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# 10-6 - PowerPoint PPT Presentation

## 10-6

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1. Systems of Equations 10-6 Course 3 Warm Up Problem of the Day Lesson Presentation

2. Systems of Equations 10-6 3V = h 1 C – S A 3 t Course 3 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

3. Systems of Equations 10-6 Course 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

4. Systems of Equations 10-6 Course 3 Learn to solve systems of equations.

5. Systems of Equations 10-6 Course 3 Insert Lesson Title Here Vocabulary system of equations solution of a system of equations

6. Systems of Equations 10-6 Course 3 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.

7. Systems of Equations 10-6 ? ? 5(1) + 2 = 7 1 – 3(2) = 11 Course 3 Additional Example 1A: Identifying Solutions of a System of Equations Determine if the ordered pair is a solution of the system of equations below. 5x + y = 7 x – 3y = 11 A. (1, 2) 5x + y = 7 x – 3y = 11 Substitute for x and y. 7 = 7 –5  11   The ordered pair (1, 2) is not a solution of the system of equations.

8. Systems of Equations 10-6 ? 2 – 3(–3) = 11 ? 5(2) + –3 = 7 Course 3 Additional Example 1B: Identifying Solutions of a System of Equations Determine if the ordered pair is a solution of the system of equations below. 5x + y = 7 x – 3y = 11 B. (2, –3) 5x + y = 7 x – 3y = 11 Substitute for x and y. 7 = 7 11 = 11   The ordered pair (2, –3) is a solution of the system of equations.

9. Systems of Equations 10-6 ? ? 5(20) + (3) = 7 20 – 3(3) = 11 Course 3 Additional Example 1C: Identifying Solutions of a System of Equations Determine if the ordered pair is a solution of the system of equations below. 5x + y = 7 x – 3y = 11 C. (20, 3) 5x + y = 7 x – 3y = 11 Substitute for x and y. 103  7  11 = 11  The ordered pair (20, 3) is not a solution of the system of equations.

10. Systems of Equations 10-6 ? ? 4(1) + 2 = 8 1 – 4(2) = 12 Course 3 Try This: Example 1A Determine if each ordered pair is a solution of the system of equations below. 4x + y = 8 x – 4y = 12 A. (1, 2) 4x + y = 8 x – 4y = 12 Substitute for x and y. 6  8 –7  12   The ordered pair (1, 2) is not a solution of the system of equations.

11. Systems of Equations 10-6 ? 2 – 4(–3) = 12 ? 4(2) + –3 = 8 Course 3 Try This: Example 1B Determine if each ordered pair is a solution of the system of equations below. 4x + y = 8 x – 4y = 12 B. (2, –3) 4x + y = 8 x – 4y = 12 Substitute for x and y. 5  8 14  12   The ordered pair (2, –3) is not a solution of the system of equations.

12. Systems of Equations 10-6 ? 1 – 4(4) = 12 ? 4(1) + 4 = 8 Course 3 Try This: Example 1C Determine if each ordered pair is a solution of the system of equations below. 4x + y = 8 x – 4y = 12 C. (1, 4) 4x + y = 8 x – 4y = 12 Substitute for x and y. 8 = 8  –15  12  The ordered pair (1, 4) is not a solution of the system of equations.

13. Systems of Equations 10-6 Helpful Hint When solving systems of equations, remember to find values for all of the variables. Course 3

14. Systems of Equations 10-6 Course 3 Additional Example 2: Solving Systems of Equations y = x – 4 Solve the system of equations. y = 2x – 9 y=y y= x – 4 y =2x – 9 x – 4 = 2x – 9 Solve the equation to find x. x – 4 = 2x – 9 – x– x Subtract x from both sides. –4 = x – 9 + 9+ 9 Add 9 to both sides. 5 = x

15. Systems of Equations 10-6 ? ? 1 = 5 – 4 1 = 2(5) – 9 Course 3 Additional Example 2 Continued To find y, substitute 5 for x in one of the original equations. y = x – 4 = 5 – 4 = 1 The solution is (5, 1). Check: Substitute 5 for x and 1 for y in each equation. y = x – 4 y = 2x – 9 1 = 1  1 = 1 

16. Systems of Equations 10-6 Course 3 Try This: Example 2 y = x – 5 Solve the system of equations. y = 2x – 8 y=y y= x – 5 y =2x – 8 x – 5 = 2x – 8 Solve the equation to find x. x – 5 = 2x – 8 – x– x Subtract x from both sides. –5 = x – 8 + 8+ 8 Add 8 to both sides. 3 = x

17. Systems of Equations 10-6 ? ? –2 = 3 – 5 –2 = 2(3) – 8 Course 3 Try This: Example 2 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). Check: Substitute 3 for x and –2 for y in each equation. y = x – 5 y = 2x – 8 –2 = –2  –2 = –2 

18. Systems of Equations 10-6 Course 3 To solve a general system of two equations with two variables, you can solve both equations for x or both for y.

19. Systems of Equations 10-6 Course 3 Additional Example 3A: Solving Systems of Equations Solve the system of equations. A. x + 2y = 8 x – 3y = 13 Solve both equations for x. x + 2y = 8 x – 3y = 13 –2y–2y+ 3y+ 3y x = 8 – 2y x = 13 + 3y 8 – 2y = 13 + 3y Add 2y to both sides. + 2y+ 2y 8 = 13 + 5y

20. Systems of Equations 10-6 = –5 5 5y 5 Course 3 Additional Example 3A Continued 8 = 13 + 5y Subtract 13 from both sides. –13–13 –5 = 5y Divide both sides by 5. –1 = y x = 8 – 2y = 8 – 2(–1)Substitute –1 for y. = 8 + 2 = 10 The solution is (10, –1).

21. Systems of Equations 10-6 Helpful Hint You can choose either variable to solve for. It is usually easiest to solve for a variable that has a coefficient of 1. Course 3

22. Systems of Equations 10-6 = – –3 –3y –3 3x –3 –3 Course 3 Additional Example 3B: Solving Systems of Equations Solve the system of equations. B. 3x – 3y = -3 2x + y = -5 Solve both equations for y. 3x – 3y = –3 2x + y = –5 –3x–3x–2x–2x –3y = –3 – 3xy = –5 – 2x y = 1 + x 1 + x = –5 – 2x

23. Systems of Equations 10-6 –6 3x 3 3 = Course 3 Additional Example 3B Continued 1 + x = –5 – 2x Add 2x to both sides. + 2x+ 2x 1 + 3x = –5 Subtract 1 from both sides. –1–1 3x = –6 Divide both sides by 3. x = –2 y = 1 + x Substitute –2 for x. = 1 + –2 = –1 The solution is (–2, –1).

24. Systems of Equations 10-6 Course 3 Try This: Example 3A Solve the system of equations. A. x + y = 5 3x + y = –1 Solve both equations for y. x + y = 5 3x + y = –1 –x–x– 3x– 3x y = 5 – x y = –1 – 3x 5 – x = –1 – 3x Add x to both sides. + x+ x 5 = –1 – 2x

25. Systems of Equations 10-6 Course 3 Try This: Example 3A Continued 5 = –1 – 2x + 1+ 1 Add 1 to both sides. 6 = –2x –3 = x Divide both sides by –2. y = 5 – x = 5 – (–3)Substitute –3 for x. = 5 + 3 = 8 The solution is (–3, 8).

26. Systems of Equations 10-6 Course 3 Try This: Example 3B Solve the system of equations. B. x + y = –2 –3x + y = 2 Solve both equations for y. x + y = –2 –3x + y = 2 – x– x+ 3x+ 3x y = –2 – xy = 2 + 3x –2 – x = 2 + 3x

27. Systems of Equations 10-6 Course 3 Try This: Example 3B Continued –2 – x = 2 + 3x Add x to both sides. + x+ x –2 = 2 + 4x Subtract 2 from both sides. –2–2 –4 = 4x Divide both sides by 4. –1 = x y = 2 + 3x Substitute –1 for x. = 2 + 3(–1) = –1 The solution is (–1, –1).

28. Systems of Equations 10-6 1 2 ( , 2) Course 3 Insert Lesson Title Here Lesson Quiz 1. Determine if the ordered pair (2, 4) is a solution of the system. y = 2x; y = –4x + 12 Solve each system of equations. 2.y = 2x + 1; y = 4x 3. 6x – y = –15; 2x + 3y = 5 4. Two numbers have a sum of 23 and a difference of 7. Find the two numbers. yes (–2,3) 15 and 8