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Warm Up Problem of the Day Lesson Presentation Lesson Quizzes
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Solve for the indicated variable. 1. P = R – C for R 2. V = Ah for A
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 1 3 = A 3V h C – S t Rt + S = C
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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
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Sunshine State Standards
MA.8.A.1.3 Use tables, graphs, and models to represent, analyze, and solve real-world problems related to systems of linear equations.
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When solving systems of equations, remember to find values for all of the variables.
Caution!
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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 are equal to each other. y = 4x – 6 y = x + 3 4x – 6 = x + 3
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Additional Example 1A Continued
Solve the equation to find x. 4x – 6 = x + 3 – x – x Subtract x from both sides. 3x – 6 = Add 6 to both sides. 3x Divide both sides by 3. = x = To find y, substitute 3 for x in one of the original equations. y = x + 3 = = 6 The solution is (3, 6).
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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.
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Check It Out: Example 1A Solve each system of equations. y = x – 5 y = 2x – 8 x – 5 = 2x – 8 y = x – 5 3 = x y = (3) – 5 = –2 The solution is (3, –2).
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Check It Out: Example 1B Solve each system of equations. y = 2x y = x + 6 2x = x + 6 y = 2x x = 6 y = 2(6) = 12 The solution is (6, 12).
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When equations in a system are not already solve for one variable, you can solve both equations for x or both for y.
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Additional Example 2A: Solving Systems of Equations by Solving for a Variable
Solve the system of equations. x + 4y = – x – 3y = 11 Solve both equations for x. x + 4y = – x – 3y = 11 –4y –4y y y x = –10 – 4y x = y –10 – 4y = y –10 – 4y = y Subtract 3y from both sides. – 3y – 3y –10 – 7y = 11
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Additional Example 2A Continued
–10 – 7y = 11 Add 10 to both sides. – 7y Divide both sides by –7. –7 = – 7 y = –3 x = y = (–3) Substitute –3 for y. = 11 + –9 = 2 The solution is (2, –3).
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You can solve for either variable
You can solve for either variable. It is usually easiest to solve for a variable that has a coefficient of 1. Helpful Hint
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Additional Example 2B: Solving Systems of Equations by Solving for a Variable
Solve the system of equations. –2x + 10y = – x – 5y = 4 Solve both equations for x. –2x + 10y = – x – 5y = 4 –10y –10y y y –2x = –8 – 10y x = 4 + 5y = – –8 –2 10y –2x 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.
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Check It Out: Example 2A Solve each system of equations. 2x + y = 0 2x + 3y = 8 −2x + 8 3 y = −2x and y = −2x + 8 3 –2x = ; x –2 2x + y = 0 2(−2) + y = 0 y = 4 The solution is (–2, 4).
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Check It Out: Example 2B Solve the system of equations. y = x –1 –3x + 3y = 4 3x + 4 3 y = x − 1 and y = 3x + 4 3 x – 1 = −3 ≠ 4 There are no solutions.
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Lesson Quizzes Standard Lesson Quiz Lesson Quiz for Student Response Systems
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( , 2) Lesson Quiz Solve each system of equations. 1. y = 5x + 10
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) 1 2 (–2, 3) 15 and 8
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Lesson Quiz for Student Response Systems
1. Solve the given system of equations. y = 11x + 20 y = –2 + 11x A. (2, 2) B. (1, 1) C. (1, –1) D. no solution
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Lesson Quiz for Student Response Systems
2. Solve the given system of equations. 4x + y = 11 2x + 3y = –7 A. (4, –5) B. (4, 5) C. (2, –5) D. (2, 5) 21
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Lesson Quiz for Student Response Systems
3. Two numbers have a sum of 37 and a difference of 17. Identify the two numbers. A. –27 and –10 B. –27 and 10 C. 27 and 10 D. 27 and –10 22
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