Graphing Absolute Value Equations Unit 7 Section 1...

Graphing Absolute Value

Equations

Unit 7 Section 1 How do I make one of those V

graphs?

Absolute value

transformation

Vocabulary: • Translation: shifts the parent function graph

horizontally and vertically.

• Reflection: it creates a mirror image of the

parent-function graph across the line of

reflection.

Parent Function y = │x

Vertical Movements?

• Absolute value rules:

– If you have minus outside the absolute value,

you move down on the coordinate plane.

y =│x│- 2

– If you have addition outside the absolute value,

you move up on the coordinate plane.

y =│x│+ 2

Lets practice! • y = │x│ - 4

• y = │x│+ 3

Your turn!

• y = │x│ + 7

• y = │x│ - 2

Horizontal Movements?

• Absolute value special rules:

– If you have minus inside the absolute value,

you move to the right on the coordinate plane.

y =│x - 2 │

– If you have addition inside the absolute value,

you move to the left on the coordinate plane.

y =│x + 2 │

Lets Practice!

• y =│x - 5│

• y =│x + 6│

Your Turn!

• y =│x + 4│

• y =│x - 1│

What about y = │x - 4 │+ 5

What about y = -│x - 2│

y = -│x │

Same rule, but now it makes it look down

Reflection of the

Parent-function

Practice Time!!!

1) y = │x - 1│

2) y = │x +3│

3) y = │x │ + 5

4) y = │x│ - 4

5) y = │x + 3│ + 3

6) y = │x - 1│+ 2

7) y = -│x - 6│

8) y = -│x │- 4

9) │x│+ y = - 3

10) y = -│x + 1│ - 1

Review

Verbal Comprehension

Analysis • Graph the functions y = │x │ and y = │x -

3 │ together on the same coordinate plane.

• What effect does the 3 have on the graph?

• Graph the functions y = │x │ and y = │x

│- 3 on the same coordinate plane.

• What effect does the 3 have on the graph?

• Graph the functions y = │x │ and y = - │x

│ together on a coordinate plane.

• What effect does the negative sign have on

the graph?

Practice!!

7-1 Pgs. 327-328 #1-3 all,

#4-34 even

Solving Systems by Graphing

Section 7-2

Vocabulary

• System of Linear Equations

• Solutions of a System of Linear Equations

Definition

• System of Linear Equations - a set of two or more linear

equations containing two or more variables.

– Example:

• Solution of a System of Linear Equations – is an ordered

pair that satisfies each equation in the system. So, if an

ordered pair is a solution, it will make both equations true.

A system of linear equations is a grouping

of two or more linear equations where each

equation contains one or more variables.

A solution of a system of equations consists of values for the variables that satisfy each equation of the system. When we are solving a system of two linear equations containing two unknowns, we represent the solution as an ordered pair (x, y), a point.

y = – 4x – 6

y = 2x

5x + 3y = – 8

x – 4y = 7

A brace is used to remind us that we are dealing

with a system of equations.

Solutions

Tell whether the ordered pair is a solution of the given system.

(5, 2);

The ordered pair (5, 2) makes both equations true.

(5, 2) is the solution of the system.

3x – y = 13

2 – 2 0

0 3(5) – 2 13

15 – 2 13

3x – y = 13

Substitute 5 for

x and 2 for

Example: Identifying

Solutions to a System

If an ordered pair does not satisfy the first equation in

the system, there is no reason to check the other

equations.

Helpful Hint

Solutions

(–2, 2);

–x + y = 2

x + 3y = 4

Substitute –2

for x and 2

for y.

x + 3y = 4

–2 + 6 4

4 –2 + (3)2

–x + y = 2

2 –(–2) + 2

The ordered pair (–2, 2) makes one equation true, but

not the other. (–2, 2) is not a solution of the system.

Example: Identifying

Solutions to a System

(1, 3); 2x + y = 5

–2x + y = 1

2x + y = 5

2(1) + 3 5

2 + 3 5

The ordered pair (1, 3) makes both equations true.

Substitute 1 for x

and 3 for y.

–2x + y = 1

–2(1) + 3 1

–2 + 3 1

(1, 3) is the solution of the system.

Your Turn:

(2, –1); x – 2y = 4

3x + y = 6

The ordered pair (2, –1) makes one equation true, but not the other.

(2, –1) is not a solution of the system.

3x + y = 6

3(2) + (–1) 6

6 – 1 6

x – 2y = 4

2 – 2(–1) 4

2 + 2 4

Substitute 2 for x

and –1 for y.

Your Turn:

Wren and Jenni are reading the same book. Wren is on

and reads 2 pages every night. Jenni is on page 6

and reads 3 pages every night. After how many nights

will they have read the same number of pages? How

many pages will that be?

Example: Writing a

System of Equations

Understand the Problem

The answer will be the number of nights it takes for the

number of pages read to be the same for both girls.

List the important information:

Wren on page 14 Reads 2 pages a night

Jenni on page 6 Reads 3 pages a night

Example: Continued

2 Write a System of Equations

Write a system of equations, one equation to represent the

number of pages read by each girl. Let x be the number of

nights and y be the total pages read.

pages is

number

read every

night plus already

Wren y = 2 x + 14

Jenni y = 3 x + 6

Example: Continued

Solve 3

(8, 30)

Nights

Graph y = 2x + 14 and y = 3x + 6. The lines appear to intersect at

(8, 30). So, the number of pages read will be the same at 8 nights

with a total of 30 pages.

Example: Continued

Verify the Solution 4

Check (8, 30) using both equations.

After 8 nights, Wren will have read 30 pages:

After 8 nights, Jenni will have read 30 pages:

3(8) + 6 = 24 + 6 = 30

2(8) + 14 = 16 + 14 = 30

Example: Continued

Video club A charges $10 for membership and $3

per movie rental. Video club B charges $15 for

membership and $2 per movie rental. For how

many movie rentals will the cost be the same at

both video clubs? What is that cost?

Your Turn:

1 Understand the Problem

The answer will be the number of movies rented for

which the cost will be the same at both clubs.

List the important information:

• Rental price: Club A $3 Club B $2

• Membership: Club A $10 Club B $15

Continued

2 Write a System of Equations

Write a system of equations, one equation to represent the

cost of Club A and one for Club B. Let x be the number of

movies rented and y the total cost.

cost is price rentals plus

membership

fee. times

Club A y = 3 + 10

Club B y = 2 + 15

Continued

Solve 3

Graph y = 3x + 10 and y = 2x + 15. The lines appear to

intersect at (5, 25). So, the cost will be the same for 5 rentals

and the total cost will be $25.

Continued

Verify the Solution 4

Check (5, 25) using both equations.

Number of movie rentals for Club A to reach $25:

Number of movie rentals for Club B to reach $25:

2(5) + 15 = 10 + 15 = 25

3(5) + 10 = 15 + 10 = 25

Continued

All solutions of a linear equation are on its graph. To

find a solution of a system of linear equations, you

need a point that each line has in common. In other

words, you need their point of intersection.

y = 2x – 1

y = –x + 5

The point (2, 3) is where the two

lines intersect and is a solution of

both equations, so (2, 3) is the

solution of the systems.

System Solution on a Graph

Steps for Obtaining the Solution of a System of

Linear Equations by Graphing

Step 1: Graph the first equation in the system.

Step 2: Graph the second equation in the system.

Step 3: Determine the point of intersection, if any.

Step 4: Verify that the point of intersection

determined in Step 3 is a solution of the

system. Remember to check the point in both

equations.

Solving a System by

Graphing

Solve the system by graphing. Check your answer. y = x

y = –2x – 3 Graph the system.

The solution appears to be at

(–1, –1).

The solution is (–1, –1).

Substitute (–1, –1) into the

system. y = x

y = –2x – 3

(–1) (–1)

–1 –1

y = –2x – 3

(–1) –2(–1) –3

–1 2 – 3

–1 – 1

Example: System solution by

Graphing

Solve the system by graphing. Check your answer.

Additional Example 2B: Solving a System Equations by Graphing

y = x – 6

y = 1/3x + 1

y = x – 6

Graph the system.

3 x =– 1

Additional Example 2B Continued

y = x – 6

y + x = –1

+ – 1

–1 – 1

y = x – 6

The solution is

Check Substitute into the system

Check It Out! Example 2a

y = –2x – 1

y = x + 5 Graph the system.

The solution appears to be (–2, 3).

Check Substitute (–2, 3) into the system.

y = x + 5

3 –2 + 5

y = –2x – 1

3 –2(–2) – 1

3 4 – 1

3 3 The solution is (–2, 3).

y = x + 5

y = –2x – 1

Check It Out! Example 2b

2x + y = 4

Graph the system.

Rewrite the second equation in

slope-intercept form.

2x + y = 4

–2x – 2x

y = –2x + 4

The solution appears to be (3, –2).

y = –2x + 4

Check It Out! Example 2b Continued

2x + y = 4 Check Substitute (3, –2) into the system.

2x + y = 4

2(3) + (–2) 4

6 – 2 4

–2 (3) – 3

–2 1 – 3

–2 –2

The solution is (3, –2).

• There are three possible outcomes or

solutions when graphing two linear

equations in a plane.

• One point of intersection, so one solution.

• Parallel lines, so no solution.

• Same lines, so infinite # of solutions.

System Possible Solutions

IDENTIFYING THE NUMBER OF SOLUTIONS

NUMBER OF SOLUTIONS OF A LINEAR SYSTEM

Lines intersect

one solution

Lines are parallel

no solution

Lines coincide

infinitely many

solutions

(the coordinates

of every point on

the line)

CONCEPT

SUMMARY

Lines intersect one solution

Lines are parallel no solution

Lines coincide infinitely many solutions

Example: A Linear System with No Solution

Show that this linear system

has no solution.

Rewrite each equation

in slope-intercept form.

5 4 3 2 1 0 1 2 3 4 5 1

y –2 x 5 Revised Equation 1

Graph the linear system.

y –2 x 1 Revised Equation 2

2 x y 5 Equation 1

2 x y 1 Equation 2

The lines are parallel; they

have the same slope but

different y-intercepts. Parallel

lines never intersect, so the

system has no solution.

y 2x 1

y 2x 5

METHOD 1: GRAPHING

Example: A Linear System with Infinite Solutions

has infinitely many solutions.

– 2 x y 3 Equation 1

– 4 x 2y 6 Equation 2

y 2 x 3 Revised Equation 1

y 2 x 3 Revised Equation 2

Rewrite each equation

in slope-intercept form.

Graph the linear system.

From these graphs you can see that the equations represent the same line. Any point on the line is a solution.

5 4 3 2 1 0 1 2 3 4 5 1

METHOD 1: GRAPHING

– 4x 2y 6

–2x y 3

Assignment

• 7-2 Pg. 343-344 #1-23 even, 29-33 all

Solving Systems Using

Substitution

Section 7-3

Sometimes it is difficult to identify the exact solution

to a system by graphing. In that case, you can use an

algebraic method called substitution to solve a

system.

Substitution is used to reduce the system to one

equation that has only one variable. Then you can

solve this equation by the methods taught in Chapter 2.

Substitution Method

• Sometimes it is difficult to identify the exact

solution to a system by graphing.

• In this case, you can use a method called

substitution.

• Substitution is used to reduce the system to one

equation that has only one variable.

• Then you can solve this equation for the one

variable and substitute again to find the other

variable.

Solving Systems of Equations by Substitution

Step 2

Step 3

Step 4

Step 5

Step 1 Solve for one variable in at least one equation, if

necessary.

Substitute the resulting expression into the other

equation.

Solve that equation to get the value of the first

variable.

Substitute that value into one of the original equations

and solve for the other variable.

Write the values from steps 3 and 4 as an ordered pair,

(x, y), and check.

Substitution Method

Solve the system by substitution.

y = 3x

y = x – 2

Step 1 y = 3x

y = x – 2

Both equations are solved for y.

Step 2 y = x – 2

3x = x – 2

Substitute 3x for y in the second

equation.

Now solve this equation for x. Subtract x from both sides and then divide by 2.

Step 3 –x –x

2x = –2 2x = –2

x = –1

Example: Substitution

Step 4 y = 3x Write one of the original

equations.

Substitute –1 for x. y = 3(–1)

y = –3

Step 5 (–1, –3)

Check Substitute (–1, –3) into both equations in the system.

Write the solution as an ordered pair.

y = 3x

–3 3(–1)

–3 –3

y = x – 2

–3 –1 – 2

–3 –3

Example: Continued

You can substitute the value of one variable into

either of the original equations to find the value of

the other variable.

Helpful Hint

y = x + 1

4x + y = 6

Step 1 y = x + 1 The first equation is solved for y.

Step 2 4x + y = 6

4x + (x + 1) = 6 Substitute x + 1 for y in the second

equation.

Subtract 1 from both sides.

5x = 5

Step 3 –1 –1

5x = 5

5x + 1 = 6 Simplify. Solve for x.

Divide both sides by 5.

Write the second equation.

Step 4 y = x + 1 Write one of the original

equations.

Substitute 1 for x. y = 1 + 1

Step 5 (1, 2) Write the solution as an ordered pair.

Example: Continued

x + 2y = –1

x – y = 5

Step 1 x + 2y = –1 Solve the first equation for x by

subtracting 2y from both sides.

Step 2 x – y = 5

(–2y – 1) – y = 5

Substitute –2y – 1 for x in the second

equation.

–3y – 1 = 5 Simplify.

−2y −2y

x = –2y – 1

Step 3 –3y – 1 = 5

Add 1 to both sides. +1 +1

–3y = 6

–3 –3

y = –2

Solve for y.

Divide both sides by –3.

Step 4 x – y = 5

x – (–2) = 5

x + 2 = 5 –2 –2

Step 5 (3, –2)

Write one of the original equations.

Substitute –2 for y.

Example: Continued

Solve the system by substitution. Check your answer.

y = x + 3

y = 2x + 5

Both equations are solved for y. Step 1 y = x + 3

y = 2x + 5

Substitute 2x + 5 for y in the first

equation.

Solve for x. Subtract x and 5 from both

sides. –x – 5 –x – 5

x = –2

Step 3 2x + 5 = x + 3

Step 2

2x + 5 = x + 3

y = x + 3

Your Turn:

Solve the system by substitution. Check your answer.

Step 4 y = x + 3 Write one of the original

equations.

Substitute –2 for x. y = –2 + 3

Step 5 (–2, 1) Write the solution as an ordered pair.

Check Substitute (–2, 1) into both equations in the system.

y = x + 3

1 (–2) + 3

y = 2x + 5

1 2(–2) + 5 1 –4 + 5

Continued

x = 2y – 4

x + 8y = 16

The first equation is solved for x. Step 1 x = 2y – 4

Substitute 2y – 4 for x in the second

equation.

Simplify. Then solve for y.

(2y – 4) + 8y = 16

x + 8y = 16 Step 2

Step 3 10y – 4 = 16 Add 4 to both sides. +4 +4

10y = 20

10y 20

10 10 = Divide both sides by 10.

Your Turn:

Step 4 x + 8y = 16 Write one of the original equations.

Substitute 2 for y. x + 8(2) = 16

x + 16 = 16

– 16 –16

Simplify.

Step 5 (0, 2) Write the solution as an

ordered pair.

Continued

2x + y = –4

x + y = –7

Solve the second equation for x by

subtracting y from each side.

Substitute –y – 7 for x in the first

equation.

Distribute 2.

2(–y – 7) + y = –4

x = –y – 7 Step 2

Step 1 x + y = –7 – y – y

x = –y – 7

2(–y – 7) + y = –4

–2y – 14 + y = –4

Your Turn:

Combine like terms. Step 3

+14 +14

–y = 10

–2y – 14 + y = –4

Add 14 to each side.

–y – 14 = –4

y = –10

Step 4 x + y = –7 Write one of the original equations.

Substitute –10 for y. x + (–10) = –7

x – 10 = – 7

Continued

x – 10 = –7 Step 5

+10 +10

Add 10 to both sides.

Step 6 (3, –10) Write the solution as an

ordered pair.

Continued

When you solve one equation for a variable, you

must substitute the value or expression into the other

original equation, not the one that had just been

solved. You have to use both equations.

Caution

y + 6x = 11

3x + 2y = –5 Solve by substitution.

Solve the first equation for y by

subtracting 6x from each side. Step 1 y + 6x = 11 – 6x – 6x

y = –6x + 11

Substitute –6x + 11 for y in the

second equation.

Distribute 2 to the expression in

parentheses.

3x + 2(–6x + 11) = –5

3x + 2y = –5 Step 2

3x + 2(–6x + 11) = –5

Example: Substitution Involving

Distribution

Step 3 3x + 2(–6x) + 2(11) = –5

–9x + 22 = –5

Simplify. Solve for x.

Subtract 22 from both

sides. –9x = –27

– 22 –22

Divide both sides by –

9. –9x = –27

–9 –9

3x – 12x + 22 = –5

3x + 2(–6x + 11) = –5

Continued

Step 4 y + 6x = 11

Substitute 3 for x. y + 6(3) = 11

Subtract 18 from each side.

y + 18 = 11

–18 –18

y = –7

Step 5 (3, –7) Write the solution as an ordered

Simplify.

Write one of the original equations.

Continued

–2x + y = 8

3x + 2y = 9 Solve by substitution.

Solve the first equation for y by

adding 2x to each side.

Step 1 –2x + y = 8 + 2x +2x

y = 2x + 8

Substitute 2x + 8 for y in the second

equation. 3x + 2(2x + 8) = 9

3x + 2y = 9 Step 2

Distribute 2 to the expression in

parentheses. 3x + 2(2x + 8) = 9

Your Turn:

Step 3 3x + 2(2x) + 2(8) = 9

7x + 16 = 9

Simplify. Solve for x.

sides. 7x = –7

–16 –16

Divide both sides by 7. 7x = –7

x = –1

3x + 4x + 16 = 9

Continued

3x + 2(2x + 8) = 9

Step 4 –2x + y = 8

Substitute –1 for x. –2(–1) + y = 8

y + 2 = 8

–2 –2

Step 5 (–1, 6) Write the solution as an

ordered pair.

Subtract 2 from each side.

Simplify.

Write one of the original

equations.

Continued x = –1

Example: A Linear System with No Solution

Show that this linear system has no solution.

2 x y 5 Equation 1

2 x y 1 Equation 2

Because Equation 2 can be revised

to y –2 x 1, you can substitute –2 x 1 for y in Equation 1.

2 x y 5

2 x (–2 x 1) 5 Substitute –2 x 1 for y.

1 5 Simplify. False statement!

Once the variables are eliminated, the statement is not true regardless

of the values of x and y. This tells you the system has no solution.

Write Equation 1.

METHOD: SUBSTITUTION

Substitute 2 x + 3 for y in Equation 2.

– 4 x 2y 6 Equation 2 METHOD: SUBSTITUTION

You can solve Equation 1 for y.

– 4 x 2( 2 x + 3) 6

– 4 x + 4x 6 6 Simplify.

6 6 True statement!

The variables are eliminated and you are left with a statement that is true

regardless of the values of x and y. This result tells you that the linear system

Example: A Linear System with Many Solutions

y 2 x + 3

Your Turn:

Solve the system.

9 3 30

30 = 30 (true statement)

Infinite Solutions

6 = -9 (false statement)

No Solution

Assignment

• 7-3 Pg. 349-351 #1 & 2, 6-32 even, #45

Elimination

Section 7-4

Another method for solving systems of equations is elimination. Like substitution, the goal of elimination is to get one equation that has only one variable. To do this by elimination, you add the two equations in the system together.

Remember that an equation stays balanced if you add equal amounts to both sides. So, if 5x + 2y = 1, you can add 5x + 2y to one side of an equation and 1 to the other side and the balance is maintained.

Elimination Method

Since –2y and 2y have opposite coefficients, the y-term is eliminated. The result is one equation that has only one variable: 6x = –18.

When you use the elimination method to solve a system of linear equations, align all like terms in the equations. Then determine whether any like terms can be eliminated because they have opposite coefficients.

Elimination Method

Solving Systems of

Equations • So far, we have solved systems using

graphing and substitution. These notes show

how to solve the system algebraically using

ELIMINATION with addition and

subtraction.

• Elimination is easiest when the equations are

in standard form.

Step 1: Put the equations in Standard Form.

Step 2: Determine which

variable to eliminate.

Step 3: Add or subtract the

equations.

Step 4: Plug back in to find the

other variable.

Step 5: Check your solution.

Standard Form: Ax + By = C

Look for variables that have the

same coefficient.

Solve for the variable.

Substitute the value of the variable

into the equation.

Substitute your ordered pair into

BOTH equations.

Elimination Procedure

x + y = 5

3x – y = 7

Step 1: Put the equations in

Standard Form.

They already are!

The y’s have the same

coefficient.

equations.

Add to eliminate y.

x + y = 5

(+) 3x – y = 7

4x = 12

Example: Elimination by

Addition

other variable.

x + y = 5

(3) + y = 5

(3, 2)

(3) + (2) = 5

3(3) - (2) = 7

The solution is (3, 2). What do you think the answer would

be if you solved using substitution?

x + y = 5

3x – y = 7

Example: Continued

y + 3x = –2

2y – 3x = 14 Solve by elimination.

Write the system so that like

terms are aligned. 2y – 3x = 14

y + 3x = –2

Add the equations to

eliminate the x-terms. 3y + 0 = 12

3y = 12 Simplify and solve for y.

Your Turn:

y + 3x = –2 Write one of the original

equations.

4 + 3x = –2 Substitute 4 for y. Subtract 4 from both sides. –4 –4

3x = –6

Divide both sides by 3. 3x = –6 3 3

x = –2 Write the solution as an

ordered pair. (–2, 4)

Continued

3x – 4y = 10

x + 4y = –2 Solve by elimination.

3x – 4y = 10 Write the system so that like

terms are aligned. Add the equations to

eliminate the y-terms.

4x = 8 Simplify and solve for x.

x + 4y = –2

4x + 0 = 8

Divide both sides by 4. 4x = 8

Your Turn:

x + 4y = –2 Write one of the original

equations.

2 + 4y = –2 Substitute 2 for x.

–2 –2

4y = –4

4y –4

4 4 y = –1

(2, –1)

Continued

More Elimination

• When two equations each contain the same

term, you can subtract one equation from

the other to solve the system.

• To subtract an equation add the opposite of

each term.

4x + y = 7

4x – 2y = -2

Standard Form. They already are!

The x’s have the same

coefficient.

equations.

Subtract to eliminate x.

4x + y = 7

(-) 4x – 2y = -2

3y = 9

Remember to

“keep-change-

change”

Example: Elimination by

Subtracting

other variable.

4x + y = 7

4x + (3) = 7

4x = 4

(1, 3)

4(1) + (3) = 7

4(1) - 2(3) = -2

4x + y = 7

4x – 2y = -2

Example: Continued

3x + 3y = 15

–2x + 3y = –5 Solve by elimination.

3x + 3y = 15

–(–2x + 3y = –5)

3x + 3y = 15 + 2x – 3y = +5

Add the opposite of each

term in the second

equation.

Eliminate the y term.

Simplify and solve for x.

5x + 0 = 20

5x = 20

Your Turn:

equations. Substitute 4 for x.

3x + 3y = 15

3(4) + 3y = 15

12 + 3y = 15

–12 –12 3y = 3

Simplify and solve for y.

Write the solution as an

ordered pair.

(4, 1)

Continued

2x + y = –5

2x – 5y = 13 Solve by elimination.

Add the opposite of each

term in the second

equation.

–(2x – 5y = 13)

2x + y = –5

–2x + 5y = –13

Eliminate the x term.

0 + 6y = –18

6y = –18

y = –3

Your Turn:

equations. 2x + y = –5

2x + (–3) = –5 Substitute –3 for y.

Add 3 to both sides.

2x – 3 = –5

2x = –2 Simplify and solve for x.

x = –1

ordered pair.

(–1, –3)

Continued

Assignment

• 7-4 Part 1 Pg. 356-358: #2-8 even

Elimination

Section 7-4 Part 2

Further Elimination

• In Part 1 of this lesson, you was that to

eliminate a variable, its coefficients must

have a sum or difference of zero.

• In some cases, you will first need to

multiply one or both of the equations by a

number so that one variable has opposite

coefficients, so you can add or subtract to

eliminate the variable.

Step 1: Put the equations in Standard Form.

Step 3: Multiply the equations

and solve.

other variable.

Standard Form: Ax + By = C

Look for variables that have the

same coefficient.

Solve for the variable.

Substitute the value of the variable

into the equation.

Substitute your ordered pair into

BOTH equations.

Further Elimination Procedure

2x + 2y = 6

3x – y = 5

Standard Form.

They already are!

None of the coefficients are the

Find the least common multiple

of each variable.

LCM = 6x, LCM = 2y

Which is easier to obtain?

(you only have to multiply

the bottom equation by 2)

Example: Multiplying One

Equation

other variable.

2(2) + 2y = 6

4 + 2y = 6

2y = 2

2x + 2y = 6

3x – y = 5

and solve.

Multiply the bottom equation by 2

2x + 2y = 6

(2)(3x – y = 5)

8x = 16

2x + 2y = 6

(+) 6x – 2y = 10

Example: Continued

(2, 1)

2(2) + 2(1) = 6

3(2) - (1) = 5

2x + 2y = 6

3x – y = 5

Solving with multiplication adds one more step

to the elimination process.

Example: Continued

x + 4y = 7

4x – 3y = 9

of each variable.

LCM = 4x, LCM = 12y

(you only have to multiply

the top equation by -4 to make

them inverses)

Example: Multiplying One

Equation

x + 4y = 7

4x – 3y = 9

other variable.

x + 4(1) = 7

x + 4 = 7

and solve.

Multiply the top equation by -4

(-4)(x + 4y = 7)

4x – 3y = 9)

-4x – 16y = -28

(+) 4x – 3y = 9

-19y = -19

Example: Continued

(3, 1)

(3) + 4(1) = 7

4(3) - 3(1) = 9

x + 4y = 7

4x – 3y = 9

Example: Continued

x + 2y = 11

–3x + y = –5

Solve the system by elimination.

Multiply each term in the

second equation by –2 to

get opposite y-coefficients.

x + 2y = 11

–2(–3x + y = –5)

x + 2y = 11

+(6x –2y = +10) Add the new equation to

the first equation. 7x + 0 = 21

7x = 21

Your Turn:

equations. x + 2y = 11

Substitute 3 for x. 3 + 2y = 11

Subtract 3 from each side. –3 –3

2y = 8

ordered pair.

(3, 4)

Continued

3x + 2y = 6

–x + y = –2

3x + 2y = 6

3(–x + y = –2)

3x + 2y = 6

+(–3x + 3y = –6)

0 + 5y = 0

Multiply each term in the second

equation by 3 to get opposite

x-coefficients.

Add the new equation to

the first equation.

Simplify and solve for y. 5y = 0

Your Turn:

equations. –x + y = –2

Substitute 0 for y. –x + 3(0) = –2

–x + 0 = –2

–x = –2

ordered pair. (2, 0)

Continued

3x + 4y = -1

4x – 3y = 7

of each variable.

LCM = 12x, LCM = 12y

Either! I’ll pick y because the signs

are already opposite.

Example: Multiplying Both

Equations

3x + 4y = -1

4x – 3y = 7

other variable.

3(1) + 4y = -1

3 + 4y = -1

4y = -4

y = -1

and solve.

Multiply both equations

(3)(3x + 4y = -1)

(4)(4x – 3y = 7)

9x + 12y = -3

(+) 16x – 12y = 28

25x = 25

Example: Continued

(1, -1)

3(1) + 4(-1) = -1

4(1) - 3(-1) = 7

3x + 4y = -1

4x – 3y = 7

Example: Continued

–5x + 2y = 32

2x + 3y = 10

2(–5x + 2y = 32)

5(2x + 3y = 10)

Multiply the first equation

by 2 and the second

equation by 5 to get

opposite x-coefficients –10x + 4y = 64

+(10x + 15y = 50) Add the new equations.

Simplify and solve for y. 19y = 114

Your Turn:

equations. 2x + 3y = 10

Substitute 6 for y. 2x + 3(6) = 10

Subtract 18 from both sides. –18 –18

2x = –8

2x + 18 = 10

x = –4 Simplify and solve for x.

ordered pair.

(–4, 6)

Continued

2x + 5y = 26

–3x – 4y = –25

3(2x + 5y = 26)

+(2)(–3x – 4y = –25)

Multiply the first equation

by 3 and the second

equation by 2 to get

opposite x-coefficients 6x + 15y = 78

+(–6x – 8y = –50) Add the new equations.

Simplify and solve for y. y = 4

0 + 7y = 28

Your Turn:

equations. 2x + 5y = 26

Substitute 4 for y. 2x + 5(4) = 26

ordered pair. (3, 4)

2x + 20 = 26 –20 –20

2x = 6

sides.

Your Turn:

CONCEPT

SUMMARY

Lines intersect one solution

Lines are parallel no solution

Lines coincide infinitely many solutions

Identifying The Number of

Solutions

• If both variable terms are eliminated as you

solve a system of equations, the answer is

either no solution or infinite solutions.

– No solution: get a false statement when solving

the system.

– Infinite solutions: get a true statement when

solving the system.

– 4 x 2y 6 Equation 2 METHOD: Elimination

You can multiply Equation 1 by –2.

4x – 2y – 6 Multiply Equation 1 by –2.

– 4 x 2y 6 Write Equation 2.

0 0 Add Equations. True statement!

The variables are eliminated and you are left with a statement that is true

A Linear System with Infinite Solutions

has no solution.

2 x y 5 Equation 1

2 x y 1 Equation 2

You can multiply Equation 1 by –1.

-2 x - y -5 Multiply Equation 1 by –1.

2 x y 1 Write Equation 2.

0 - 4 Add Equations. False statement!

The variables are eliminated and you are left with a statement that is false

has no solution.

A Linear System with No Solution

METHOD: Elimination

Your Turn:

Solve the systems using elimination.

12 8 20

False Statement

No Solution

True Statement

Infinite Solutions

Summary

Summary of Methods for Solving Systems

7.3 The Elimination Method

Substitution

The value of one variable is

known and can easily be

substituted into the other

equation.

6x + y = 10

y = 5 Example

Suggested

Method

Elimination

eliminate ‘y’ 5

Add the two equations

2x – 5y = –20

4x + 5y = 14 Example

Suggested

Method

Elimination

9a – 2b = –11

8a + 4b = 25 Example

Suggested

Method

Why eliminate ‘b’ 4

Multiply first equation by 2

Add the equations

Assignment

• 7-4 Part 2 Pg. 356-358 : #10-14 even, 18-28

even and #30

Lesson 7-5 Graphing Linear Inequalities

•Graph a linear inequality

in two variables

•Model a real life situation

with a linear inequality.

Graphing an Inequality in Two Variables

Graph x < 2

Step 1: Start by graphing the line x =

Now what points

would give you less

than 2?

Since it has to be x < 2

we shade everything to

the left of the line.

Graphing a Linear Inequality

Sketch a graph of y 3

Graphing a Linear Inequality

Sketch a graph of y < -3

Some Helpful Hints •If the sign is > or < the line is

dashed

•If the sign is or the line will be

When dealing with just x and y.

•If the sign > or the shading

either goes up or to the right

•If the sign is < or the shading

either goes down or to the left

Using What We Know Sketch a graph of x + y < 3

Step 1: Put into

slope intercept form

y <-x + 3

Step 2: Graph the

line y = -x + 3

Using What We Know Sketch a graph of 2x + 3y < 6

Step 1: Put into

y < -2/3 x + 2

Step 2: Graph the

line y = -2/3 x + 2

Using What We Know Sketch a graph of y > 2x - 3

Step 1: Put into

It already is!

Step 2: Graph the

line y = 2x - 3

When dealing with slanted lines

•If it is > or then you shade above

•If it is < or then you shade below

the line

Look at the two graphs. Determine the following:

A. The equation of each line.

B. How the graphs are similar.

C. How the graphs are different.

A. The equation of each line is y = x + 3.

B. The lines in each graph are the same and represent all of

the solutions to the equation y = x + 3.

C. The graph on the right is shaded above the line and this

means that all of these points are solutions as well.

Pick a point from the shaded region

and test that point in the equation y =

x + 3.

Point: (-4, 5)

This is incorrect. Five is

greater than or equal to

negative 1.

5 1 5 1 or

If a solid line is used, then the equation would be 5 -1.

If a dashed line is used, then the equation would be 5 > -1.

The area above the line is shaded.

Pick a point from the shaded region

and test that point in the equation y =

-x + 4.

Point: (1, -3)

This is incorrect. Negative

three is less than or equal to 3.

3 3 3 3 or

If a solid line is used, then the equation would be -3 3.

If a dashed line is used, then the equation would be -3 < 3.

The area below the line is shaded.

1. Write the inequality in slope-intercept form.

2. Use the slope and y-intercept to plot two points.

3. Draw in the line. Use a solid line for less than or equal to ()

or greater than or equal to (). Use a dashed line for less than

(<) or greater than (>).

4. Pick a point above the line or below the line. Test that point in

the inequality. If it makes the inequality true, then shade the

region that contains that point. If the point does not make the

inequality true, shade the region on the other side of the line.

5. Systems of inequalities – Follow steps 1-4 for each inequality.

Find the region where the solutions to the two inequalities

would overlap and this is the region that should be shaded.

Graph the following linear system of inequalities.

Use the slope and y-intercept to

plot two points for the first

inequality.

Draw in the line. For use a

solid line.

Pick a point and test it in the

inequality. Shade the

appropriate region.

3 2y x

2 4 Point (0,0)

0 2(0) - 4

0 -4The region above the line should be

shaded.

Now do the same for the second

inequality.

plot two points for the second

inequality.

Draw in the line. For < use a

dashed line.

Pick a point and test it in the

inequality. Shade the

appropriate region.

Point (-2,-2)

-2 (-2) + 2

-2 < 8

The region below the line should be

shaded.

The solution to this system of

inequalities is the region where

the solutions to each inequality

overlap. This is the region above

or to the left of the green line and

below or to the left of the blue

Shade in that region.

Graph the following linear systems of inequalities.

1. y x

y Use the slope and y-intercept to

plot two points for the first

inequality.

Draw in the line.

Shade in the appropriate

region.

plot two points for the second

inequality.

Draw in the line.

Shade in the appropriate

region.

The final solution is the region

where the two shaded areas

overlap (purple region).

Graphing Absolute Value Equations Unit 7 Section 1...

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