1. 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 – y = 6x + 1
y = –x y = 6x – 4
6. y = 2x – y = x + 5
6x – 3y = y = –3x + 5
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2. Solving Systems by Graphing
ALGEBRA 1 LESSON 7-1
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
–6x  15 –3
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3. 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.
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4. 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 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.
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5. 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 =
non-member =
T(c) c
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6. 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.
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7. 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.
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8. 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
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9. Solving Systems by Graphing
ALGEBRA 1 LESSON 7-1
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
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