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Chapter 3 Section 1 Graphing Systems of Equations

Chapter 3

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Chapter 3. Section 1 Graphing Systems of Equations. Graph each equation. 1. y = 3 x – 2 2. y = – x 3. y = – x + 4 Graph each equation. Use one coordinate plane for all three graphs. 4. 2 x – y = 1 5. 2 x – y = –1 6. x + 2 y = 2. 1 2. Lesson Preview p. 116. - PowerPoint PPT Presentation

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Page 1: Chapter 3

Chapter 3

Section 1

Graphing Systems of Equations

Page 2: Chapter 3

Lesson Preview p. 116Graph each equation.

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

Graph each equation. Use one coordinate plane for all three graphs.

4. 2x – y = 1 5. 2x – y = –1 6. x + 2y = 2

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Page 3: Chapter 3

Lesson Preview Answers

1. y = 3x – 2 2. y = –xslope = 3 slope = –1y-intercept = –2 y-intercept = 0

3. y = – x + 4 4. 2x – y = 1

slope = – –y= –2x + 1

y-intercept = 4 y = 2x – 1

5. 2x – y = –1 6. x + 2y = 2–y = –2x – 1 2y = –x + 2y = 2x + 1 y = – x + 1

12 1

2

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Page 4: Chapter 3

Graphing Systems of Equations

Systems of Equations – a set of two or more equations that use the same variable

Linear system – a system where the graph of each equation in a system of two variables is a lineA brace is used to keep the equation system

together

32

3

xy

xy

Page 5: Chapter 3

Solutions to Systems of Linear Equations A solution of a system of equations is a set

of values for the variables that makes all the equations true.

One way to solve linear systems is by graphing -the solution is the point(s) where the graphs intersect

Page 6: Chapter 3

Solve the system by graphing.

x + 3y = 23x + 3y = –6

Check: Show that (–4, 2) makes both equations true.

Graph the equations and find the intersection. The solution appears to be (–4, 2).

x + 3y = 2 3x + 3y = –6(–4) + 3(2) 2 3(–4) + 3(2) –6

(–4) + 6 2 –12 + 6 –6 2 = 2 –6 = –6

3

2

3

1

23

23

xy

xy

yx

2

633

633

xy

xy

yxHere are the equations solved for slope intercept form:

Page 7: Chapter 3

Check Understanding p. 117

Solve by graphing. Check your solution.

2

52

yx

yx

2

52

xy

xygraph

Step1:

Graph the equations (slope-intercept form is nice) Step 2: Locate where the

two graphs intersect

Step 3: Name the coordinate point

(1,3)

Step 4: Check in both equations for accuracy.

55

532

53)1(2

52

yx

22

231

2

yx

Page 8: Chapter 3

Classifying Systems -by the number of solutions (graphing)

Independent – the system has one solution (only one coordinate pair will work for all equations). These lines intersect in one point.

Page 9: Chapter 3

Dependent – the system has no unique solutions (there is more than one coordinate pair that will work). These are graphed as coinciding lines.

Page 10: Chapter 3

Inconsistent – the system has no solution (no coordinate pair will work). These lines are parallel.

Page 11: Chapter 3

Classify the System

Independent

One Solution

Dependent

No Unique Solution

(All points of line are solutions)

Inconsistent

No Solution

Page 12: Chapter 3

Classifying Systems without Graphing-comparing slopes and intercepts

Independent – Different slopes

m ≠

Dependent – Equal slopes Equal y-intercepts

m = b =

Inconsistent – Equal slopes Different y-intercepts

m = b ≠

Step 1: Solve for slope intercept form

Step 2: Write clearly m= and b= for each equation

Step 3: Compare slopes and intercepts

Page 13: Chapter 3

Classify the system without graphing.

Since the slopes are the same, compare the y-intercepts.

y = 3x + 2–6x + 2y = 4

y = 3x + 2 –6x + 2y = 4Rewrite in slope-intercept form. y = 3x + 2

m = 3, b = 2 Find the slope and y-intercept. m = 3, b = 2

Since the y-intercepts are the same, the lines are dependent.

Page 14: Chapter 3

Homework