# 3.1 solve linear systems by graphing note: definitely need graph paper for your notes today

TRANSCRIPT

3.1 Solve Linear Systems by Graphing

Note: Definitely need graph paper for your notes today

System of Two Linear Equations

• Also called a linear system

• Simply two linear equations

• Solution of the system: an ordered pair that satisfies both equations

EXAMPLE 1 Solve a system graphically

Graph the linear system and estimate the solution. Then check the solution algebraically.

4x + y = 8

2x – 3y = 18

Equation 1

Equation 2

SOLUTION

Begin by graphing both equations, as shown at the right. From the graph, the lines appear to intersect at (3, – 4). You can check this algebraically as follows.

EXAMPLE 1 Solve a system graphically

Equation 1 Equation 2

4x + y = 8

4(3) + (– 4) 8=?

=?12 – 4 8

8 = 8

2x – 3y = 18

=?2(3) – 3( – 4) 18

=?6 + 12 18

18 = 18

The solution is (3, – 4).

SOLUTION

GUIDED PRACTICE for Example 1

Graph the linear system and estimate the solution. Then check the solution algebraically.1. 3x + 2y = – 4

x + 3y = 13x + 2y = – 4x + 3y = 1

Equation 1Equation 2

Begin by graphing both equations, as shown at the right. From the graph, the lines appear to intersect at (–2, 1). You can check this algebraically as follows.

GUIDED PRACTICE for Example 1

Equation 1 Equation 23x + 2y = –4

=?–6 + 2 –4

x + 3y = 1

–2 + 3 1=?

1 = 1

The solution is (–2, 1).

=?3(–2) + 2(1) –4

–4 = –4

(–2 ) + 3( 1) 1=?

SOLUTION

GUIDED PRACTICE for Example 1

Graph the linear system and estimate the solution. Then check the solution algebraically.

Equation 1Equation 2

2. 4x – 5y = – 102x – 7y = 4

2x – 7y = 44x – 5y = – 10

Begin by graphing both equations, as shown at the right. From the graph, the lines appear to intersect at (–5, –2). You can check this algebraically as follows.

GUIDED PRACTICE for Example 1

Equation 1 Equation 2 2x –7y = 4

–10 + 14 4=?

4 = 4

The solution is (–5, –2).

–10 = –10

4x – 5y = –10=?4(–5) – 5(–2 ) –10 2(–5) – 7(–2 ) 4=?

=?–20 + 10 –10

SOLUTION

GUIDED PRACTICE for Example 1

Graph the linear system and estimate the solution. Then check the solution algebraically.

Equation 1Equation 2

3. 8x – y = 83x + 2y = – 16

8x – y = 83x + 2y = – 16

Begin by graphing both equations, as shown at the right. From the graph, the lines appear to intersect at (0, –8). You can check this algebraically as follows.

GUIDED PRACTICE for Example 1

Equation 1 Equation 2

=? 0 + 8 8 0 – 16 –16=?

–16 = –16

The solution is (0, –8).

8(0) – (–8 ) 8=?

8 = 8

8x – y = 8 3x + 2y = –163(0) + 2(–8) –16=?

How to do it with a graphing calculator

Classifying Systems page 154

• Consistent: at least one solution

• Inconsistent: no solution (lines never intersect)

Types of Consistent Systems

• Independent- exactly one solution (graphs have one intersection point)

• Dependent- infinitely many solutions (two graphs are the same line)

Keep in Mind…

• If it’s inconsistent- parallel lines, neither independent or dependent

• If it’s consistent- either one intersection point or two equations that represent the same line- classify as either independent or dependent

EXAMPLE 2 Solve a system with many solutions

Solve the system. Then classify the system as consistent and independent,consistent and dependent, or inconsistent.

4x – 3y = 8

8x – 6y = 16

Equation 1

Equation 2

SOLUTION

The graphs of the equations are the same line. So, each point on the line is a solution, and the system has infinitely many solutions. Therefore, the system is consistent and dependent.

If it is an inconsistent system (no solution)

• The lines are parallel– The lines have the same _______.

EXAMPLE 3 Solve a system with no solution

Solve the system. Then classify the system as consistent and independent,consistent and dependent, or inconsistent.

2x + y = 4

2x + y = 1

Equation 1

Equation 2

SOLUTION

The graphs of the equations are two parallel lines.Because the two lines have no point of intersection, the system has no solution. Therefore, the system is inconsistent.

Guided Practice 4 – 6

EXAMPLE 4 Standardized Test Practice

SOLUTION

Equation 1 (Option A)

y = 1 x + 30

Equation 2 (Option B)

EXAMPLE 4 Standardized Test Practice

y = x2.5

To solve the system, graph the equations y = x + 30 and y = 2.5x, as shown at the right.

EXAMPLE 4 Standardized Test Practice

Notice that you need to graph the equations only in the first quadrant because only nonnegative values of x and y make sense in this situation.

The lines appear to intersect at about the point (20, 50). You can check this algebraically as follows.

Equation 1 checks.

Equation 2 checks.

50 = 20 + 30

50 = 2.5(20)

ANSWERThe total costs are equal after 20 rides.

The correct answer is B.

Extra Word Problem Example

• A soccer league offers two options for membership plans. Option A incluces an initial fee of $40 and costs $5 for each game played. Option B costs $10 for each game played. After how many games will the total cost of the two options be the same?

Review Questions

• How do you solve a linear system by graphing?

• How can you tell just by looking at the system that the same graph is shown twice?

Homework

• 1, 2, 6-11, 15, 16, 20 – 25, 28-31, 35 – 39