Solving Systems by Graphing ALGEBRA 1 LESSON 7-1 (For help, go to Lessons 2-4 and 6-2.) Graph each pair of equations on the same coordinate plane. 4. y = 3x – 6 5. y = 6x + 1 y = –x + 2 y = 6x – 4 6. y = 2x – 5 7. y = x + 5 6x – 3y = 15 y = –3x + 5 7-1

Solving Systems by Graphing ALGEBRA 1 LESSON 7-1 4. y = 3x – 6 5. y = 6x + 1 y = –x + 2 y = 6x – 4 6. y = 2x – 5 7. y = x + 5 6x – 3y = 15 y = –3x + 5 –3y = –6x – 15 y = y = 2x – 5 Solutions –6x  15 –3 7-1

Solving Systems by Graphing ALGEBRA 1 LESSON 7-1 Solve by graphing. Check your solutions. y = 2x + 1 y = 3x – 1 y = 2x + 1 The slope is 2. The y-intercept is 1. y = 3x – 1 The slope is 3. The y-intercept is –1. Graph both equations on the same coordinate plane. 7-1

Solving Systems by Graphing ALGEBRA 1 LESSON 7-1 (continued) Find the point of intersection. The lines intersect at (2, 5), so (2, 5) is the solution of the system. y = 2x + 1 y = 3x – 1 5 2(2) + 1 Substitute (2, 5) for (x, y). 5 3(2) – 1 5 4 + 1 5 6 – 1 5 = 5 5 = 5 Check: See if (2, 5) makes both equations true. 7-1

Solving Systems by Graphing ALGEBRA 1 LESSON 7-1 Suppose you plan to start taking an aerobics class. Non-members pay $4 per class while members pay $10 a month plus an additional $2 per class. After how many classes will the cost be the same? What is that cost? Define: Let = number of classes. Let = total cost of the classes. c T(c) Relate: cost is membership plus cost of classes fee attended Write: member = 10 + 2 non-member = 0 + 4 T(c) c 7-1

Solving Systems by Graphing ALGEBRA 1 LESSON 7-1 (continued) Method 1: Using paper and pencil. T(c) = 2c + 10 The slope is 2. The intercept on the vertical axis is 10. T(c) = 4c The slope is 4. The intercept on the vertical axis is 0. Graph the equations. T(c) = 2c + 10 T(c) = 4c The lines intersect at (5, 20). After 5 classes, both will cost $20. 7-1

Solving Systems by Graphing ALGEBRA 1 LESSON 7-1 Solve by graphing. y = 3x + 2 y = 3x – 2 Graph both equations on the same coordinate plane. y = 3x – 2 The slope is 3. The y-intercept is –2. y = 3x + 2 The slope is 3. The y-intercept is 2. The lines are parallel. There is no solution. 7-1

Solving Systems by Graphing ALGEBRA 1 LESSON 7-1 Solve by graphing. 3x + 4y = 12 y = – x + 3 3 4 Graph both equations on the same coordinate plane. y = – x + 3 The slope is – . The y-intercept is 3. 3 4 3x + 4y = 12 The y-intercept is 3. The x-intercept is 4. The graphs are the same line. The solutions are an infinite number of ordered pairs (x, y), such that y = – x + 3. 3 4 7-1

Solving Systems by Graphing ALGEBRA 1 LESSON 7-1 Solve by graphing. 1. y = –x – 2 2. y = –x + 3 3. y = 3x + 2 y = x + 3 y = 2x – 6 6x – 2y = –4 4. 2x – 3y = 9 5. –2x + 4y = 12 y = x – 5 – x + y = –3 2 3 (3, 1) (3, 0) Infinitely many solutions 1 2 (6, 1) no solution 7-1