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Solving Systems by Graphing

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Presentation on theme: "Solving Systems by Graphing"— Presentation transcript:

1 Solving Systems by Graphing
ALGEBRA 1 LESSON 9-1 (For help, go to Lessons 2-4 and 6-2.) Solve each equation. 1. 2n + 3 = 5n – – 4z = 2z – q – 12 = 3q + 23 Graph each pair of equations on the same coordinate plane. 4. y = 3x – y = 6x + 1 y = –x y = 6x – 4 6. y = 2x – y = x + 5 6x – 3y = y = –3x + 5 9-1

2 Solving Systems by Graphing
ALGEBRA 1 LESSON 9-1 n + 3 = 5n – – 4z = 2z – 13 2n – 2n + 3 = 5n – 2n – – 4z + 4z = 2z + 4z – 13 3 = 3n – = 6z – 13 5 = 3n 21 = 6z 1 = n 3 = z q – 12 = 3q + 23 8q – 3q – 12 = 3q – 3q + 23 5q – 12 = 23 5q = 35 q = 7 Solutions 2 3 1 2 9-1

3 Solving Systems by Graphing
ALGEBRA 1 LESSON 9-1 4. y = 3x – y = 6x + 1 y = –x y = 6x – 4 y = 2x – y = x + 5 6x – 3y = y = –3x + 5 –3y = –6x – 15 y = y = 2x – 5 Solutions (continued) –6x  15 –3 9-1

4 Solving Systems by Graphing
ALGEBRA 1 LESSON 9-1 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. 9-1

5 Solving Systems by Graphing
ALGEBRA 1 LESSON 9-1 (continued) Find the point of intersection. The lines intersect at (2, 5), so (2, 5) is the solution of the system. y = 2x y = 3x – 1 5 2(2) + 1 Substitute (2, 5) for (x, y) (2) – 1 – 1 5 = = 5 Check: See if (2, 5) makes both equations true. 9-1

6 Solving Systems by Graphing
ALGEBRA 1 LESSON 9-1 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 = non-member = T(c) c 9-1

7 Solving Systems by Graphing
ALGEBRA 1 LESSON 9-1 (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. 9-1

8 Solving Systems by Graphing
ALGEBRA 1 LESSON 9-1 (continued) Method 2: Using a graphing calculator. First rewrite the equations using x and y. T(c) = 2c y = 2x + 10 T(c) = 4c y = 4x Then graph the equations using a graphing calculator. Set an appropriate range. Then graph the equations. Use the key to find the coordinates of the intersection point. The lines intersect at (5, 20). After 5 classes, both will cost $20. 9-1

9 Solving Systems by Graphing
ALGEBRA 1 LESSON 9-1 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. 9-1

10 Solving Systems by Graphing
ALGEBRA 1 LESSON 9-1 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 9-1

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


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