1. 11-6 Systems of Equations Course 3 Warm Up Warm Up Problem of the Day Problem of the Day Lesson Presentation Lesson Presentation

2. Warm Up Solve for the indicated variable. 1. P = R – C for R 2. V = Ah for A 3. R = for C R = P + C Rt + S = C Course 3 11-6 Systems of Equations 1 3 C – S t = A 3V3V h

3. Problem of the Day At an audio store, stereos have 2 speakers and home-theater systems have 5 speakers. There are 30 sound systems with a total of 99 speakers. How many systems are stereo systems and how many are home-theater systems? 17 stereo systems, 13 home-theater systems Course 3 11-6 Systems of Equations

4. Learn to solve systems of equations. Course 3 11-6 Systems of Equations

5. Vocabulary system of equations solution of a system of equations Insert Lesson Title Here Course 3 11-6 Systems of Equations

6. A system of equations is a set of two or more equations that contain two or more variables. A solution of a system of equations is a set of values that are solutions of all of the equations. If the system has two variables, the solutions can be written as ordered pairs. Each system has: a) “1” solution b) “infinitely many” solutions c) “no solutions” Course 3 11-6 Systems of Equations

7. Course 3 11-6 Systems of Equations all When solving systems of equations, remember to find values for all of the variables. Caution!

8. Example 1A: Solving Systems of Equations Solve the system of equations. y = 4x – 6 y = x + 3 y = 4x – 6y = x + 3 The expressions x + 3 and 4x – 6 both equal y. So by the Transitive Property they equal each other. 4x – 6 = x + 3 Course 3 11-6 Systems of Equations

9. Example 1A Continued To find y, substitute 3 for x in one of the original equations. y = x + 3 = 3 + 3 = 6 The solution is (3, 6). Course 3 11-6 Systems of Equations Solve the equation to find x. 4x – 6 = x + 3 – x – xSubtract x from both sides. 3x  6 = 3 3x 9  6 Add 6 to both sides. 3  = 3 x = 3 Divide both sides by 3.

10. Check It Out: Example 1A Solve the system of equations. y = x – 5 y = 2x – 8 y = x – 5y = 2x – 8 x – 5 = 2x – 8 Course 3 11-6 Systems of Equations The expressions x – 5 and 2x – 8 both equal y. So by the Transitive Property they equal each other.

11. Check It Out: Example 1A Continued To find y, substitute 3 for x in one of the original equations. y = x – 5 = 3 – 5 = –2 The solution is (3, –2). Course 3 11-6 Systems of Equations Solve the equation to find x. x – 5 = 2x – 8 –2x Subtract 2x from both sides. -x – 5 = – 8 -x = -3 Divide by -1. + 5 Add 5 to both sides.

12. The system of equations has no solution. Course 3 11-6 Systems of Equations 2x + 9 = 8 + 2x – 2x – 2x Transitive Property Subtract 2x from both sides. 9 ≠ 8 Example 1B: Solving Systems of Equations y = 2x + 9 y = 8 + 2x

13. When both variables are eliminated and the two constants remaining are not equal, the system of equations has no solution. Course 3 11-6 Systems of Equations 3x + 7 = 6 + 3x – 3x – 3x Transitive Property Subtract 3x from both sides. 7 ≠ 6 Check It Out: Example 1B y = 3x + 7 y = 6 + 3x

14. To solve a general system of two equations with two variables, you can: *Solve both equations for x or both for y. Course 3 11-6 Systems of Equations

15. Example 2A: Solving Systems of Equations by Solving for a Variable Solve the system of equations. x + 4y = -10 x – 3y = 11 y y  3y  3y Solve both equations for x. x = -10  4y x = 11 + 3y y y -10 = 11 + 7y Add 4y to both sides. Course 3 11-6 Systems of Equations -10 - 4y = 11 + 3y

16. Example 2A Continued –11 -21 7y Subtract 11 from both sides. Divide both sides by 7. 7 = 7 x = 11 + 3y = 11 + 3(3)Substitute –3 for y. = 11 + –9 = 2 The solution is (2, –3). Course 3 11-6 Systems of Equations -10 = 11 + 7y 3 = y

17. Check It Out: Example 2A Solve the system of equations. x + y = 5 3x + y = –1 –x –x – 3x – 3x Solve both equations for y. y = 5 – x y = –1 – 3x 5 – x = –1 – 3x + x 5 = –1 – 2x Add x to both sides. Course 3 11-6 Systems of Equations

18. Check It Out: Example 2A Continued 5 = –1 – 2x + 1 6 = –2x Add 1 to both sides. Divide both sides by –2. –3 = x y = 5 – x = 5 – (–3)Substitute –3 for x. = 5 + 3 = 8 The solution is (–3, 8). Course 3 11-6 Systems of Equations

19. Course 3 11-6 Systems of Equations You can solve for either variable. It is usually easiest to solve for a variable that has a coefficient of 1. Helpful Hint

20. Example 2B: Solving Systems of Equations by Solving for a Variable Solve the system of equations. –2x + 10y = –8 x – 5y = 4 –10y –10y +5y +5y Solve both equations for x. –2x = –8 – 10y x = 4 + 5y = – –8 –2 10y –2 –2x –2 x = 4 + 5y 4 + 5y = 4 + 5y Course 3 11-6 Systems of Equations  5y Subtract 5y from both sides. 4 = 4 Since 4 = 4 is always true, the system of equations has an infinite number of solutions.

21. Check It Out: Example 2B Solve the system of equations. x + y = –2 –3x + y = 2 – x – x + 3x + 3x Solve both equations for y. y = –2 – x y = 2 + 3x –2 – x = 2 + 3x Course 3 11-6 Systems of Equations

22. + x Add x to both sides. –2 = 2 + 4x –2 –4 = 4x –2 – x = 2 + 3x Subtract 2 from both sides. Divide both sides by 4. –1 = x y = 2 + 3x = 2 + 3(–1) = –1 Substitute –1 for x. The solution is (–1, –1). Course 3 11-6 Systems of Equations Check It Out: Example 2B Continued

23. Lesson Quiz Solve each system of equations. 1. y = 5x + 10 y = –7 + 5x 2. y = 2x + 5 y = 3x + 2 3. 6x – y = –15 2x + 3y = 5 4. Two numbers have a sum of 23 and a difference of 7. Find the two numbers. Insert Lesson Title Here (–2,3) 15 and 8 ( 3, 11 ) Course 3 11-6 Systems of Equations no solution