holt algebra 1 6-1 solving systems by graphing warm up evaluate each expression for x = 1 and y...

Holt Algebra 1

6-1 Solving Systems by Graphing

Warm UpEvaluate each expression for x = 1 and y =–3.

1. x – 4y 2. –2x + y

Write each expression in slope-

intercept form.

3. y – x = 1

4. 2x + 3y = 6

5. 0 = 5y + 5x

13 –5

y = x + 1

y = x + 2

y = –x

Holt Algebra 1


Identify solutions of linear equations in two variables.

Solve systems of linear equations in two variables by graphing.

Objectives

Holt Algebra 1


A system of linear equations is a set of two or more linear equations containing two or more variables. A solution of a system of linear equations with two variables 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.

Holt Algebra 1


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

Example 1A: Identifying Systems of Solutions

(5, 2);

The ordered pair (5, 2) makes both equations true.(5, 2) is the solution of the system.

Substitute 5 for x and 2 for y in each equation in the system.

3x – y = 13

2 – 2 00 0

0 3(5) – 2 13

15 – 2 13

13 13

3x – y 13

Holt Algebra 1


If an ordered pair does not satisfy the first equation in the system, there is no reason to check the other equations.

Helpful Hint

Holt Algebra 1


Example 1B: Identifying Systems of Solutions


(–2, 2);x + 3y = 4–x + y = 2

–2 + 3(2) 4

x + 3y = 4

–2 + 6 44 4

–x + y = 2

–(–2) + 2 24 2

Substitute –2 for x and 2 for y in each equation in the system.

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

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

Holt Algebra 1


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.

Holt Algebra 1


Sometimes it is difficult to tell exactly where the lines cross when you solve by graphing. It is good to confirm your answer by substituting it into both equations.

Helpful Hint

Holt Algebra 1


Solve the system by graphing. Check your answer.Example 2A: Solving a System Equations by Graphing

y = xy = –2x – 3 Graph the system.

The solution appears to be at (–1, –1).

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

CheckSubstitute (–1, –1) into the system.

y = x

y = –2x – 3

• (–1, –1)

y = x

(–1) (–1)

–1 –1

y = –2x – 3

(–1) –2(–1) –3

–1 2 – 3–1 – 1

Holt Algebra 1


Solve the system by graphing. Check your answer.Example 2B: Solving a System Equations by Graphing

y = x – 6

Rewrite the second equation in slope-intercept form.

y + x = –1Graph using a calculator and then use the intercept command.

y = x – 6

y + x = –1

− x − x

y =

Holt Algebra 1


Solve the system by graphing. Check your answer.Example 2B Continued

Check Substitute into the system.

y = x – 6

The solution is .

+ – 1

–1

–1

–1 – 1

y = x – 6

– 6

Holt Algebra 1


Example 3: Problem-Solving Application

Wren and Jenni are reading the same book. Wren is on page 14 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?

Holt Algebra 1


1 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 3 Continued

Holt Algebra 1


2 Make a Plan

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.

Totalpages is

number read

everynight plus

already read.

Wren y = 2 x + 14

Jenni y = 3 x + 6

Example 3 Continued

Holt Algebra 1


Solve3

Example 3 Continued

(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.

Holt Algebra 1


Look Back4

Check (8, 30) using both equations.

Number of days for Wren to read 30 pages.

Number of days for Jenni to read 30 pages.

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

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

Example 3 Continued

Holt Algebra 1

6-2 Solving Systems by Substitution

Warm-UpTell whether the ordered pair is a solution of the given system.

1. (–3, 1) 2. (2, –4)

3. Joy has 5 collectable stamps and will buy 2 more each month. Ronald has 25 collectable stamps and will sell 3 each month. After how many months will they have the same number of stamps? How many will that be?

Holt Algebra 1


Solve linear equations in two variables by substitution.

Objective

Holt Algebra 1


Sometimes it is difficult to identify the exact solution to a system by graphing. In this case, you can use a method called substitution.

The goal when using substitution is 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.

Holt Algebra 1


Solving Systems of Equations by Substitution

Step 2

Step 3

Step 4

Step 5

Step 1Solve 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.

Write the values from steps 3 and 4 as an ordered pair, (x, y), and check.

Holt Algebra 1


Solve the system by substitution.

Example 1A: Solving a System of Linear Equations by Substitution

y = 3x

y = x – 2

Step 1 y = 3xy = x – 2

Both equations are solved for y.

Step 2 y = x – 23x = x – 2

Substitute 3x for y in the second equation.

Solve for x. Subtract x from both sides and then divide by 2.

Step 3 –x –x2x = –22x = –22 2x = –1

Holt Algebra 1



Example 1A Continued

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

Holt Algebra 1



Example 1B: Solving a System of Linear Equations by Substitution

y = x + 1

4x + y = 6

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

Step 2 4x + y = 64x + (x + 1) = 6

Substitute x + 1 for y in the second equation.

Subtract 1 from both sides. Step 3 –1 –1

5x = 5 5 5

x = 1

5x = 5

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

Divide both sides by 5.

Holt Algebra 1



Example1B Continued

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

Substitute 1 for x. y = 1 + 1y = 2

Step 5 (1, 2)

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


y = x + 1

2 1 + 12 2

4x + y = 6

4(1) + 2 66 6

Holt Algebra 1


Sometimes you substitute an expression for a variable that has a coefficient. When solving for the second variable in this situation, you can use the Distributive Property.

Holt Algebra 1


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.

Caution

Holt Algebra 1


Example 2: Using the Distributive Property

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 parenthesis.

3x + 2(–6x + 11) = –5

3x + 2y = –5 Step 2

3x + 2(–6x + 11) = –5

Holt Algebra 1


Step 3

Example 2 Continued

y + 6x = 11


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

x = 3

3x – 12x + 22 = –5

Holt Algebra 1


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 pair.

Simplify.

Example 2 Continued

y + 6x = 11


Write one of the original equations.

Holt Algebra 1


Example 2: Consumer Economics Application

Jenna is deciding between two cell-phone plans. The first plan has a $50 sign-up fee and costs $20 per month. The second plan has a $30 sign-up fee and costs $25 per month. After how many months will the total costs be the same? What will the costs be? If Jenna has to sign a one-year contract, which plan will be cheaper? Explain.

Write an equation for each option. Let t represent the total amount paid and m represent the number of months.

Holt Algebra 1


Example 2 Continued

Total paid is

sign-up fee plus

paymentamount

for eachmonth.

Option 1 t = $50 + $20 m

Option 2 t = $30 + $25 m

Step 1 t = 50 + 20mt = 30 + 25m

Both equations are solved for t.

Step 2 50 + 20m =30 + 25m Substitute 50 + 20m for t in the second equation.

Holt Algebra 1


Step 3 50 + 20m = 30 + 25m Solve for m. Subtract 20m from both sides.–20m – 20m

50 = 30 + 5m Subtract 30 from both sides. –30 –30

20 = 5m Divide both sides by 5.


Step 4 t = 30 + 25m

t = 30 + 25(4)

t = 30 + 100

t = 130

Substitute 4 for m.Simplify.

Example 2 Continued

5 5

m = 4

20 = 5m

Holt Algebra 1


Step 5 (4, 130)Write the solution as an

ordered pair.

In 4 months, the total cost for each option would be the same $130.

Jenna should choose the first plan because it costs $290 for the year and the second plan costs $330.

Example 2 Continued

Option 1: t = 50 + 20(12) = 290

Option 2: t = 30 + 25(12) = 330

If Jenna has to sign a one-year contract, which plan will be cheaper? Explain.

Holt Algebra 1

6-3 Solving Systems by Elimination

Warm-Up

Holt Algebra 1


Solve systems of linear equations in two variables by elimination. Compare and choose an appropriate method for solving systems of linear equations.

Objectives

Holt Algebra 1


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.

Holt Algebra 1


Solving Systems of Equations by Elimination

Step 1 Write the system so that like terms are aligned.

Step 2 Eliminate one of the variables and solve for the other variable.

Step 3Substitute the value of the variable into one of the original equations and solve for the other variable.

Step 4Write the answers from Steps 2 and 3 as an ordered pair, (x, y), and check.

Holt Algebra 1


Example 1: Elimination Using Addition

3x – 4y = 10x + 4y = –2

Solve by elimination.

Step 1 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 = –24x + 0 = 8Step 2

Divide both sides by 4.4x = 84 4x = 2

Holt Algebra 1


Example 1 Continued

Step 3 x + 4y = –2 Write one of the original equations.

2 + 4y = –2 Substitute 2 for x.–2 –2

4y = –4

4y –44 4y = –1

Step 4 (2, –1)

Subtract 2 from both sides.



Holt Algebra 1


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.

Holt Algebra 1


2x + y = –52x – 5y = 13

Solve by elimination.

Example 2: Elimination Using Subtraction

Add the opposite of each term in the second equation.

Step 1–(2x – 5y = 13)

2x + y = –5

2x + y = –5–2x + 5y = –13

Eliminate the x term.Simplify and solve for y.

0 + 6y = –18 Step 26y = –18

y = –3

Holt Algebra 1


Example 2 Continued


Step 3 2x + y = –5

2x + (–3) = –5Substitute –3 for y.

Add 3 to both sides.2x – 3 = –5

+3 +3

2x = –2 Simplify and solve for x.

x = –1


Step 4 (–1, –3)

Holt Algebra 1


Remember to check by substituting your answer into both original equations.

Remember!

Holt Algebra 1


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. This will be the new Step 1.

Holt Algebra 1


x + 2y = 11–3x + y = –5

Solve the system by elimination.

Example 3A: Elimination Using Multiplication First

Multiply each term in the second equation by –2 to get opposite y-coefficients.

x + 2y = 11Step 1 –2(–3x + y = –5)

x + 2y = 11+(6x –2y = +10) Add the new equation to

the first equation.7x + 0 = 21Step 2 7x = 21

x = 3

Simplify and solve for x.

Holt Algebra 1




Step 3 x + 2y = 11

Substitute 3 for x. 3 + 2y = 11Subtract 3 from each side.–3 –3

2y = 8y = 4

Simplify and solve for y.


Step 4 (3, 4)

Holt Algebra 1


–5x + 2y = 32

2x + 3y = 10


Example 3B: Elimination Using Multiplication First

Step 1 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

y = 6

Step 2

Holt Algebra 1


Example 3B Continued


Step 3 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.

Step 4 Write the solution as an ordered pair.

(–4, 6)

Holt Algebra 1


Check It Out! Example 3b


2x + 5y = 26

–3x – 4y = –25

Step 1 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 = 40 + 7y = 28Step 2

Holt Algebra 1


Check It Out! Example 3b Continued


Step 3 2x + 5y = 26

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

Simplify and solve for x.

Step 4 Write the solution as an ordered pair.

(3, 4) x = 3

2x + 20 = 26–20 –20

2X = 6


Holt Algebra 1


Example 4: Application

Paige has $7.75 to buy 12 sheets of felt and card stock for her scrapbook. The felt costs $0.50 per sheet, and the card stock costs $0.75 per sheet. How many sheets of each can Paige buy?

Write a system. Use f for the number of felt sheets and c for the number of card stock sheets.

0.50f + 0.75c = 7.75 The cost of felt and card stock totals $7.75.

f + c = 12 The total number of sheets is 12.

Holt Algebra 1


Example 4 Continued

Step 1 0.50f + 0.75c = 7.75

+ (–0.50)(f + c) = 12

Multiply the second equation by –0.50 to get opposite f-coefficients.

0.50f + 0.75c = 7.75+ (–0.50f – 0.50c = –6)

Add this equation to the first equation to eliminate the f-term.

Simplify and solve for c.

Step 2 0.25c = 1.75

c = 7

Step 3 f + c = 12

Substitute 7 for c.f + 7 = 12

–7 –7f = 5



Holt Algebra 1



Step 4 (7, 5)

Paige can buy 7 sheets of card stock and 5 sheets of felt.

Example 4 Continued

Holt Algebra 1


Holt Algebra 1

6-4 Solving Special Systems

Warm-Up

Solve each system by elimination.

1.

2.

3.

(2, –3)

(11, 3)

(–3, –7)

2x + y = 253y = 2x – 13

x = –2y – 4–3x + 4y = –18

–2x + 3y = –153x + 2y = –23

4. Harlan has $44 to buy 7 pairs of socks. Athletic socks cost $5 per pair. Dress socks cost $8 per pair. How many pairs of each can Harlan buy?4 pairs of athletic socks and 3 pairs of dress socks

Holt Algebra 1


Solve special systems of linear equations in two variables.

Classify systems of linear equations and determine the number of solutions.

Objectives

Holt Algebra 1


In Lesson 6-1, you saw that when two lines intersect at a point, there is exactly one solution to the system. Systems with at least one solution are called consistent.

When the two lines in a system do not intersect they are parallel lines. There are no ordered pairs that satisfy both equations, so there is no solution. A system that has no solution is an inconsistent system.

Holt Algebra 1


Example 1: Systems with No Solution

Solve y = x – 4

Method 1 Compare slopes and y-intercepts.

y = x – 4 y = 1x – 4 Write both equations in slope-intercept form.

–x + y = 3 y = 1x + 3

–x + y = 3

The lines are parallel because they have the same slope and different y-intercepts.

This system has no solution so it is an inconsistent system.

Holt Algebra 1


If two linear equations in a system have the same graph, the graphs are coincident lines, or the same line. There are infinitely many solutions of the system because every point on the line represents a solution of both equations.

Holt Algebra 1


Solve y = 3x + 2

3x – y + 2= 0

Example 2A: Systems with Infinitely Many Solutions

Method 1 Compare slopes and y-intercepts.

y = 3x + 2 y = 3x + 2 Write both equations in slope-intercept form. The lines have the same slope and the same y-intercept.

3x – y + 2= 0 y = 3x + 2

If this system were graphed, the graphs would be the same line. There are infinitely many solutions.

Holt Algebra 1


Solve y = 3x + 2

3x – y + 2= 0

Method 2 Solve the system algebraically. Use the elimination method.

y = 3x + 2 y − 3x = 2

3x − y + 2= 0 −y + 3x = −2

Write equations to line up like terms.

Add the equations.

True. The equation is an identity.

0 = 0

There are infinitely many solutions.


Holt Algebra 1


0 = 0 is a true statement. It does not mean the system has zero solutions or no solution.

Caution!

Holt Algebra 1


Holt Algebra 1


Example 3A: Classifying Systems of Linear Equations

Solve3y = x + 3

x + y = 1

Classify the system. Give the number of solutions.

Write both equations in slope-intercept form.

3y = x + 3 y = x + 1

x + y = 1 y = x + 1 The lines have the same slope and the same y-intercepts. They are the same.

The system is consistent and dependent. It has infinitely many solutions.

Holt Algebra 1


Example 3B: Classifying Systems of Linear equations

Solvex + y = 5

4 + y = –x

Classify the system. Give the number of solutions.

x + y = 5 y = –1x + 5

4 + y = –x y = –1x – 4

Write both equations in slope-intercept form.

The lines have the same slope and different y-intercepts. They are parallel.

The system is inconsistent. It has no solutions.

Holt Algebra 1

6-5 Solving Linear Inequalities

Warm UpGraph each inequality.1. x > –5 2. y ≤ 0

3. Write –6x + 2y = –4 in slope-intercept form, and graph. y = 3x – 2

Holt Algebra 1


Graph and solve linear inequalities in two variables.

Objective

Holt Algebra 1


A linear inequality is similar to a linear equation, but the equal sign is replaced with an inequality symbol. A solution of a linear inequality is any ordered pair that makes the inequality true.

Holt Algebra 1


Tell whether the ordered pair is a solution of the inequality.

Example 1A: Identifying Solutions of Inequalities

(–2, 4); y < 2x + 1

Substitute (–2, 4) for (x, y).

y < 2x + 1

4 2(–2) + 1

4 –4 + 14 –3<

(–2, 4) is not a solution.

Holt Algebra 1


Tell whether the ordered pair is a solution of the inequality.

Example 1B: Identifying Solutions of Inequalities

(3, 1); y > x – 4

Substitute (3, 1) for (x, y).

y > x − 4

1 3 – 4

1 – 1>

(3, 1) is a solution.

Holt Algebra 1


A linear inequality describes a region of a coordinate plane called a half-plane. All points in the region are solutions of the linear inequality. The boundary line of the region is the graph of the related equation.

Holt Algebra 1


Holt Algebra 1


Graphing Linear Inequalities

Step 1 Solve the inequality for y (slope-intercept form).

Step 2Graph the boundary line. Use a solid line for ≤ or ≥. Use a dashed line for < or >.

Step 3Shade the half-plane above the line for y > or ≥. Shade the half-plane below the line for y < or y ≤. Check your answer.

Holt Algebra 1


Graph the solutions of the linear inequality.

Example 2B: Graphing Linear Inequalities in Two Variables

5x + 2y > –8

Step 1 Solve the inequality for y.

5x + 2y > –8 –5x –5x

2y > –5x – 8

y > x – 4

Step 2 Graph the boundary line Use a dashed line for >.

y = x – 4.

Holt Algebra 1


Step 3 The inequality is >, so shade above the line.


Graph the solutions of the linear inequality.5x + 2y > –8

Holt Algebra 1



Substitute ( 0, 0) for (x, y) because it is not on the boundary line.

The point (0, 0) satisfies the inequality, so the graph is correctly shaded.

Check

y > x – 4

0 (0) – 4

0 –40 –4>

Graph the solutions of the linear inequality.5x + 2y > –8

Holt Algebra 1


The point (0, 0) is a good test point to use if it does not lie on the boundary line.

Helpful Hint

Holt Algebra 1



Example 2C: Graphing Linear Inequalities in two Variables

4x – y + 2 ≤ 0


4x – y + 2 ≤ 0

–y ≤ –4x – 2

–1 –1

y ≥ 4x + 2

Step 2 Graph the boundary line y ≥= 4x + 2. Use a solid line for ≥.

Holt Algebra 1


Step 3 The inequality is ≥, so shade above the line.

Example 2C Continued


4x – y + 2 ≤ 0

Holt Algebra 1


Example 2C Continued

Substitute ( –3, 3) for (x, y) because it is not on the boundary line.

The point (–3, 3) satisfies the inequality, so the graph is correctly shaded.

Check

3 4(–3)+ 2 3 –12 + 2

3 ≥ –10

y ≥ 4x + 2

Holt Algebra 1


Check It Out! Example 2a


4x – 3y > 12


4x – 3y > 12 –4x –4x

–3y > –4x + 12

y < – 4

Step 2 Graph the boundary line y = – 4.

Use a dashed line for <.

Holt Algebra 1


Check It Out! Example 2a Continued

Step 3 The inequality is <, so shade below the line.


4x – 3y > 12

Holt Algebra 1


Check It Out! Example 2a Continued

Substitute ( 1, –6) for (x, y) because it is not on the boundary line.

The point (1, –6) satisfies the inequality, so the graph is correctly shaded.

Check

y < – 4

–6 (1) – 4 –6 – 4

–6 <


4x – 3y > 12

Holt Algebra 1


Check It Out! Example 2c


Step 1 The inequality is already solved for y.

Step 3 The inequality is ≥, so shade above the line.

Step 2 Graph the boundary

line . Use a solid line for

≥.

=

Holt Algebra 1


Check It Out! Example 2c Continued

Check

y ≥ x + 1

0 (0) + 1

0 0 + 1

0 ≥ 1

A false statement means that the half-plane containing

(0, 0) should NOT be shaded. (0, 0) is not one of the solutions, so the graph is shaded correctly.

Graph the solutions of the linear inequality.Substitute (0, 0) for (x, y) because it

is not on the boundary line.

Holt Algebra 1


Ada has at most 285 beads to make jewelry. A necklace requires 40 beads, and a bracelet requires 15 beads.

Example 3a: Application

Write a linear inequality to describe the situation.

Let x represent the number of necklaces and y the number of bracelets.

Write an inequality. Use ≤ for “at most.”

Holt Algebra 1


Example 3a Continued

Necklacebeads

braceletbeadsplus

is atmost

285beads.

40x + 15y ≤ 285

Solve the inequality for y.

40x + 15y ≤ 285–40x –40x

15y ≤ –40x + 285Subtract 40x from

both sides.


Holt Algebra 1


Example 3b

b. Graph the solutions.

=

Step 1 Since Ada cannot make a

negative amount of jewelry, the

system is graphed only in

Quadrant I. Graph the boundary

line . Use a solid line

for ≤.

Holt Algebra 1


b. Graph the solutions.

Step 2 Shade below the line. Ada can only make whole numbers of jewelry. All points on or below the line with whole number coordinates are the different combinations of bracelets and necklaces that Ada can make.

Example 3b Continued

Holt Algebra 1


c. Give two combinations of necklaces and bracelets that Ada could make.

Example 3c

Two different combinations of jewelry that Ada could make with 285 beads could be 2 necklaces and 8 bracelets or 5 necklaces and 3 bracelets.

(2, 8)

(5, 3)

Holt Algebra 1


Write an inequality to represent the graph.

Example 4A: Writing an Inequality from a Graph

y-intercept: 1; slope:

Write an equation in slope-intercept form.

The graph is shaded above a dashed boundary line.

Replace = with > to write the inequality

Holt Algebra 1


Write an inequality to represent the graph.

Example 4B: Writing an Inequality from a Graph

y-intercept: –5 slope:

Write an equation in slope-intercept form.

The graph is shaded below a solid boundary line.

Replace = with ≤ to write the inequality

Holt Algebra 1

6-6 Solving Systems of Linear Inequalities

Warm-Up

1. You can spend at most $12.00 for drinks at a picnic. Iced tea costs $1.50 a gallon, and lemonade costs $2.00 per gallon. Write an inequality to describe the situation. Graph the solutions, describe reasonable solutions, and then give two possible combinations of drinks you could buy.

2. Write an inequality to represent the graph.

Holt Algebra 1


Graph and solve systems of linear inequalities in two variables.

Objective

Holt Algebra 1


A system of linear inequalities is a set of two or more linear inequalities containing two or more variables. The solutions of a system of linear inequalities consists of all the ordered pairs that satisfy all the linear inequalities in the system.

Holt Algebra 1



Example 1A: Identifying Solutions of Systems of Linear Inequalities

(–1, –3); y ≤ –3x + 1

y < 2x + 2

y ≤ –3x + 1

–3 –3(–1) + 1–3 3 + 1–3 4≤

(–1, –3) (–1, –3)

–3 –2 + 2–3 0<

–3 2(–1) + 2

y < 2x + 2

(–1, –3) is a solution to the system because it satisfies both inequalities.

Holt Algebra 1


Check It Out! Example 1b


(0, 0); y > –x + 1 y > x – 1

y > –x + 1

0 –1(0) + 10 0 + 10 1>

(0, 0) (0, 0)

0 –1≥ 0 0 – 1

y > x – 1

(0, 0) is not a solution to the system because it does not satisfy both inequalities.

Holt Algebra 1


System of InequalitiesGraph the following inequalities on the same

graph: y < 2

x ≥ -1

y > x -2

The Graph of the

system is the overlap, or

intersection, of the 3

half-planes shown.

Holt Algebra 1


A Triangular Solution RegionA solution of a system of inequalities

is an ordered pair that is a solution

of each inequality

in the

system.

(2,1) is a solution because:

y < 21 < 2

x ≥ -12 ≥ -1

y > x -2 1 > 2 -2

(2,1)

Holt Algebra 1


Solution Region Between Parallel Lines

What is the system of

inequalities represented by this graph?

The graph of one inequality is the

half-plane below

the line y = 3

The graph of the other inequality is the half-plane

above the line y = 1

Holt Algebra 1


Solution Region Between Parallel Lines

The shaded region of the graph is the horizontal

band that lies between

the two horizontal

lines, but not on the lines.

The solution of this system of inequalities

can be represented

by this compound inequality:1 < y < 3

Holt Algebra 1


Example 2A: Solving a System of Linear Inequalities by Graphing

Graph the system of linear inequalities. Give two ordered pairs that are solutions and two that are not solutions.

y ≤ 3

y > –x + 5

y ≤ 3 y > –x + 5

Graph the system.

(8, 1) and (6, 3) are solutions.

(–1, 4) and (2, 6) are not solutions.

(6, 3)

(8, 1)

(–1, 4)(2, 6)

Holt Algebra 1


Example 2B: Solving a System of Linear Inequalities by Graphing

Graph the system of linear inequalities. Give two ordered pairs that are solutions and two that are not solutions.

–3x + 2y ≥ 2

y < 4x + 3

–3x + 2y ≥ 2 Write the first inequality in slope-intercept form.

2y ≥ 3x + 2

Holt Algebra 1


y < 4x + 3

Graph the system.


(2, 6) and (1, 3) are solutions.

(0, 0) and (–4, 5) are not solutions.

(2, 6)

(1, 3)

(0, 0)

(–4, 5)

Holt Algebra 1


In Lesson 6-4, you saw that in systems of linear equations, if the lines are parallel, there are no solutions. With systems of linear inequalities, that is not always true.

Holt Algebra 1


Graph the system of linear inequalities.

Example 3A: Graphing Systems with Parallel Boundary Lines

y ≤ –2x – 4 y > –2x + 5

This system has no solutions.

Holt Algebra 1



Example 3B: Graphing Systems with Parallel Boundary Lines

y > 3x – 2 y < 3x + 6

The solutions are all points between the parallel lines but not on the dashed lines.

Holt Algebra 1



y > x + 1 y ≤ x – 3

Check It Out! Example 3a

This system has no solutions.

Holt Algebra 1


Example 4: Application

In one week, Ed can mow at most 9 times and rake at most 7 times. He charges $20 for mowing and $10 for raking. He needs to make more than $125 in one week. Show and describe all the possible combinations of mowing and raking that Ed can do to meet his goal. List two possible combinations.

Earnings per Job ($)

Mowing

Raking

20

10

Holt Algebra 1


Example 4 Continued

Step 1 Write a system of inequalities.

Let x represent the number of mowing jobs and y represent the number of raking jobs.

x ≤ 9

y ≤ 7

20x + 10y > 125

He can do at most 9 mowing jobs.

He can do at most 7 raking jobs.

He wants to earn more than $125.

Holt Algebra 1


Step 2 Graph the system.

The graph should be in only the first quadrant because the number of jobs cannot be negative.

Solutions

Example 4 Continued

Holt Algebra 1


Step 3 Describe all possible combinations. All possible combinations represented by ordered pairs of whole numbers in the solution region will meet Ed’s requirement of mowing, raking, and earning more than $125 in one week. Answers must be whole numbers because he cannot work a portion of a job.

Step 4 List the two possible combinations.Two possible combinations are:

7 mowing and 4 raking jobs 8 mowing and 1 raking jobs

Example 4 Continued

Holt Algebra 1


An ordered pair solution of the system need not have whole numbers, but answers to many application problems may be restricted to whole numbers.

Helpful Hint

Holt Algebra 1


Lesson Quiz: Part Iy < x + 2 5x + 2y ≥ 10

1. Graph .

Give two ordered pairs that are solutions and two that are not solutions.

Possible answer: solutions: (4, 4), (8, 6); not solutions: (0, 0), (–2, 3)

Holt Algebra 1


Lesson Quiz: Part II

2. Dee has at most $150 to spend on restocking dolls and trains at her toy store. Dolls cost $7.50 and trains cost $5.00. Dee needs no more than 10 trains and she needs at least 8 dolls. Show and describe all possible combinations of dolls and trains that Dee can buy. List two possible combinations.

Holt Algebra 1


Solutions

Lesson Quiz: Part II Continued

Reasonable answers must be whole numbers. Possible answer: (12 dolls, 6 trains) and (16 dolls, 4 trains)

holt algebra 1 6-1 solving systems by graphing warm up evaluate each expression for x = 1 and y...

Documents

graphing y

x slide

system equations

system of linear equations

systems of linear equations

graphing example

holt algebra

given system