1 topic 5.1.1 the graphing method. 2 california standard: 9.0 students solve a system of two linear...

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Topic 5.1.1Topic 5.1.1

The Graphing MethodThe Graphing Method

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The Graphing MethodThe Graphing Method

California Standard:9.0 Students solve a system of two linear equations in two variables algebraically and are able to interpret the answer graphically. Students are able to solve a system of two linear inequalities in two variables and to sketch the solution sets.

What it means for you:You’ll solve systems of linear equations by graphing the lines and working out where they intersect.

Key Words:• system of linear equations• simultaneous equations

Topic5.1.1

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The Graphing MethodThe Graphing Method

In Section 4.5 you graphed two inequalities to find the region of points that satisfied both inequalities.

Plotting two linear equations on a graph involves fewer steps, and it means you can show the joint solution to the equations graphically.

Topic5.1.1

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The Graphing MethodThe Graphing Method

Systems of Linear Equations

A system of linear equations consists of two or more linear equations in the same variables.

Topic5.1.1

For example: 3x + 2y = 7 and x – 3y = –5 form a

system of linear equations in two variables — x and y.

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The Graphing MethodThe Graphing MethodTopic5.1.1

The solution of a system of linear equations in two variables is a pair of values like x and y, or (x, y), that satisfies each of the equations in the system.

For example, x = 1, y = 2 or (1, 2) is the solution of the

system of equations 3x + 2y = 7 and x – 3y = –5 , since it

satisfies both equations:

3x + 2y = 73(1) + 2(2) = 7

3 + 4 = 7

3x + 2y = 73(1) + 2(2) = 7

3 + 4 = 7

x – 3y = –51 – 3(2) = –5

1 – 6 = –5

x – 3y = –51 – 3(2) = –5

1 – 6 = –5

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The Graphing MethodThe Graphing MethodTopic5.1.1

Equations in a system are often called simultaneous equations because any solution has to satisfy the equations simultaneously (at the same time).

The equations can’t be solved independently of one another.

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The Graphing MethodThe Graphing Method

Solving Systems of Equations by Graphing

Topic5.1.1

A system of two linear equations can be solved graphically, by graphing both equations in the same coordinate plane.

Every point on the line of an equation is a solution of that equation.

The point at which the two lines cross lies on both lines and so is the solution of both equations.

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The Graphing MethodThe Graphing Method

The solution of a system of linear equations in two variables is the point of intersection (x, y) of their graphs.

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Point of intersection

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The Graphing MethodThe Graphing Method

Example 1

Solve this system of equations by graphing:

Solution follows…

Topic5.1.1

2x – 3y = 7

–2x + y = –1

Step 1: Graph both equations in the same coordinate plane.

Solution

Line of first equation: Line of second equation: 2x – 3y = 7

–2x + y = –1

The line goes through the points (2, –1) and (–1, –3).

The line goes through the points (0, –1) and (1, 1).

y = 2x –1

3y = 2x – 7

y = x – 2

3

7

3

x y

2 –1

–1 –3

x y

0 –1

1 1

First find some points that lie on each of the lines.

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The Graphing MethodThe Graphing Method

Example 1

Solution continues…

Now you can draw the graph:

Solution (continued)

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Step 2: Read off the coordinates of the point of intersection.

(–1, –3)

(1, 1)

(2, –1)(0, –1)

y = 2x –1

y = x – 23

73

The point of intersection is (–1, –3). (–1, –3)

Solve this system of equations by graphing: 2x – 3y = 7

–2x + y = –1

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The Graphing MethodThe Graphing Method

Example 1

Solution (continued)

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Step 3: Check whether your coordinates give true statements when they are substituted into each equation.

So x = –1, y = –3 is the solution of the system of equations.

2x – 3y = 7 2(–1) – 3(–3) = 7 7 = 7 — True statement

–2x + y = –1 –2(–1) + (–3) = –1 –1 = –1 — True statement

Solve this system of equations by graphing: 2x – 3y = 7

–2x + y = –1

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The Graphing MethodThe Graphing Method

Guided Practice

Solution follows…

Solve each system of equations in Exercises 1–2by graphing on x- and y-axes spanning from –6 to 6.

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2. y + x = 3 and 3y – x = 51. y + x = 2 and y = x + 22

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y + x = 2

2

3y = x + 2

y + x = 3

3y – x = 5

(0, 2) (1, 2)

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The Graphing MethodThe Graphing Method

Guided Practice

Solution follows…

3. y = x – 3 and y + 2x = 3

Solve each system of equations in Exercises 3–4by graphing on x- and y-axes spanning from –6 to 6.

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4. y – x = 1 and y + x = –33

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2

y + 2x = 3

y = x – 3

y – x = 13

2

y + x = –31

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(2, –1) (–2, –2)

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The Graphing MethodThe Graphing Method

Guided Practice

Solution follows…

5. y – x = 3 and y + x = –1

Solve each system of equations in Exercises 5–6by graphing on x- and y-axes spanning from –6 to 6.

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6. 2y – x = –6 and y + x = –31

2

y – x = 3

y + x = –1

(–2, 1)

y + x = –31

22y – x = –6

(0, –3)

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The Graphing MethodThe Graphing Method

Solution follows…

Solve each system of equations in Exercises 1–2by graphing on x- and y-axes spanning from –6 to 6.

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2. x + y = 0 and y = –2x1. 2x + y = 7 and y = x + 1

2x + y = 7

y = x + 1

(2, 3)

Independent Practice

x + y = 0

y = –2x

(0, 0)

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The Graphing MethodThe Graphing Method

Solution follows…

Solve each system of equations in Exercises 3–4by graphing on x- and y-axes spanning from –6 to 6.

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4. x – y = 4 and x + 4y = –13. y = –3 and x – y = 2

y = –3

x – y = 2

(–1, –3)

Independent Practice

x – y = 4

x + 4y = –1

(3, –1)

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The Graphing MethodThe Graphing Method

Solution follows…

Solve each system of equations in Exercises 5–6by graphing on x- and y-axes spanning from –6 to 6.

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6. y = –x and y = 4x5. 2y + 4x = 4 and y = –x + 3

2y + 4x = 4

y = –x + 3

(–1, 4)

Independent Practice

y = –xy = 4x

(0, 0)

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The Graphing MethodThe Graphing Method

Solution follows…

Determine the solution to the systems of equations graphed in Exercises 7 and 8.

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8. 7.

y = –x

Independent Practice

y = 2x – 7

x + y = –4

y = x – 41

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(3, –3) (2, –3)

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The Graphing MethodThe Graphing Method

Solution follows…

Solve each system of equations in Exercises 9–10by graphing on x- and y-axes spanning from –6 to 6.

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10. y = 2x – 1 and x + y = 89. x – y = 6 and x + y = 0

x + y = 0

Independent Practice

y = 2x – 1

x + y = 8

x – y = 6

(3, 5)(3, –3)

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The Graphing MethodThe Graphing Method

Solution follows…

Solve each system of equations in Exercises 11–12by graphing on x- and y-axes spanning from –6 to 6.

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12. x – y = 0 and x + y = 811. 4x – 3y = 0 and 4x + y = 16

4x – 3y = 04x + y = 16

(3, 4)

Independent Practice

x – y = 0

x + y = 8

(4, 4)

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The Graphing MethodThe Graphing Method

Solution follows…

Solve each system of equations in Exercises 13–14by graphing on x- and y-axes spanning from –6 to 6.

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14. x – y = 1 and x + y = –313. y = –x + 6 and x – y = –4

y = –x + 6x – y = –4

(1, 5)

Independent Practice

x – y = 1 x + y = –3

(–1, –2)

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The Graphing MethodThe Graphing Method

Solution follows…

Solve each system of equations in Exercises 15–16by graphing on x- and y-axes spanning from –6 to 6.

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16. 2x + y = –8 and 3x + y = –1315. x + y = 1 and x – 2y = 1

x + y = 1

x – 2y = 1

Independent Practice

2x + y = –8

3x + y = –13

(1, 0) (–5, 2)

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Round UpRound Up

There’s something very satisfying about taking two long linear equations and coming up with just a one-coordinate-pair solution.

You should always substitute your solution back into the original equations, to check that you’ve got the correct answer.

The Graphing MethodThe Graphing MethodTopic5.1.1