Lesson 3 1 Solving Systems of Equations by

Lesson 3 -1 Solving Systems of Equations by Graphing Lesson 3 -2 Solving Systems of Equations Algebraically Lesson 3 -3 Solving Systems of Inequalities by Graphing Lesson 3 -4 Linear Programming Lesson 3 -5 Solving Systems of Equations in Three Variables

Five-Minute Check (over Chapter 2) Main Ideas and Vocabulary California Standards Example 1: Solve the System of Equations by Completing a Table Example 2: Solve by Graphing Example 3: Break-Even Point Analysis Example 4: Intersecting Lines Example 5: Same Line Example 6: Parallel Lines Concept Summary

• Solve systems of linear equations by graphing. • Determine whether a system of linear equations is consistent and independent, consistent and dependent, or inconsistent. • system of equations • consistent • independent • dependent

Standard 2. 0 Students solve systems of linear equations and inequalities (in two or three variables) by substitution, with graphs, or with matrices. (Key)

Solve the System of Equations by Completing a Table Solve the system of equations by completing a table. x+y=3 – 2 x + y = – 6 Solve for y in each equation. x+y = 3 y = –x + 3 – 2 x + y = – 6 y = 2 x – 6

Solve the System of Equations by Completing a Table Use a table to find the solution that satisfies both equations. Answer: The solution to the system is (3, 0).

What is the solution of the system of equations? x+y=2 x – 3 y = – 6 A. (1, 1) A. A B. (0, 2) B. B C. (2, 0) C. C D. (– 4, 6) D. D

Solve by Graphing Solve the system of equations by graphing. x – 2 y = 0 x+y=6 Write each equation in slope-intercept form. The graphs appear to intersect at (4, 2).

Solve by Graphing Check Substitute the coordinates into each equation. x – 2 y = 0 ? x+y =6 ? 4 – 2(2) = 0 4+2 =6 0=0 6=6 Original equations Replace x with 4 and y with 2. Simplify. Answer: The solution of the system is (4, 2).

Which graph shows the solution to the system of equations below? x + 3 y = 7 x–y = 3 A. A A. C. B. B C. C D. D B. D.

Break-Even Point Analysis SALES A service club is selling copies of its holiday cookbook to raise funds for a project. The printer’s set -up charge is $200, and each book costs $2 to print. The cookbooks will sell for $6 each. How many must the club sell before it makes a profit? Let x = the number of cookbooks, and let y = the number of dollars. Cost of books is cost per book plus set-up charge. y = 2 x + 200

Break-Even Point Analysis Income from books y is price per book times number of books. = 6 x Answer: The graphs intersect at (50, 300). This is the break -even point. If the group sells fewer than 50 books, they will lose money. If the group sells more than 50 books, they will make a profit.

The student government is selling candy bars. It cost $1 for each candy bar plus a $60 set-up fee. The group will sell the candy bars for $2. 50 each. How many do they need to sell to break even? A. 0 A. A B. 40 B. B C. 60 C. C D. 80 D. D

Intersecting Lines Graph the system of equations and describe it as consistent and independent, consistent and dependent, or inconsistent. x–y=5 x + 2 y = – 4 Write each equation in slope-intercept form.

Intersecting Lines Answer: The graphs of the equations intersect at (2, – 3). Since there is one solution to this system, this system is consistent and independent.

Graph the system of equations below. What type of system of equations is shown? x+y=5 2 x = y – 11 A. A A. consistent and independent B. B B. consistent and dependent C. C C. consistent D. D D. none of the above

Same Line Graph the system of equations and describe it as consistent and independent, consistent and dependent, or inconsistent. 9 x – 6 y = – 6 6 x – 4 y = – 4 Write each equation in slope-intercept form. Since the equations are equivalent, their graphs are the same line.

Same Line Answer: Any ordered pair representing a point on that line will satisfy both equations. So, there are infinitely many solutions. This system is consistent and dependent.

Graph the system of equations below. What type of system of equations is shown? x+y=3 2 x = – 2 y + 6 A. A A. consistent and independent B. B B. consistent and dependent C. C C. inconsistent D. D D. none of the above

Parallel Lines Graph the system of equations and describe it as consistent and independent, consistent and dependent, or inconsistent. 15 x – 6 y = 0 5 x – 2 y = 10 Write each equation in slope-intercept form.

Parallel Lines Answer: The lines do not intersect. Their graphs are parallel lines. So, there are no solutions that satisfy both equations. This system is inconsistent.

Graph the system of equations below. What type of system of equations is shown? y = 3 x + 2 – 6 x + 2 y = 10 A. A A. consistent and independent B. B B. consistent and dependent C. C C. inconsistent D. D D. none of the above

Five-Minute Check (over Lesson 3 -1) Main Ideas and Vocabulary California Standards Example 1: Solve by Using Substitution Example 2: Standards Example: Solve by Substitution Example 3: Solve by Using Elimination Example 4: Multiply, Then Use Elimination Example 5: Inconsistent System

• Solve systems of linear equations by using substitution. • Solve systems of linear equations by using elimination. • substitution method • elimination method

Standard 2. 0 Students solve systems of linear equations and inequalities (in two or three variables) by substitution, with graphs, or with matrices. (Key)

Solve by Using Substitution Use substitution to solve the system of equations. x + 4 y = 26 x – 5 y = – 10 Solve the first equation for x in terms of y. x + 4 y = 26 x = 26 – 4 y First equation Subtract 4 y from each side.

Solve by Using Substitution Substitute 26 – 4 y for x in the second equation and solve for y. x – 5 y = – 10 26 – 4 y – 5 y = – 10 – 9 y = – 36 y=4 Second equation Substitute 26 – 4 y for x. Subtract 26 from each side. Divide each side by – 9.

Solve by Using Substitution Now substitute the value for y in either of the original equations and solve for x. x + 4 y = 26 x + 4(4) = 26 x + 16 = 26 x = 10 First equation Replace y with 4. Simplify. Subtract 16 from each side. Answer: The solution of the system is (10, 4).

Solve the system of equations using substitution. What is the solution to the system of equations? x – 3 y = 2 x + 7 y = 12 A. (1, 5) B. A. A B. B C. C C. (8, 2) D. (5, 1) D. D

Solve by Substitution Lancaster Woodworkers Furniture Store builds two types of wooden outdoor chairs. A rocking chair sells for $265 and an Adirondack chair with footstool sells for $320. The books show that last month, the business earned $13, 930 for the 48 outdoor chairs sold. How many of each chair were sold? Read the Item You are asked to find the number of each type of chair sold.

Solve by Substitution Solve the Item Step 1 Define variables and write the system of equations. Let x represent the number of rocking chairs sold and y represent the number of Adirondack chairs sold. x + y = 48 265 x + 320 y = 13, 930 The total number of chairs sold was 48. The total amount earned was $13, 930.

Solve by Substitution Step 2 Solve one of the equations for one of the variables in terms of the other. Since the coefficient of x is 1, solve the first equation for x in terms of y. x + y = 48 x = 48 – y First equation Subtract y from each side.

Solve by Substitution Step 3 Substitute 48 – y for x in the second equation. 265 x + 320 y = 13, 930 265(48 – y) + 320 y = 13, 930 12, 720 – 265 y + 320 y = 13, 930 55 y = 1210 y = 22 Second equation Substitute 48 – y for x. Distributive Property Simplify. Divide each side by 55.

Solve by Substitution Step 4 Now find the value of x. Substitute the value for y into either equation. x + y = 48 x + 22 = 48 x = 26 First equation Replace y with 22. Subtract 22 from each side. Answer: They sold 26 rocking chairs and 22 Adirondack chairs.

AMUSEMENT PARKS At Amy’s Amusement Park, tickets sell for $24. 50 for adults and $16. 50 for children. On Sunday, the amusement park made $6405 from selling 330 tickets. How many of each kind of ticket was sold? A. A A. 210 adult; 120 children B. B B. 120 adult; 210 children C. C C. 300 children; 30 adult D. D D. 300 children; 30 adult

Solve by Using Elimination Use the elimination method to solve the system of equations. x + 2 y = 10 x+y=6 In each equation, the coefficient of x is 1. If one equation is subtracted from the other, the variable x will be eliminated. x + 2 y = 10 (–)x + y = 6 y= 4 Subtract the equations.

Solve by Using Elimination Now find x by substituting 4 for y in either original equation. x+y =6 Second equation x+4 =6 Replace y with 4. x =2 Subtract 4 from each side. Answer: The solution is (2, 4).

Use the elimination method to solve the system of equations. What is the solution to the system? x + 3 y = 5 x + 5 y = – 3 A. (2, – 1) A. A B. (17, – 4) B. B C. (2, 1) C. C D. no solution D. D

Multiply, Then Use Elimination Use the elimination method to solve the system of equations. 2 x + 3 y = 12 5 x – 2 y = 11 Multiply the first equation by 2 and the second equation by 3. Then add the equations to eliminate the y variable. 2 x + 3 y = 12 Multiply by 2. 5 x – 2 y = 11 Multiply by 3. 4 x + 6 y = 24 (+)15 x – 6 y = 33 19 x = 57 x =3

Multiply, Then Use Elimination Replace x with 3 and solve for y. 2 x + 3 y = 12 2(3) + 3 y = 12 6 + 3 y = 12 3 y = 6 y=2 First equation Replace x with 3. Multiply. Subtract 6 from each side. Divide each side by 3. Answer: The solution is (3, 2).

Use the elimination method to solve the system of equations. What is the solution to the system of equations? x + 3 y = 7 2 x + 5 y = 10 A. A B. (1, 2) B. B C. (– 5, 4) C. C D. no solution D. D

Inconsistent System Use the elimination method to solve the system of equations. – 3 x + 5 y = 12 6 x – 10 y = – 21 Use multiplication to eliminate x. – 3 x + 5 y = 12 6 x – 10 y = – 21 Multiply by 2. – 6 x + 10 y = 24 (+)6 x – 10 y = – 21 0= 3 Answer: Since there are no values of x and y that will make the equation 0 = 3 true, there are no solutions for the system of equations.

Use the elimination method to solve the system of equations. What is the solution to the system of equations? 2 x + 3 y = 11 – 4 x – 6 y = 20 A. (1, 3) A. A B. (– 5, 0) B. B C. (2, – 2) C. C D. no solution D. D

Five-Minute Check (over Lesson 3 -2) Main Ideas and Vocabulary California Standards Example 1: Intersecting Regions Example 2: Separate Regions Example 3: Write and Use a System of Inequalities Example 4: Find Vertices

• Solve systems of inequalities by graphing. • Determine the coordinates of the vertices of a region formed by the graph of a system of inequalities. • system of inequalities

Standard 2. 0 Students solve systems of linear equations and inequalities (in two or three variables) by substitution, with graphs, or with matrices. (Key)

Intersecting Regions A. Solve the system of inequalities by graphing. y ≥ 2 x – 3 y < –x + 2 Solution of y ≥ 2 x – 3 → Regions 1 and 2 Solution of y < –x + 2 → Regions 2 and 3

Intersecting Regions A. Answer: The intersection of these regions is Region 2, which is the solution of the system of inequalities. Notice that the solution is a region containing an infinite number of ordered pairs.

Intersecting Regions B. Solve the system of inequalities by graphing. y ≤ –x + 1 │x + 1│< 3 The inequality │x + 1│< 3 can be written as x + 1 < 3 and x + 1 > – 3, or x < 2 and x > – 4.

Intersecting Regions Graph all of the inequalities on the same coordinate plane and shade the region or regions that are common to all. Answer: Animation: Use Graphing to Solve the System of Inequalities

A. Solve the system of inequalities by graphing. What is the solution to the system of inequalities below? y ≤ 3 x – 3 y>x+1 A. B. 1. 2. 3. 4. C. D. A B C D

B. Solve the system of inequalities by graphing. What is the solution to the system of inequalities below? y ≥ – 2 x – 3 │x + 2│< 1 A. B. 1. 2. 3. 4. C. D. A B C D

Separate Regions Solve the system of inequalities by graphing. Graph both inequalities. The graphs do not overlap, so the solutions have no points in common. Answer: The solution set is Ø.

Solve the system of inequalities by graphing. A. C. B. D. 1. 2. 3. 4. A B C D

Write and Use a System of Inequalities MEDICINE Medical professionals recommend that patients have a cholesterol level below 200 milligrams per deciliter (mg/d. L) of blood and a triglyceride level below 150 mg/d. L. Write and graph a system of inequalities that represents the range of cholesterol levels and triglyceride levels for patients. Let c represent the cholesterol levels in mg/d. L. It must be less than 200 mg/d. L. Since cholesterol levels cannot be negative, we can write this as 0 ≤ c < 200. Let t represent the triglyceride levels in mg/d. L. It must be less than 150 mg/d. L. Since triglyceride levels also cannot be negative, we can write this as 0 ≤ t < 150.

Write and Use a System of Inequalities Graph all of the inequalities. Any ordered pair in the intersection of the graphs is a solution of the system. Answer: 0 ≤ c < 200 0 ≤ t < 150

SAFETY The speed limits while driving on the highway are different for trucks and cars. Cars must drive between 45 and 65 miles per hour, inclusive. Trucks are required to drive between 40 and 55 miles per hour, inclusive. Let c represent the range of speeds for cars and t represent the range of speeds for trucks. Which graph represents this situation?

A. C. B. D. 1. 2. 3. 4. A B C D

Find Vertices Find the coordinates of the vertices of the figure formed by 2 x – y ≥ – 1, x + y ≤ 4, and x + 4 y ≥ 4. Graph each inequality. The intersection of the graphs forms a triangle. Answer: The vertices of the triangle are at (0, 1), (4, 0), and (1, 3).

Find the coordinates of the vertices of the figure formed by x + 2 y ≥ 1, x + y ≤ 3, and – 2 x + y ≤ 3. A. (– 1, 0), (0, 3), and (5, A. – 2) A B. (– 1, 0), (0, 3), and (4, B. – 2) B C. (– 1, 1), (0, 3), and (5, C. – 2) C D. (0, 3), (5, – 2), and (1, D. 0) D

Five-Minute Check (over Lesson 3 -3) Main Ideas and Vocabulary California Standards Example 1: Bounded Region Example 2: Unbounded Region Key Concept: Linear Programming Procedure Example 3: Linear Programming

• Find the maximum and minimum values of a function over a region. • Solve real-world problems using linear programming. • constraints • feasible region • bounded • vertex • unbounded • linear programming

Reinforcement of Standard 2. 0 Students solve systems of linear equations and inequalities (in two or three variables) by substitution, with graphs, or with matrices. (Key)

Bounded Region Graph the following system of inequalities. Name the coordinates of the vertices of the feasible region. Find the maximum and minimum values of the function f(x, y) = 3 x – 2 y for this region. x≤ 5 y≤ 4 x+y≥ 2 Step 1 Graph the inequalities. The polygon formed is a triangle with vertices at (– 2, 4), (5, – 3), and (5, 4).

Bounded Region Step 2 Use a table to find the maximum and minimum values of f(x, y). Substitute the coordinates of the vertices into the function. ← minimum ← maximum Answer: The vertices of the feasible region are (– 2, 4), (5, – 3), and (5, 4). The maximum value is 21 at (5, – 3). The minimum value is – 14 at (– 2, 4).

Graph the following system of inequalities. What are the maximum and minimum values of the function f(x, y) = 4 x – 3 y for the feasible region of the graph? x≤ 4 y≤ 5 x+y≥ 6 A. maximum: f(4, 5) = 5 A. minimum: f(1, 5) = – 11 B. maximum: f(4, 2) = 10 B. minimum: f(1, 5) = – 11 C. maximum: f(4, 2) = 10 C. minimum: f(4, 5) = 5 D. maximum: f(1, 5) = – 11 D. minimum: f(4, 2) = 10 A B C D

Unbounded Region Graph the following system of inequalities. Name the coordinates of the vertices of the feasible region. Find the maximum and minimum values of the function f(x, y) = 2 x + 3 y for this region. –x + 2 y ≤ 2 x – 2 y ≤ 4 x + y ≥ – 2 Graph the system of inequalities. There are only two points of intersection, (– 2, 0) and (0, – 2).

Unbounded Region The minimum value is – 6 at (0, – 2). Although f(– 2, 0) is – 4, it is not the maximum value since there are other points that produce greater values. For example, f(2, 1) is 7 and f(3, 1) is 10. It appears that because the region is unbounded, f(x, y) has no maximum value. Answer: The vertices are at (– 2, 0) and (0, – 2). There is no maximum value. The minimum value is – 6 at (0, – 2).

Graph the following system of inequalities. What are the maximum and minimum values of the function f(x, y) = x + 2 y for the feasible region of the graph? x + 3 y ≤ 6 –x – 3 y ≤ 9 2 y – x ≥ – 6 A. maximum: no maximum minimum: f(6, 0) = 6 A. A B. maximum: f(6, 0) = 6 minimum: f(0, – 3) = – 6 B. B C. maximum: f(6, 0) = 6 minimum: no minimum C. C D. maximum: no maximum minimum: f(0, – 3) = – 6 D. D

Linear Programming Procedure Step 1 Define the variables. Step 2 Write a system of inequalities. Step 3 Graph the system of inequalities. Step 4 Find the coordinates of the vertices of the feasible region. Step 5 Write a function to be maximized or minimized. Step 6 Substitute the coordinates of the vertices into the function. Step 7 Select the greatest or least result. Answer the problem.

Linear Programming LANDSCAPING A landscaping company has crews who mow lawns and prune shrubbery. The company schedules 1 hour for mowing jobs and 3 hours for pruning jobs. Each crew is scheduled for no more than 2 pruning jobs per day. Each crew’s schedule is set up for a maximum of 9 hours per day. On the average, the charge for mowing a lawn is $40 and the charge for pruning shrubbery is $120. Find a combination of mowing lawns and pruning shrubs that will maximize the income the company receives per day from one of its crews.

Linear Programming Step 1 Define the variables. m = the number of mowing jobs p = the number of pruning jobs

Linear Programming Step 2 Write a system of inequalities. Since the number of jobs cannot be negative, m and p must be nonnegative numbers. m ≥ 0, p ≥ 0 Mowing jobs take 1 hour. Pruning jobs take 3 hours. There are 9 hours to do the jobs. 1 m + 3 p ≤ 9 There are no more than 2 pruning jobs a day. p≤ 2

Linear Programming Step 3 Graph the system of inequalities.

Linear Programming Step 4 Find the coordinates of the vertices of the feasible region. From the graph, the vertices are at (0, 2), (3, 2), (9, 0), and (0, 0). Step 5 Write the function to be maximized. The function that describes the income is f(m, p) = 40 m + 120 p. We want to find the maximum value for this function.

Linear Programming Step 6 Substitute the coordinates of the vertices into the function. Step 7 Select the greatest amount.

Linear Programming Answer: The maximum values are 360 at (3, 2) and 360 at (9, 0). This means that the company receives the most money with 3 mowings and 2 prunings or 9 mowings and 0 prunings. Animation: Linear Programming

LANDSCAPING A landscaping company has crews who rake leaves and mulch. The company schedules 2 hours for mulching jobs and 4 hours for raking jobs. Each crew is scheduled for no more than 2 raking jobs per day. Each crew’s schedule is set up for a maximum of 8 hours per day. On the average, the charge for raking a lawn is $50 and the charge for mulching is $30.

What is a combination of raking leaves and mulching that will maximize the income the company receives per day from one of its crews? A. 0 mulching; 2 raking A. A B. 4 mulching; 0 raking B. B C. 0 mulching; 4 raking C. C D. 2 mulching; 0 raking D. D

Five-Minute Check (over Lesson 3 -4) Main Ideas and Vocabulary California Standards Key Concept: System of Equations in Three Variables Example 1: One Solution Example 2: Infinitely Many Solutions Example 3: No Solution Example 4: Real-World Example: Write and Solve a System of Equations

• Solve systems of linear equations in three variables. • Solve real-world problems using systems of linear equations in three variables. • ordered triple

Standard 2. 0 Students solve systems of linear equations and inequalities (in two or three variables) by substitution, with graphs, or with matrices. (Key)

System of Equations in Three Variables

One Solution Solve the system of equations. 5 x + 3 y + 2 z = 2 2 x + y – z = 5 x + 4 y + 2 z = 16 Step 1 Use elimination to make a system of two equations in two variables. 5 x + 3 y + 2 z = 2 2 x + y – z = 5 Multiply by 2. 5 x + 3 y + 2 z = 2 First equation (+)4 x + 2 y – 2 z = 10 Second equation 9 x + 5 y = 12 Add to eliminate z.

One Solution 5 x + 3 y + 2 z = 2 (–) x + 4 y + 2 z = 16 4 x – y = – 14 First equation Third equation Subtract to eliminate z. Notice that the z terms in each equation have been eliminated. The result is two equations with the same two variables x and y.

One Solution Step 2 Solve the system of two equations. 9 x + 5 y = 12 4 x – y = – 14 (+) 20 x – 5 y = – 70 Multiply by 5 29 x = – 58 Add to eliminate y. x = – 2 Divide by 29.

One Solution Substitute – 2 for x in one of the two equations with two variables and solve for y. 4 x – y = – 14 Equation with two variables 4(– 2) – y = – 14 – 8 – y = – 14 y=6 Replace x with – 2. Multiply. Simplify. The result is x = – 2 and y = 6.

One Solution Step 3 Solve for z using one of the original equations with three variables. 2 x + y – z = 5 Original equation with three variables 2(– 2) + 6 – z = 5 – 4 + 6 – z = 5 z = – 3 Replace x with – 2 and y with 6. Multiply. Simplify. Answer: The solution is (– 2, 6, – 3). You can check this solution in the other two original equations.

What is the solution to the system of equations shown below? 2 x + 3 y – 3 z = 16 x + y + z = – 3 x – 2 y – z = – 1 A. A B. (– 3, – 2, 2) B. B C. (1, 2, – 6) C. C D. (– 1, 2, – 4) D. D

Infinite Solutions Solve the system of equations. 2 x + y – 3 z = 5 x + 2 y – 4 z = 7 6 x + 3 y – 9 z = 15 Eliminate y in the first and third equations. Multiply by 3. 2 x + y – 3 z = 5 6 x + 3 y – 9 z = 15 (–)6 x + 3 y – 9 z = 15 0= 0

Infinite Solutions The equation 0 = 0 is always true. This indicates that the first and third equations represent the same plane. Check to see if this plane intersects the second plane. Multiply by 6. x + 2 y – 4 z = 7 6 x + 3 y – 9 z = 15 6 x + 12 y – 24 z = 42 (–)6 x + 3 y – 9 z = 15 9 y – 15 z = 27 Divide by the GCF, 3. 3 y – 5 z = 9 Answer: The planes intersect in a line. So, there is an infinite number of solutions.

What is the solution to the system of equations shown below? x + y – 2 z = 3 – 3 x – 3 y + 6 z = – 9 2 x + y – z = 6 A. (1, 2, 0) A. A B. (2, 2, 0) B. B C. infinite number of C. solutions D. no solution C D. D

No Solution Solve the system of equations. 3 x – y – 2 z = 4 6 x + 4 y + 8 z = 11 9 x + 6 y + 12 z = – 3 Eliminate x in the last two equations. Multiply by 3. 6 x + 4 y + 8 z = 11 18 x + 12 y + 24 z = 33 9 x + 6 y + 12 z = – 3 (–) 18 x + 12 y + 24 z = – 6 Multiply by 2. 0 = 39 Answer: The equation 0 = 39 is never true. So, there is no solution of this system.

What is the solution to the system of equations shown below? 3 x + y – z = 5 – 15 x – 5 y + 5 z = 11 x+y+z=2 A. (0, 6, 1) A. A B. (1, 0, – 2) B. B C. infinite number of. C. solutions C D. no solution D. D

Write and Solve a System of Equations SPORTS There are 49, 000 seats in a sports stadium. Tickets for the seats in the upper level sell for $25, the ones in the middle level cost $30, and the ones in the bottom level are $35 each. The number of seats in the middle and bottom levels together equals the number of seats in the upper level. When all of the seats are sold for an event, the total revenue is $1, 419, 500. How many seats are there in each level? Explore Read the problem and define the variables. u = number of seats in the upper level m = number of seats in the middle level b = number of seats in the bottom level

Write and Solve a System of Equations Plan There are 49, 000 seats. u + m + b = 49, 000 When all the seats are sold, the revenue is 1, 419, 500. Seats cost $25, $30, and $35. 25 u + 30 m + 35 b = 1, 419, 500 The number of seats in the middle and bottom levels together equal the number of seats in the upper level. m+b=u

Write and Solve a System of Equations Solve Substitute u = m + b in each of the first two equations. (m + b) + m + b = 49, 000 2 m + 2 b = 49, 000 m + b = 24, 500 Replace u with m + b. Simplify. Divide by 2. 25(m + b) + 30 m + 35 b = 1, 419, 500 Replace u with m + b. 25 m + 25 b + 30 m + 35 b = 1, 419, 500 Distributive Property 55 m + 60 b = 1, 419, 500 Simplify.

Write and Solve a System of Equations Now, solve the system of two equations in two variables. Multiply by 55. m + b = 24, 500 55 m + 60 b = 1, 419, 500 55 m + 55 b = 1, 347, 500 (–) 55 m + 60 b = 1, 419, 500 – 5 b = – 72, 000 b = 14, 400

Write and Solve a System of Equations Substitute 14, 400 for b in one of the equations with two variables and solve for m. m + b = 24, 500 m + 14, 400 = 24, 500 m = 10, 100 Equation with two variables b = 14, 400 Subtract 14, 400 from each side.

Write and Solve a System of Equations Substitute 14, 400 for b and 10, 100 for m in one of the original equations with three variables. m+b =u 10, 100 + 14, 400 = u 24, 500 = u Equation with three variables m = 10, 100, b = 14, 400 Add. Answer: There are 24, 500 upper level, 100 middle level, and 14, 400 bottom level seats.

Write and Solve a System of Equations Check to see if all the criteria are met. 24, 500 + 10, 100 + 14, 400 = 49, 000 The number of seats in the middle and bottom levels equals the number of seats in the upper level. 10, 100 + 14, 400 = 24, 500 When all of the seats are sold, the revenue is $1, 419, 500. 24, 500($25) + 10, 100($30) + 14, 400($35) = $1, 419, 500

BUSINESS The school store sells pens, pencils, and paper. The pens are $1. 25 each, the pencils are $0. 50 each, and the paper is $2 per pack. Yesterday the store sold 25 items and earned $32. The number of pens sold equaled the number of pencils sold plus the number of packs of paper sold minus 5. How many of each item did the store sell? Interactive Lab: Solving Systems of Linear Equations and Inequalities

How many of each item did the store sell? A. pens: 5; pencils: 10; A. paper: 10 A B. pens: 8; pencils: 7; B. paper: 10 B C. pens: 10; pencils: 7; C. paper: 8 C D. pens: 11; pencils: 2; D. paper: 12 D

Five-Minute Checks Image Bank Math Tools Use Graphing to Solve the System of Inequalities Linear Programming Solving Systems of Linear Equations and Inequalities

Lesson 3 -1 (over Chapter 2) Lesson 3 -2 (over Lesson 3 -1) Lesson 3 -3 (over Lesson 3 -2) Lesson 3 -4 (over Lesson 3 -3) Lesson 3 -5 (over Lesson 3 -4)

To use the images that are on the following three slides in your own presentation: 1. Exit this presentation. 2. Open a chapter presentation using a full installation of Microsoft® Power. Point® in editing mode and scroll to the Image Bank slides. 3. Select an image, copy it, and paste it into your presentation.

(over Chapter 2) Find the domain (D) and range (R) of the relation {(– 4, 1), (0, 0), (1, – 4), (2, 0), (– 2, 0)}. Determine whether the relation is a function. A. D = {– 4, – 2, 0, 1, 2}, A. R = {– 4, 0, 1}; no A B. D = {– 4, 0, 1}, B. B R = {– 4, – 2, 0, 1, 2}; no C. D = {– 4, – 2, 0, 1, 2}, C. R = {– 4, 0, 1}; yes C D. D = {– 4, 0, 1}, D. D R = {– 4, – 2, 0, 1, 2}; yes

(over Chapter 2) Find the value of f(4) for f(x) = 8 – x 2. A. 20 A. A B. 12 B. B C. – 4 C. C D. – 12 D. D

(over Chapter 2) Find the slope of the line that passes through (5, 7) and (– 1, 0). A. A A. B. C. D. B. B C. C D. D

(over Chapter 2) Write an equation in slope-intercept form for the line that has x-intercept – 3 and y-intercept 6. A. y = 2 x + 6 A. A B. y = 3 x – 6 B. B C. y = – 3 x + 6 C. C D. y = x + 3 D. D

(over Chapter 2) The Math Club is using the prediction equation y = 1. 25 x + 10 to estimate the number of members it will have, where x represents the number of years the club has been in existence. About how many members does the club expect to have in its fifth year? A. 12 A. A B. 15 B. B C. 16 C. C D. 19 D. D

(over Chapter 2) Which function has the greatest value for f(– 2)? A. f(x) = x B. f(x) = [x] C. f(x) = |x| – 1 D. f(x) = – |x| A. A B. B C. C D. D

(over Lesson 3 -1) Which choice shows a graph of the solution of this system? y = 3 x – 2 y = – 3 x + 2 A. B. C. D. 1. 2. 3. 4. A B C D

(over Lesson 3 -1) State whether the graph of the system of equations is consistent and independent, consistent and dependent, or inconsistent? 2 x + y = 6 3 y = – 6 x + 6 A. A A. consistent and independent B. B B. consistent and dependent C. C C. inconsistent D. none of the above D. D

(over Lesson 3 -1) Which equation is inconsistent with 4 x + 5 y = 30? A. A B. 5 y = – 4 x – 30 B. B C. C D. 5 y = 4 x – 30 D. D

(over Lesson 3 -2) Solve the system of equations by using substitution. x – 4 y = – 12 3 x + 2 y = 20 A. (4, 4) A. A B. (8, 5) B. B C. no solution C. C D. infinitely many solutions D. D

(over Lesson 3 -2) Solve the system of equations by using substitution. x + 2 y = 6 2 x + 4 y = 15 A. (3, 1. 5) A. A B. (3. 5, 2) B. B C. no solution C. C D. infinitely many solutions D. D

(over Lesson 3 -2) Solve the system of equations by using elimination. 2 x + 5 y = 9 – 2 x + 8 y = 4 A. (– 2, 0) A. A B. (2, 1) B. B C. no solution C. C D. infinitely many solutions D. D

(over Lesson 3 -2) Solve the system of equations by using elimination. x + 2 y = 7 14 – 4 y = 2 x A. (0, 0) A. A B. (– 5, – 6) B. B C. no solution C. C D. infinitely many D. solutions D

(over Lesson 3 -2) Two times one number minus a second number is 7. The second number is three more than the first number. What are the two numbers? A. A A. 7 and 10 B. 10 and 13 C. 13 and 16 B. B C. C D. 7 and 20 D. D

(over Lesson 3 -3) Which choice shows a graph of the solution of the system of inequalities? x ≤ – 2 y>3 A. B. C. D. 1. 2. 3. 4. A B C D

(over Lesson 3 -3) Which choice shows a graph of the solution of the system of inequalities? y ≤ 3 x + 2 y > –x A. B. C. D. 1. 2. 3. 4. A B C D

(over Lesson 3 -3) Find the coordinates of the vertices of the figure formed by the system of inequalities. x≥ 0 y≤ 0 – 3 x + y = – 6 A. (0, 0), (2, 0), (– 6, 0) A. B. (0, 0), (0, 2), (0, 6) A B. B C. (0, 0), (2, 0), (0, – 6) C. C D. (0, 0), (0, – 2), (0, – 6) D. D

(over Lesson 3 -3) Find the coordinates of the vertices of the figure formed by the system of inequalities. y≤ 3 y≥ 0 x≥ 0 2 y + 3 x ≤ 12 A. A A. (0, 0), (0, 6), (2, 3), (4, 0) B. B B. (0, 0), (3, 2), (4, 0) C. C C. (0, 0), (3, 0), (0, 6), (2, 3) D. D D. (0, 0), (0, 3), (2, 3), (4, 0)

(over Lesson 3 -3) Which point is not the solution of the system? |x| 2 y 3 A. A A. (0, 4) B. B B. (– 3, 5) C. (– 2, 5) D. (1, 4) C. C D. D

(over Lesson 3 -4) Which choice shows a graph of the system of inequalities and the coordinates of the vertices of the feasible region? 1≤x≤ 4 A. B. y≥x y ≤ 2 x + 3 C. D. 1. 2. 3. 4. A B C D

(over Lesson 3 -4) The vertices of a feasible region are (0, 0), (0, 6), (4, 2), and (5, 5). What are the maximum and minimum values of the function f(x, y) = 3 x – 5 y over this region? A. minimum: f(0, 0) = 0 A. A maximum: f(5, 5) = – 10 B. minimum: f(0, 0) = 0 B. B maximum: f(4, 2) = 2 C. C C. minimum: f(0, 6) = – 30 maximum: f(4, 2) = 2 D. D D. minimum: f(0, 6) = – 30 maximum: f(5, 5) = – 10

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