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Welcome to Interactive Chalkboard
Algebra 1 Interactive Chalkboard Copyright © by The McGraw-Hill Companies, Inc. Send all inquiries to: GLENCOE DIVISION Glencoe/McGraw-Hill 8787 Orion Place Columbus, Ohio Welcome to Interactive Chalkboard
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Splash Screen
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Lesson 7-1 Graphing Systems of Equations Lesson 7-2 Substitution
Lesson 7-3 Elimination Using Addition and Subtraction Lesson 7-4 Elimination Using Multiplication Lesson 7-5 Graphing Systems of Inequalities Contents
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Example 1 Number of Solutions Example 2 Solve a System of Equations
Example 3 Write and Solve a System of Equations Lesson 1 Contents
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Answer: Since the graphs of and are parallel, there are no solutions.
Use the graph to determine whether the system has no solution, one solution, or infinitely many solutions. Answer: Since the graphs of and are parallel, there are no solutions. Example 1-1a
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Use the graph to determine whether the system has no solution, one solution, or infinitely many solutions. Answer: Since the graphs of and are intersecting lines, there is one solution. Example 1-1a
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Use the graph to determine whether the system has no solution, one solution, or infinitely many solutions. Answer: Since the graphs of and coincide, there are infinitely many solutions. Example 1-1a
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Answer: infinitely many
Use the graph to determine whether each system has no solution, one solution, or infinitely many solutions. a. b. c. Answer: one Answer: no solution Answer: infinitely many Example 1-1b
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Answer: Graph the system of equations. Then determine whether the system has no solution, one solution, or infinitely many solutions. If the system has one solution, name it. The graphs of the equations coincide. There are infinitely many solutions of this system of equations. Example 1-2a
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Graph the system of equations
Graph the system of equations. Then determine whether the system has no solution, one solution, or infinitely many solutions. If the system has one solution, name it. Answer: The graphs of the equations are parallel lines. Since they do not intersect, there are no solutions of this system of equations. Example 1-2a
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Graph the system of equations
Graph the system of equations. Then determine whether the system has no solution, one solution, or infinitely many solutions. If the system has one solution, name it. a. Answer: one; (0, 3) Example 1-2b
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Graph the system of equations
Graph the system of equations. Then determine whether the system has no solution, one solution, or infinitely many solutions. If the system has one solution, name it. b. Answer: no solution Example 1-2b
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Bicycling Tyler and Pearl went on a 20-kilometer bike ride that lasted 3 hours. Because there were many steep hills on the bike ride, they had to walk for most of the trip. Their walking speed was 4 kilometers per hour. Their riding speed was 12 kilometers per hour. How much time did they spend walking? Words You have information about the amount of time spent riding and walking. You also know the rates and the total distance traveled. Variables Let the number of hours they rode and the number of hours they walked. Write a system of equations to represent the situation. Example 1-3a
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Equations r + w = 3 12r + 4w = 20 The number of hours riding plus
the number of hours walking equals the total number of hours of the trip. r + w = 3 The distance traveled riding plus the distance traveled walking equals the total distance of the trip. 12r + 4w = 20 Example 1-3a
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Graph the equations and .
The graphs appear to intersect at the point with the coordinates (1, 2). Check this estimate by replacing r with 1 and w with 2 in each equation. Example 1-3a
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Answer: Tyler and Pearl walked for 3 hours.
Check Answer: Tyler and Pearl walked for 3 hours. Example 1-3a
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number of weeks amount of money saved
Alex and Amber are both saving money for a summer vacation. Alex has already saved $100 and plans to save $25 per week until the trip. Amber has $75 and plans to save $30 per week. In how many weeks will Alex and Amber have the same amount of money? Answer: 5 weeks number of weeks amount of money saved Example 1-3b
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End of Lesson 1
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Example 1 Solve Using Substitution
Example 2 Solve for One Variable, Then Substitute Example 3 Dependent System Example 4 Write and Solve a System of Equations Lesson 2 Contents
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Use substitution to solve the system of equations.
Since substitute 4y for x in the second equation. Second equation Simplify. Combine like terms. Divide each side by 15. Simplify. Example 2-1a
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Use to find the value of x.
First equation Simplify. Answer: The solution is (20, 5). Example 2-1a
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Use substitution to solve the system of equations.
Answer: (1, 2) Example 2-1b
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Use substitution to solve the system of equations.
Solve the first equation for y since the coefficient of y is 1. First equation Subtract 4x from each side. Simplify. Example 2-2a
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Find the value of x by substituting for y in the second equation.
Distributive Property Combine like terms. Add 36 to each side. Simplify. Divide each side by 10. Simplify. Example 2-2a
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Substitute 5 for x in either equation to find the value of y.
Answer: The solution is (5, –8). The graph verifies the solution. First equation Simplify. Subtract 20 from each side. Example 2-2a
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Use substitution to solve the system of equations.
Answer: (–3, 2) Example 2-2b
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Use substitution to solve the system of equations.
Solve the second equation for y. Second equation Subtract x from each side. Simplify. Substitute for y in the first equation. First equation Distributive Property Simplify. Example 2-3a
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The statement is false. This means there are no solutions of the system of equations. This is true because the slope-intercept form of both equations show that the equations have the same slope, but different y-intercepts. That is, the graphs of the lines are parallel. Answer: no solution Example 2-3a
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Use substitution to solve the system of equations.
Answer: infinitely many solutions Example 2-3b
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Gold Gold is alloyed with different metals to make it hard enough to be used in jewelry. The amount of gold present in a gold alloy is measured in 24ths called karats. 24-karat gold is or 100% gold. Similarly, 18- karat gold is or 75% gold. How many ounces of 18-karat gold should be added to an amount of 12-karat gold to make 4 ounces of 14-karat gold? Example 2-4a
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The system of equations is and Use substitution to solve this system.
Let the number of ounces of 18-karat gold and the number of ounces of 12-karat gold. Use the table to organize the information. Ounces of Gold 4 y x Total Ounces 14-karat gold 12-karat gold 18-karat gold The system of equations is and Use substitution to solve this system. Example 2-4a
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Subtract y from each side.
First equation Subtract y from each side. Simplify. Second equation Distributive Property Example 2-4a
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Subtract 3 from each side.
Combine like terms. Subtract 3 from each side. Simplify. Multiply each side by –4. Simplify. Example 2-4a
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Subtract from each side.
First equation Subtract from each side. Simplify. Answer: ounces of the 18-karat gold and ounces of the 12-karat gold should be used. Example 2-4a
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Answer: 5mL of 10% solution, 5mL of 40% solution
Chemistry Mikhail needs a 10 milliliters of 25% HCl (hydrochloric acid) solution for a chemistry experiment. There is a bottle of 10% HCl solution and a bottle of 40% HCl solution in the lab. How much of each solution should he use to obtain the required amount of 25% HCl solution? Answer: 5mL of 10% solution, 5mL of 40% solution Example 2-4b
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End of Lesson 2
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Example 1 Elimination Using Addition
Example 2 Write and Solve a System of Equations Example 3 Elimination Using Subtraction Example 4 Elimination Using Subtraction Lesson 3 Contents
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Use elimination to solve the system of equations.
Since the coefficients of the x terms, –3 and 3, are additive inverses, you can eliminate the x terms by adding the equations. Write the equation in column form and add. Notice that the x value is eliminated. Divide each side by –2. Simplify. Example 3-1a
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Now substitute –15 for y in either equation to find the value of x.
First equation Replace y with –15. Simplify. Add 60 to each side. Simplify. Divide each side by –3. Simplify. Answer: The solution is (–24, –15). Example 3-1a
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Use elimination to solve the system of equations.
Answer: (2, 1) Example 3-1b
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Let x represent the first number and y represent the second number.
Four times one number minus three times another number is 12. Two times the first number added to three times the second number is 6. Find the numbers. Let x represent the first number and y represent the second number. Four times one number minus three times another number is 12. 4x – 3y = 12 Two times the first number added to three times the second number is 6. 2x + 3y = 6 Example 3-2a
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Use elimination to solve the system.
Write the equation in column form and add. Notice that the y value is eliminated. Divide each side by 6. Simplify. Example 3-2a
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Now substitute 3 for x in either equation to find the value of y.
First equation Replace x with 3. Simplify. Subtract 12 from each side. Simplify. Divide each side by –3. Simplify. Answer: The numbers are 3 and 0. Example 3-2a
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Four times one number added to another number is –10
Four times one number added to another number is –10. Three times the first number minus the second number is –11. Find the numbers. Answer: –3, 2 Example 3-2b
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Use elimination to solve the system of equations.
Since the coefficients of the x terms, 4 and 4, are the same, you can eliminate the x terms by subtracting the equations. Write the equation in column form and subtract. Notice that the x value is eliminated. Divide each side by 5. Simplify. Example 3-3a` `
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Now substitute 2 for y in either equation to find the value of x.
Second equation Simplify. Add 6 to each side. Simplify. Divide each side by 4. Simplify. Answer: The solution is (6, 2). Example 3-3a` `
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Use elimination to solve the system of equations.
Answer: The solution is (2, –6). Example 3-3b
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Multiple-Choice Test Item
If and what is the value of y? A (3, –8) B 3 C –8 D (–8, 3) Read the Test Item You are given a system of equations, and you are asked to find the value of y. Example 3-4a
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Write the equation in column form and subtract.
Solve the Test Item You can eliminate the y terms by subtracting one equation from the other. Write the equation in column form and subtract. Notice that the y value is eliminated. Divide each side by 14. Simplify. Example 3-4a
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Now substitute 3 for x in either equation to solve for y.
First equation Simplify. Subtract 24 from each side. Simplify. Notice that B is the value of x and A is the solution of the system of equations. However, the question asks for the value of y. Answer: C Example 3-4a
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Multiple-Choice Test Item
If and what is the value of x? A 4 B (4, –4) C (–4, 4) D –4 Answer: D Example 3-4b
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End of Lesson 3
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Example 1 Multiply One Equation to Eliminate
Example 2 Multiply Both Equations to Eliminate Example 3 Determine the Best Method Example 4 Write and Solve a System of Equations Lesson 4 Contents
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Use elimination to solve the system of equations.
Multiply the first equation by –2 so the coefficients of the y terms are additive inverses. Then add the equations. Multiply by –2. Add the equations. Divide each side by –1. Simplify. Example 4-1a
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Now substitute 9 for x in either equation to find the value of y.
First equation Simplify. Subtract 18 from each side. Simplify. Answer: The solution is (9, 5). Example 4-1a
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Use elimination to solve the system of equations.
Answer: (5, 1) Example 4-1b
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Use elimination to solve the system of equations.
Method 1 Eliminate x. Multiply by 3. Multiply by –4. Add the equations. Divide each side by 29. Simplify. Example 4-2a
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Now substitute 4 for y in either equation to find x.
First equation Simplify. Subtract 12 from each side. Simplify. Divide each side by 4. Simplify. Answer: The solution is (–1, 4). Example 4-2a
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Method 2 Eliminate y. Multiply by 5. Multiply by 3. Add the equations.
Divide each side by 29. Simplify. Example 4-2a
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Now substitute –1 for x in either equation.
First equation Simplify. Add 4 to each side. Simplify. Divide each side by 3. Simplify. Answer: The solution is (–1, 4), which matches the result obtained with Method 1. Example 4-2a
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Use elimination to solve the system of equations.
Answer: (4, –1) Example 4-2b
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For an exact solution, an algebraic method is best.
Determine the best method to solve the system of equations. Then solve the system. For an exact solution, an algebraic method is best. Since neither the coefficients for x nor the coefficients for y are the same or additive inverses, you cannot use elimination using addition or subtraction. Since the coefficient of the x term in the first equation is 1, you can use the substitution method. You could also use the elimination method using multiplication. Example 4-3a
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The following solution uses substitution.
First equation Subtract 5y from each side. Simplify. Example 4-3a
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Distributive Property
Second equation Distributive Property Combine like terms. Subtract 12 from each side. Simplify. Divide each side by –22. Simplify. Example 4-3a
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Subtract 5 from each side.
First equation Simplify. Subtract 5 from each side. Simplify. Answer: The solution is (–1, 1). Example 4-3a
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Determine the best method to solve the system of equations
Determine the best method to solve the system of equations. Then solve the system. Answer: The best method to use is elimination using subtraction because the coefficient of y is the same in both equations; (3, 5). Example 4-3b
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Transportation A fishing boat travels 10 miles downstream in 30 minutes. The return trip takes the boat 40 minutes. Find the rate of the boat in still water. Let the rate of the boat in still water. Let the rate of the current. Use the formula rate time distance, or Since the rate is miles per hour, write 30 minutes as hour and 40 minutes as hour. Example 4-4a
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10 Upstream Downstream d t r
This system cannot easily be solved using substitution. It cannot be solved by just adding or subtracting the equations. The best way to solve this system is to use elimination using multiplication. Since the problem asks for b, eliminate c. Example 4-4a
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Answer: The rate of the boat is 17.5 mph.
Multiply by . Multiply by . Add the equations. Multiply each side by Simplify. Answer: The rate of the boat is 17.5 mph. Example 4-4a
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Transportation A helicopter travels 360 miles with the wind in 3 hours
Transportation A helicopter travels 360 miles with the wind in 3 hours. Te return trip against the wind takes the helicopter 4 hours. Find the rate of the helicopter in still air. Answer: mph Example 4-4b
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End of Lesson 4
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Example 1 Solve by Graphing Example 2 No Solution
Example 3 Use a System of Inequalities to Solve a Problem Example 4 Use a System of Inequalities Lesson 5 Contents
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Solve the system of inequalities by graphing.
Answer: The solution includes the ordered pairs in the intersection of the graphs of and The region is shaded in green. The graphs and are boundaries of this region. The graph is dashed and is not included in the graph of . The graph of is included in the graph of Example 5-1a
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Solve the system of inequalities by graphing.
Answer: Example 5-1b
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Solve the system of inequalities by graphing.
Answer: The graphs of and are parallel lines. Because the two regions have no points in common, the system of inequalities has no solution. Example 5-2a
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Solve the system of inequalities by graphing.
Answer: Ø Example 5-2b
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Service A college service organization requires that its members maintain at least a 3.0 grade point average, and volunteer at least 10 hours a week. Graph these requirements. Words The grade point average is at least 3.0. The number of volunteer hours is at least 10 hours. Variables If the grade point average and the number of volunteer hours, the following inequalities represent the requirements of the service organization. Example 5-3a
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Inequalities The grade point average is at least 3.0.
The number of volunteer hours is at least 10. Answer: The solution is the set of all ordered pairs whose graphs are in the intersection of the graphs of these inequalities. Example 5-3a
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The senior class is sponsoring a blood drive
The senior class is sponsoring a blood drive. Anyone who wishes to give blood must be at least 17 years old and weigh at least 110 pounds. Graph these requirements. Answer: Example 5-3b
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Employment Jamil mows grass after school but his job only pays $3 an hour. He has been offered another job as a library assistant for $6 per hour. Because of school, his parents allow him to work 15 hours per week. How many hours can Jamil mow grass and work in the library and still make at least $60 per week? Let the number of hours spent mowing grass and the number of hours spent working in the library. Since g and both represent a number of days, neither can be a negative number. The following system of inequalities can be used to represent the conditions of this problem. Example 5-4a
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The solution is the set of all ordered pairs whose graphs are in the intersection of the graphs of these inequalities. Only the portion of the region in the first quadrant is used since and . Answer: Any point in the region is a possible solution. For example (2, 10) is a point in the region. Jamil could mow grass for 2 hours and work in the library for 10 hours during the week. Example 5-4a
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number of hours baby sitting number of hours working as a cashier
Emily works no more than 20 hours per week at two jobs. Her baby-sitting job pays $3 an hour and her job as a cashier at the bookstore pays $5 per hour. How many hours can Emily work at each job to earn at least $80 per week? Answer: number of hours baby sitting number of hours working as a cashier Example 5-4b
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End of Lesson 5
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Explore online information about the information introduced in this chapter.
Click on the Connect button to launch your browser and go to the Algebra 1 Web site. At this site, you will find extra examples for each lesson in the Student Edition of your textbook. When you finish exploring, exit the browser program to return to this presentation. If you experience difficulty connecting to the Web site, manually launch your Web browser and go to Algebra1.com
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