ALGEBRA 1

Lesson 6-1 Warm-Up

ALGEBRA 1

“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.

ALGEBRA 1

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

ALGEBRA 1

Solve by graphing. Check your solution.

y = 2x + 1y = 3x – 1

Graph both equations on the same coordinate plane.

y = 2x + 1 The slope is 2. The y-intercept is 1.y = 3x – 1 The slope is 3. The y-intercept is –1.

Solving Systems by GraphingLESSON 6-1

Additional Examples

ALGEBRA 1

(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 y = 3x – 1

5 2(2) + 1 Substitute (2, 5) for (x, y). 5 3(2) – 1

5 4 + 1 5 6 – 1

5 = 5 5 = 5

Solving Systems by GraphingLESSON 6-1

Additional Examples

ALGEBRA 1

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)

Solving Systems by GraphingLESSON 6-1

Additional Examples

C

ALGEBRA 1

The system is C(n) = 4n (or y = 4x) C(n) = 10 + 2n (or y = 2x + 10)

Words: cost is membership plus cost of classes fee attended

nC(n)Equation: member = 10 + 2

C(n) nnon-member = 0 + 4

(continued)

Solving Systems by GraphingLESSON 6-1

Additional Examples

ALGEBRA 1

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?

C(a) = 2a + 10 The slope is 2. The intercept on the vertical axis is 10.

C(a) = 2a + 10

Graph the equations.

The lines intersect at (5, 20).

Part 1: Find the solutions by graphing.

C(a) = 4a The slope is 4. The intercept on the vertical axis is 0.

C(a) = 4a

Solving Systems by GraphingLESSON 6-1

Additional Examples

ALGEBRA 1

Part 2: Interpret the solution.

(continued)

Solving Systems by GraphingLESSON 6-1

Additional Examples

The lines intersect at (5, 20) so the cost will be $20 after 5 classes.

ALGEBRA 1

“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

Graph both equations on the same coordinate plane.

2x + 4y = 8 The y-intercept is 2 and the x-intercept is 4.

y = - x + 2 The slope s - . The y-intercept is 2.

The graphs are the same line, so there are an infinite number of solution

12

12

12

ALGEBRA 1

Solve by graphing. y = 3x + 2

y = 3x – 2

Graph both equations on the same coordinate plane.

The lines are parallel. There is no solution.

y = 3x + 2 The slope is 3. The y-intercept is 2.

y = 3x – 2 The slope is 3. The y-intercept is –2.

Solving Systems by GraphingLESSON 6-1

Additional Examples

ALGEBRA 1

Solve by graphing.3x + 4y = 12

y = – x + 3

Graph both equations on the same coordinate plane.

The graphs are the same line. The solutions are

an infinite number of ordered pairs (x, y), such

that y = – x + 3.34

34

3x + 4y = 12 The y-intercept is 3.The x-intercept is 4.

y = – x + 3 The slope is – . The y-intercept is 3.34

34

Solving Systems by GraphingLESSON 6-1

Additional Examples

ALGEBRA 1

1. y = –x – 2 2. y = –x + 3 3. y = 3x + 2 y = x + 3 y = 2x – 6 6x – 2y = –4

4. 2x – 3y = 9 5. –2x + 4y = 12

y = x – 5 – x + y = –3

23

12

(3, 1) (3, 0) infinitely many solutions

(6, 1) no solution

Solve by graphing.

Solving Systems by GraphingLESSON 6-1

Lesson Quiz