1. Lesson 6-1 Warm-Up

2. “Solving Systems by Graphing” (6-1)
What is a “system of linear equations”? What is the “solution of the system of linear equations”? How do you find the solution for a system of linear equations”?
System of Linear Equations: two or more lines (linear equation) that are on the same graph Solution of the System of Linear Equations: a point that all of the lines in a system of equations have in common (in other words, all of the lines in the system of linear equations cross at that point) Method 1: Make a Graph: Graph all of the lines and find the point where all of the lines cross (the point of intersection) Example: Solve y = 2x – 3 and y = x – 1 Graph both equations on the same coordinate plane. Find the point of intersection. The lines intersect at (2,1), so (2,1) is the solution to the system.

3. “Solving Systems by Graphing” (6-1) (5-3)

4.   Solve by graphing. Check your solution. y = 2x + 1 y = 3x – 1
Solving Systems by Graphing
LESSON 6-1
Additional Examples
Solve by graphing. Check your solution. 1
y = 2x + 1 2
y = 3x – 1 5 2 1 3 1
Slope = 2 =
Slope = 3 = 2
y-intercept = 1
y-intercept = -1 1 3 2  1 3  1
Step 1: Graph both equations on the same coordinate plane.
Step 2: Find the point of intersection.
The lines intersect at (2, 5),
so (2, 5) is the solution of the system.

5. Step 3: Check: See if (2, 5) makes both equations true.
Solving Systems by Graphing
LESSON 6-1
Additional Examples
(continued)
Step 3: Check: See if (2, 5) makes both equations true.
y = 2x y = 3x – 1
5 2(2) Substitute (2, 5) for (x, y) (2) – 1
– 1
5 = = 5

6. Suppose you plan to start taking an aerobics class.
Solving Systems by Graphing
LESSON 6-1
Additional Examples
Suppose you plan to start taking an aerobics class.
Non-members pay $4 per class while members pay a $10 fee plus an additional $2 per class. Write a system of equations that models these plans.
Define: Let = cost of one class
Let = number of classes.
Let = total cost of the classes. n
C(n) C

7. Words: cost is membership plus cost of classes fee attended
Solving Systems by Graphing
LESSON 6-1
Additional Examples
(continued)
Words: cost is membership plus cost of classes
fee attended n
C(n)
Equation: member =
non-member =
The system is C(n) = 4n (or y = 4x)
C(n) = n (or y = 2x + 10)

8. Solving Systems by Graphing
LESSON 6-1
Additional Examples
The system below models the cost of taking an aerobics class as a function of the number of classes. Find the solution of the system by graphing. What does the solution mean in terms of the situation?
Part 1: Find the solutions by graphing.  1
C(a) = 2a The slope is 2. The y-intercept is 10. 2 4
C(a) = 4a The slope is 4. The y-intercept is 0. 2 2 1
Graph the equations. 4 2 1  2
The lines intersect at (5, 20). 4 1 4 1 4 
Part 2: Interpret the solution.
The lines intersect at (5, 20) so the cost will be $20 after 5 classes.

9. “Solving Systems by Graphing” (6-1)
When does a system of linear equations have no solutions? When does a system of linear equations have an infinite number of solutions?
A system of linear equations has no solutions when the two (or more) lines are parallel to one another (in other words, there are no points of intersection). Example: Solve by graphing: y = -2x + 1 and y = -2x – 1 Graph both equations on the same coordinate plane. y = -2x + 1 The slope s -2. The y-intercept is 1 y = -2x - 1 The slope s -2. The y-intercept is -1. The lines are parallel, so there is no solution. A system of linear equations has an “infinite many solutions” when the graphs of the equation are the same line (in other words, every point on the line is a solution). Example: Solve by graphing:2x + 4y = 8 and y = - x + 2 2x + 4y = 8 The y-intercept is 2 and the x-intercept is 4. y = - x + 2 The slope is - .. The y-intercept is 2. The graphs are the same line, so there are an infinite number of solution 1 2   1 2 1 2

10. Graph both equations on the same coordinate plane.
Solving Systems by Graphing
LESSON 6-1
Additional Examples
Solve by graphing. y = 3x + 2
y = 3x – 2
Graph both equations on the same coordinate plane.
y = 3x The slope is 3. The y-intercept is 2.
y = 3x – The slope is 3. The y-intercept is –2.
The lines are parallel. There is no solution.

11.   Solve by graphing. 3x + 4y = 12 y = – x + 3
Solving Systems by Graphing
LESSON 6-1
Additional Examples
Solve by graphing. 3x + 4y = 12
y = – x + 3 3 4
Graph both equations on the same coordinate plane.
3x + 4y = 12 The y-intercept is 3.
The x-intercept is 4. 
y = – x + 3 The slope is – . The y-intercept is 3. 3 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

12.   Solve by graphing. 3x + 4y = 12 y = – x + 3
Solving Systems by Graphing
LESSON 6-1
Additional Examples
Solve by graphing. 3x + 4y = 12
y = – x + 3 3 4
Graph both equations on the same coordinate plane.
3x + 4y = 12 The y-intercept is 3.
The x-intercept is 4. 
y = – x + 3 The slope is . This means
“down 3, right 4” stairsteps.
The y-intercept is also 3. 3 4 -3 -3 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

13. infinitely many solutions
Solving Systems by Graphing
LESSON 6-1
Lesson Quiz
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
Solve by graphing. 2 3
(3, 1)
(3, 0)
infinitely many solutions 1 2
(6, 1)
no solution