systems of equations and inequalities module 9. solving linear equations by graphing lesson 9.1 –...

Systems of Equations and

InequalitiesModule 9

Solving Linear Equations by Graphing Lesson 9.1 – Page 277

Vocabulary

• System of linear equations – also called a linear system, consists of two or more linear equations that have the same variables.

• Solution of a system of linear equations – an ordered pair that satisfies all of the equations in the system.

Systems of linear equations can be solved using various methods:

• Graphing• Substitution• Adding or subtracting• Multiplying

In this lesson, we will solve systems of linear equations by graphing.

Example 1

Solve the system of linear equations below by graphing. Check your answer.

2x – y = 2 x + 2y = 6

You can use the x- and y-intercepts of each function to graph each line.

2x – y = 2

x-intercept y-intercept

2x – 0 = 2 2(0) – y = 22x = 2 - y = 2

= =

x = 1 y = -2

Now graph the points (1, 0) and (0, -2). Draw a line through the two points.

x + 2y = 6

x-intercept y-interceptx + 2(0) = 6 (0) + 2y = 6

x = 6 2y = 6

=

y = 3

Now graph the points (6, 0) and (0, 3). Draw a line through the two points.

Find the point of intersection.

The two lines intersect at the point (2, 2).

You can check to see if the solution (2, 2) makes both equations true.

2x – y = 22(2) – 2 Replace x with 2 and y with 2.4 – 2 22 = 2

x + 2y = 62 + 2(2) 6 Replace x with 2 and y with 2.2 + 4 66 = 6

The solution to the system of linear equations 2x – y = 2 x + 2y = 6

is (2, 2).

Your Turn

Now you can try. Do Your Turn #1 on page 278.

Your Turn Answer

Classification of Systems of Linear Equations

Example 2a

Use the graph to solve the system of linear equations. Classify each system.

4x + 2y = 8

2x + y = -4

Remember, there are two ways to graph lines. You can find the x- and y-intercepts to graph the lines. Or you can rewrite the equations in slope-intercept form (y = mx + b). Then use the slope and y-intercept to graph the lines.

4x + 2y = 8 can be rewritten as y = -2x + 4. The slope is -2 and the y-intercept is 4.

2x + y = - 4 can be rewritten as y = -2x – 4. The slope is -2 and the y-intercept is -4.

Both lines have the same slope, but different y-intercepts. The system is inconsistent.

The lines do not intersect, therefore there is no solution.

Example 2b

Use the graph to solve the system of linear equations. Classify each system.

2x - y = -6

4x - 2y = -12

2x – y = -6 can be rewritten as y = 2x + 6. The slope is 2 and the y-intercept is 6.

4x - 2y = - 12 can be rewritten as y = 2x + 6. The slope is 2 and the y-intercept is 6.

Both lines have the same slope and same y-intercept. The system is consistent and dependent.

The lines overlap, therefore there are infinitely many solutions.

Your Turn

Do Your Turn #3 and 4 on page 279.

Your Turn Answers

Estimating a Solution by Graphing

You can estimate a solution for a linear system by graphing and then check your estimate to determine if it is an appropriate solution.

Example 3

Estimate the solution for the linear system by graphing.

x – y = 3

x + 2y = 4

x – y = 3

The x-intercept is 3 and the y-intercept is -3.

x + 2y = 4

The x-intercept is 4 and the y-intercept is 2.

Find the point of intersection. The two lines are to intersect at the point (3, ).

Check if (3, ) is an approximate solution.

x – y = 3

3 - 3 Substitute the x and y values.

3 = 3

x + 2y = 4

3 + 2() 4 Substitute the x and y values.

+ 4

4

4 = 4

Solving Linear Systems by SubstitutionLesson 9.2 – Page 285

Vocabulary

• Substitution method – used to solve systems of liner equations by solving an equation for one variable and then substituting the resulting expression for that variable in another equation.

The steps for the substitution method are as follows:

1. Solve one of the equations for one of its variables.

2. Substitute the expression from Step 1 into the other equation and solve for the other variable.

3. Substitute the value from Step 2 into either original equation and solve to find the value of the variable in Step 1.

When solving systems of linear equations by substitution, you can have one solution, no solution or infinitely many solutions.

Example 1 – One Solution

Solve the system of linear equations by substitution.

y – x = 3

3x – 2y = 5

Step 1: Solve y – x = 3 for x.

y – x = 3 - y - y Subtract y

from both sides. - x = - y + 3

= + Divide by -1.

x = y – 3

Step 2: Substitute the resulting expression into the other equation by substitution.

3x – 2y = 53(y - 3) – 2y = 5 Replace x with y – 3.(3)y - (3)3 – 2y = 5 Use the Distributive Property3y – 9 – 2y = 53y – 2y – 9 = 5 Combine like terms.

y – 9 = 5 + 9 + 9 Add 9 to both sides.y = 14

Step 3: Substitute the value of y into one of the original equations. Then solve for x.

y – x = 33x – 2y = 5

y – x = 3 14 – x = 3 Substitute y with 14.

-14 -14 Subtract 14

- x = -11

= Divide by -1.

x = 11

OR

3x – 2y = 53x – 2(14) = 5 Substitute y with 14.

3x – 28 = 5 + 28 +28 Add 28.

3x = 33

= Divide by 3.

x = 11

The solution for the system of linear equations

y – x = 33x – 2y = 5

is (11, 14).

You can now check your solution.

The solution is (11, 14).

y – x = 314 – 11 3 Substitute x with 11 and y with 14. 3 = 3 3x – 2y = 53(11) – 2(14) 5 Substitute x with 11 and y with 14. 33 - 28 5 5 = 5

Example 2 – No Solution

Solve the system of linear equations by substitution.

2y – x = 1

-0.5x + y = 5

Step 1: Solve for x.

2y – x = 1-2y -2y Subtract 2y from both sides. - x = -2y + 1

= + Divide by -1.

x = 2y – 1

Step 2: Substitute the resulting expression into the other equation by substitution.

-0.5x + y = 5-0.5(2y – 1) + y = 5 Substitute x with 2y –

1.(-0.5)2y - (-0.5)1+ y = 5 Use the Distributive Property.-1y + 0.5 + y = 5-1y + 1y + 0.5 = 5 Combine like terms.0.5

Since 0.5 is not equal to 5, there is no solution.

Example 3 – Infinitely Many Solutions

Solve the system of linear equations by substitution.

2y – 2x = 6

-x + y = 3

Step 1: Solve for x.

-x + y = 3 - y - y-x = - y + 3

= +

x = y - 3