Warm Up Problem of the Day Lesson Presentation Lesson Quizzes

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

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

Learn to solve systems of equations.

Vocabulary system of equations solution of a system of equations

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.

When solving systems of equations, remember to find values for all of the variables. Caution!

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

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. 3x 9 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).

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.

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

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).

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.

To solve a general system of two equations with two variables, you can solve both equations for x or both for y.

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

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

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

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 – 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.

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

Check It Out: Example 2A Continued + 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).

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 – x y = 2 + 3x –2 – x = 2 + 3x

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).

Lesson Quizzes Standard Lesson Quiz Lesson Quiz for Student Response Systems

( , 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

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

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) 26

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 27