1 / 28

10-6

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

amaya-mejia
Télécharger la présentation

10-6

An Image/Link below is provided (as is) to download presentation Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author. Content is provided to you AS IS for your information and personal use only. Download presentation by click this link. While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server. During download, if you can't get a presentation, the file might be deleted by the publisher.

E N D

Presentation Transcript

  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

More Related