holt algebra 1 6-1 solving systems by graphing warm up evaluate each expression for x = 1 and y...
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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
6-1 Solving Systems by Graphing
Identify solutions of linear equations in two variables.
Solve systems of linear equations in two variables by graphing.
Objectives
Holt Algebra 1
6-1 Solving Systems by Graphing
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
6-1 Solving Systems by Graphing
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
6-1 Solving Systems by Graphing
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
6-1 Solving Systems by Graphing
Example 1B: Identifying Systems of Solutions
Tell whether the ordered pair is a solution of the given system.
(–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
6-1 Solving Systems by Graphing
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
6-1 Solving Systems by Graphing
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
6-1 Solving Systems by Graphing
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
6-1 Solving Systems by Graphing
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
6-1 Solving Systems by Graphing
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
6-1 Solving Systems by Graphing
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
6-1 Solving Systems by Graphing
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
6-1 Solving Systems by Graphing
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
6-1 Solving Systems by Graphing
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
6-1 Solving Systems by Graphing
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
6-2 Solving Systems by Substitution
Solve linear equations in two variables by substitution.
Objective
Holt Algebra 1
6-2 Solving Systems by Substitution
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
6-2 Solving Systems by Substitution
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
6-2 Solving Systems by Substitution
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
6-2 Solving Systems by Substitution
Solve the system by substitution.
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
6-2 Solving Systems by Substitution
Solve the system by substitution.
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
6-2 Solving Systems by Substitution
Solve the system by substitution.
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.
Write the solution as an ordered pair.
y = x + 1
2 1 + 12 2
4x + y = 6
4(1) + 2 66 6
Holt Algebra 1
6-2 Solving Systems by Substitution
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
6-2 Solving Systems by Substitution
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
6-2 Solving Systems by Substitution
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
6-2 Solving Systems by Substitution
Step 3
Example 2 Continued
y + 6x = 11
3x + 2y = –5 Solve by substitution.
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
6-2 Solving Systems by Substitution
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
3x + 2y = –5 Solve by substitution.
Write one of the original equations.
Holt Algebra 1
6-2 Solving Systems by Substitution
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
6-2 Solving Systems by Substitution
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
6-2 Solving Systems by Substitution
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.
Write one of the original equations.
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
6-2 Solving Systems by Substitution
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
6-3 Solving Systems by Elimination
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
6-3 Solving Systems by Elimination
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
6-3 Solving Systems by Elimination
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
6-3 Solving Systems by Elimination
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
6-3 Solving Systems by Elimination
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.
Divide both sides by 4.
Write the solution as an ordered pair.
Holt Algebra 1
6-3 Solving Systems by 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.
Holt Algebra 1
6-3 Solving Systems by Elimination
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
6-3 Solving Systems by Elimination
Example 2 Continued
Write one of the original equations.
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
Write the solution as an ordered pair.
Step 4 (–1, –3)
Holt Algebra 1
6-3 Solving Systems by Elimination
Remember to check by substituting your answer into both original equations.
Remember!
Holt Algebra 1
6-3 Solving Systems by Elimination
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
6-3 Solving Systems by Elimination
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
6-3 Solving Systems by Elimination
Example 3A Continued
Write one of the original equations.
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.
Write the solution as an ordered pair.
Step 4 (3, 4)
Holt Algebra 1
6-3 Solving Systems by Elimination
–5x + 2y = 32
2x + 3y = 10
Solve the system by elimination.
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
6-3 Solving Systems by Elimination
Example 3B Continued
Write one of the original equations.
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
6-3 Solving Systems by Elimination
Check It Out! Example 3b
Solve the system by elimination.
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
6-3 Solving Systems by Elimination
Check It Out! Example 3b Continued
Write one of the original equations.
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
Subtract 20 from both sides.
Holt Algebra 1
6-3 Solving Systems by Elimination
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
6-3 Solving Systems by Elimination
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
Subtract 7 from both sides.
Write one of the original equations.
Holt Algebra 1
6-3 Solving Systems by Elimination
Write the solution as an ordered pair.
Step 4 (7, 5)
Paige can buy 7 sheets of card stock and 5 sheets of felt.
Example 4 Continued
Holt Algebra 1
6-3 Solving Systems by Elimination
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
6-4 Solving Special Systems
Solve special systems of linear equations in two variables.
Classify systems of linear equations and determine the number of solutions.
Objectives
Holt Algebra 1
6-4 Solving Special Systems
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
6-4 Solving Special Systems
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
6-4 Solving Special Systems
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
6-4 Solving Special Systems
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
6-4 Solving Special Systems
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.
Example 2A Continued
Holt Algebra 1
6-4 Solving Special Systems
0 = 0 is a true statement. It does not mean the system has zero solutions or no solution.
Caution!
Holt Algebra 1
6-4 Solving Special Systems
Holt Algebra 1
6-4 Solving Special Systems
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
6-4 Solving Special Systems
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
6-5 Solving Linear Inequalities
Graph and solve linear inequalities in two variables.
Objective
Holt Algebra 1
6-5 Solving Linear Inequalities
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
6-5 Solving Linear Inequalities
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
6-5 Solving Linear Inequalities
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
6-5 Solving Linear Inequalities
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
6-5 Solving Linear Inequalities
Holt Algebra 1
6-5 Solving Linear Inequalities
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
6-5 Solving Linear Inequalities
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
6-5 Solving Linear Inequalities
Step 3 The inequality is >, so shade above the line.
Example 2B Continued
Graph the solutions of the linear inequality.5x + 2y > –8
Holt Algebra 1
6-5 Solving Linear Inequalities
Example 2B Continued
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
6-5 Solving Linear Inequalities
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
6-5 Solving Linear Inequalities
Graph the solutions of the linear inequality.
Example 2C: Graphing Linear Inequalities in two Variables
4x – y + 2 ≤ 0
Step 1 Solve the inequality for y.
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
6-5 Solving Linear Inequalities
Step 3 The inequality is ≥, so shade above the line.
Example 2C Continued
Graph the solutions of the linear inequality.
4x – y + 2 ≤ 0
Holt Algebra 1
6-5 Solving Linear Inequalities
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
6-5 Solving Linear Inequalities
Check It Out! Example 2a
Graph the solutions of the linear inequality.
4x – 3y > 12
Step 1 Solve the inequality for y.
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
6-5 Solving Linear Inequalities
Check It Out! Example 2a Continued
Step 3 The inequality is <, so shade below the line.
Graph the solutions of the linear inequality.
4x – 3y > 12
Holt Algebra 1
6-5 Solving Linear Inequalities
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 <
Graph the solutions of the linear inequality.
4x – 3y > 12
Holt Algebra 1
6-5 Solving Linear Inequalities
Check It Out! Example 2c
Graph the solutions of the linear inequality.
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
6-5 Solving Linear Inequalities
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
6-5 Solving Linear Inequalities
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
6-5 Solving Linear Inequalities
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.
Divide both sides by 15.
Holt Algebra 1
6-5 Solving Linear Inequalities
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
6-5 Solving Linear Inequalities
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
6-5 Solving Linear Inequalities
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
6-5 Solving Linear Inequalities
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
6-5 Solving Linear Inequalities
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
6-6 Solving Systems of Linear Inequalities
Graph and solve systems of linear inequalities in two variables.
Objective
Holt Algebra 1
6-6 Solving Systems of Linear Inequalities
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
6-6 Solving Systems of Linear Inequalities
Tell whether the ordered pair is a solution of the given system.
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
6-6 Solving Systems of Linear Inequalities
Check It Out! Example 1b
Tell whether the ordered pair is a solution of the given system.
(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
6-6 Solving Systems of Linear Inequalities
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
6-6 Solving Systems of Linear Inequalities
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
6-6 Solving Systems of Linear Inequalities
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
6-6 Solving Systems of Linear Inequalities
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
6-6 Solving Systems of Linear Inequalities
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
6-6 Solving Systems of Linear Inequalities
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
6-6 Solving Systems of Linear Inequalities
y < 4x + 3
Graph the system.
Example 2B Continued
(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
6-6 Solving Systems of Linear Inequalities
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
6-6 Solving Systems of Linear Inequalities
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
6-6 Solving Systems of Linear Inequalities
Graph the system of linear inequalities.
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
6-6 Solving Systems of Linear Inequalities
Graph the system of linear inequalities.
y > x + 1 y ≤ x – 3
Check It Out! Example 3a
This system has no solutions.
Holt Algebra 1
6-6 Solving Systems of Linear Inequalities
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
6-6 Solving Systems of Linear Inequalities
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
6-6 Solving Systems of Linear Inequalities
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
6-6 Solving Systems of Linear Inequalities
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
6-6 Solving Systems of Linear Inequalities
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
6-6 Solving Systems of Linear Inequalities
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
6-6 Solving Systems of Linear Inequalities
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
6-6 Solving Systems of Linear Inequalities
Solutions
Lesson Quiz: Part II Continued
Reasonable answers must be whole numbers. Possible answer: (12 dolls, 6 trains) and (16 dolls, 4 trains)
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