify. Check It Out! Example 4 Continued

23. Substitute the value of x into one equation. Substitute x into one of the original equations to solve for y. 18.75 + y = 50 y = 31.25 Solve for y. The mixture will contain 18.75 lb of the Sumatra beans and 31.25 lb of the Kona beans. Check It Out! Example 4 Continued