Lesson 9.1 solving linear systems by graphing answer key

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ALGEBRA 1 LESSON 9 -1 Solving Systems by Graphing (For help, go to Lessons 2 -4 and 6 -2. ) Solve each equation. 1. 2 n + 3 = 5 n – 2 2. 8 – 4 z = 2 z – 13 3. 8 q – 12 = 3 q + 23 Graph each pair of equations on the same coordinate plane. 4. y = 3 x – 6 y = –x + 2 5. y = 6 x + 1 y = 6 x – 4 6. y = 2 x – 5 6 x – 3 y = 15 7. y = x + 5 y = – 3 x + 5 9 -1

ALGEBRA 1 LESSON 9 -1 Solving Systems by Graphing Solutions 1. 2 n + 3 = 5 n – 2 2. 2 n – 2 n + 3 = 5 n – 2 8 – 4 z + 4 z = 2 z + 4 z – 13 3 = 3 n – 2 8 = 6 z – 13 5 = 3 n 21 = 6 z 12 = n 31 = z 3 3. 8 – 4 z = 2 z – 13 2 8 q – 12 = 3 q + 23 8 q – 3 q – 12 = 3 q – 3 q + 23 5 q – 12 = 23 5 q = 35 q=7 9 -1

ALGEBRA 1 LESSON 9 -1 Solving Systems by Graphing Solutions (continued) 4. y = 3 x – 6 5. y = 6 x + 1 y = –x + 2 y = 6 x – 4 6. y = 2 x – 5 7. y = x + 5 6 x – 3 y = 15 y = – 3 x + 5 – 3 y = – 6 x – 15 y = – 6 x 15 – 3 y = 2 x – 5 9 -1

ALGEBRA 1 LESSON 9 -1 Solving Systems by Graphing Solve by graphing. Check your solutions. y = 2 x + 1 y = 3 x – 1 Graph both equations on the same coordinate plane. y = 2 x + 1 y = 3 x – 1 The slope is 2. The y-intercept is 1. The slope is 3. The y-intercept is – 1. 9 -1

ALGEBRA 1 LESSON 9 -1 Solving Systems by Graphing (continued) Find the point of intersection. The lines intersect at (2, 5), so (2, 5) is the solution of the system. Check: See if (2, 5) makes both equations true. y = 2 x + 1 5 2(2) + 1 5 4+1 5=5 Substitute (2, 5) for (x, y). 9 -1 y = 3 x – 1 5 3(2) – 1 5 6– 1 5=5

ALGEBRA 1 LESSON 9 -1 Solving Systems by Graphing Suppose you plan to start taking an aerobics class. Nonmembers 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 c = number of classes. Let T(c) = total cost of the classes. Relate: cost is membership fee plus Write: member T(c) = 10 + 2 c = 0 + 4 c non-member T(c) 9 -1 cost of classes attended

ALGEBRA 1 LESSON 9 -1 Solving Systems by Graphing (continued) Method 1: Using paper and pencil. T(c) = 2 c + 10 The slope is 2. The intercept on the vertical axis is 10. T(c) = 4 c The slope is 4. The intercept on the vertical axis is 0. Graph the equations. T(c) = 2 c + 10 T(c) = 4 c The lines intersect at (5, 20). After 5 classes, both will cost $20. 9 -1

ALGEBRA 1 LESSON 9 -1 Solving Systems by Graphing (continued) Method 2: Using a graphing calculator. First rewrite the equations using x and y. T(c) = 2 c + 10 y = 2 x + 10 T(c) = 4 c y = 4 x 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

ALGEBRA 1 LESSON 9 -1 Solving Systems by Graphing Solve by graphing. y = 3 x + 2 y = 3 x – 2 Graph both equations on the same coordinate plane. y = 3 x + 2 The slope is 3. The y-intercept is 2. y = 3 x – 2 The slope is 3. The y-intercept is – 2. The lines are parallel. There is no solution. 9 -1

ALGEBRA 1 LESSON 9 -1 Solving Systems by Graphing Solve by graphing. 3 x + 4 y = 12 y=– 3 x+3 4 Graph both equations on the same coordinate plane. 3 x + 4 y = 12 The y-intercept is 3. The x-intercept is 4. y = – 3 x + 3 The slope is – 3. The y-intercept is 3. 4 4 The graphs are the same line. The solutions are an infinite number of ordered pairs (x, y), such that y = – 3 x + 3. 4 9 -1

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