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11-6

Systems of Equations. 11-6. Course 3. Warm Up. Problem of the Day. Lesson Presentation. Systems of Equations. 11-6. 3 V. = A. 1. C – S. h. 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.

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11-6

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

  2. Systems of Equations 11-6 3V = A 1 C – S h 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 11-6 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 11-6 Course 3 Learn to solve systems of equations.

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

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

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

  9. Systems of Equations 11-6 Course 3 Additional Example 1A Continued Solve the equation to find x. 4x – 6 = x + 3 – x– x Subtract x from both sides. 3x - 6 = 3 + 6+ 6 Add 6 to both sides. 3x9 Divide both sides by 3. 3 = 3 x = 3 To find y, substitute 3 for x in one of the original equations. y = x + 3 = 3 + 3 = 6 The solution is (3, 6).

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

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

  12. Systems of Equations 11-6 Course 3 Check It Out: Example 1A Continued 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 To find y, substitute 3 for x in one of the original equations. y = x – 5 = 3 – 5 = –2 The solution is (3, –2).

  13. Systems of Equations 11-6 Course 3 Check It Out: Example 1B y = 3x + -7 y = 6 + 3x 3x + -7 = 6 + 3x Transitive Property Subtract 3x from both sides. – 3x– 3x -7 ≠ 6 The system of equations has no solution.

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

  15. Systems of Equations 11-6 Course 3 Additional Example 2A: Solving Systems of Equations by Solving for a Variable Solve the system of equations. 5x + y = 7 x – 3y = 11 Solve both equations for x. 5x + y = 7 x – 3y = 11 -y-y+ 3y+ 3y 5x = 7 - y x = 11 + 3y 5(11 + 3y)= 7 -y 55 + 15y = 7 – y Subtract 15y from both sides. - 15y- 15y 55 = 7 – 16y

  16. Systems of Equations 11-6 Course 3 Additional Example 2A Continued 55 = 7 – 16y Subtract 7 from both sides. –7–7 48-16y –16 = -16 Divide both sides by –16. -3 = y x = 11 + 3y = 11 + 3(-3)Substitute –3 for y. = 11 + –9 = 2 The solution is (2, –3).

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

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

  19. Systems of Equations 11-6 Course 3 Check It Out: Example 2A Solve the system of equations. 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

  20. Systems of Equations 11-6 Course 3 Check It Out: Example 2A 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).

  21. Systems of Equations 11-6 Course 3 Check It Out: Example 2B Solve the system of equations. 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

  22. Systems of Equations 11-6 Course 3 Check It Out: Example 2B 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).

  23. Systems of Equations 11-6 1 2 ( , 2) Course 3 Insert Lesson Title Here Lesson Quiz Solve each system of equations. 1. y = 5x + 10 y = –7 + 5x 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. no solution (–2,3) 15 and 8

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