ALGEBRA 1 LESSON 9-1
Solving Systems by Graphing
(For help, go to Lessons 2-4 and 6-2.)
Solve each equation.
1. 2n + 3 = 5n – 2
2. 8 – 4z = 2z – 13
3. 8q – 12 = 3q + 23
Graph each pair of equations on the same coordinate plane.
4. y = 3x – 6
y = –x + 2
5. y = 6x + 1
y = 6x – 4
6. y = 2x – 5
6x – 3y = 15
7. y = x + 5
y = –3x + 5
9-1
ALGEBRA 1 LESSON 9-1
Solving Systems by Graphing
Solutions
1.
2n + 3 = 5n – 2
2.
2n – 2n + 3 = 5n – 2n – 2
8 – 4z + 4z = 2z + 4z – 13
3 = 3n – 2
8 = 6z – 13
5 = 3n
21 = 6z
12 = n
1
3
3
3.
8 – 4z = 2z – 13
2
8q – 12 = 3q + 23
8q – 3q – 12 = 3q – 3q + 23
5q – 12 = 23
5q = 35
q=7
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=z
ALGEBRA 1 LESSON 9-1
Solving Systems by Graphing
Solutions (continued)
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 = –6x  15
–3
y = 2x – 5
9-1
ALGEBRA 1 LESSON 9-1
Solving Systems by Graphing
Solve by graphing. Check your solutions.
y = 2x + 1
y = 3x – 1
Graph both equations on the same coordinate plane.
y = 2x + 1
y = 3x – 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 = 2x + 1
5 2(2) + 1
5 4+1
5=5
Substitute (2, 5) for (x, y).
9-1
y = 3x – 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) = 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
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) = 2c + 10
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
ALGEBRA 1 LESSON 9-1
Solving Systems by Graphing
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
ALGEBRA 1 LESSON 9-1
Solving Systems by Graphing
Solve by graphing. 3x + 4y = 12
y=–3 x+3
4
Graph both equations on the same coordinate plane.
3x + 4y = 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
2. y = –x + 3
3. y = 3x + 2
y=2x+3
y = 2x – 6
6x – 2y = –4
(3, 1)
(3, 0)
Infinitely many
solutions
3
4. 2x – 3y = 9
5. –2x + 4y = 12
y=x–5
– 1 x + y = –3
(6, 1)
no solution
2
9-1