Algebra Meeting (Multiple topics from pre-algebra and Algebra I)

Transcription

1 Algebra Meeting (Multiple topics from pre-algebra and Algebra I) Topic This meeting s problems cover a variety of topics from pre-algebra and Algebra I. The topics include simplifying expressions, solving equations and inequalities, linear relationships and graphing. Materials Needed Copies of the Algebra problem set (Problems and answers can be viewed below. Complete solutions and a more student-friendly version of the problems with pictures and larger font are available for download from on the MCP Members Only page of the Club Program section.) Meeting Plan The problems in this problem set increase in difficulty as students progress. Given the wide range of topics that are covered, it is suggested that students work on the problems in pairs or small groups. Students who are taking or have completed algebra should be paired with students who have not. 1. Maya initially thought that 8 was a solution to 2x + 20 = 4x 4, but when she checked her work, she realized that her answer was incorrect. After substituting 8 for x, how much greater is the expression on the left side of the equals sign than the expression on the right side? School Handbook Warm-Up Three bags each contain the same number of marbles. (The weight of an empty bag would not affect the weight on the scale.) If three bags of marbles plus two more marbles balance 14 marbles, as shown, how many marbles are in each bag? School Handbook Warm-Up The function f (x) is defined by f (x) = x 2 x. What is the value of f (4)? 2007 School Competition Countdown Round #4 4. What is the value of x such that (x, 0) is a solution of the equation y = 3x 4? Express your answer as a common fraction School Handbook Workout What integer is tripled when nine is added to three-fourths of it? 2006 School Competition Target Round #7 6. A fitness center charges a membership fee plus a fixed amount per day for each day the center is used. If Claire paid $68 for 22 days of use and Adrian paid $71 for 24 days of use, how may dollars is the membership fee? 2006 School Competition Countdown Round #28 7. What number should be added to both the numerator and the dominator of 1/5 to get a fraction equivalent to 4/5? School Handbook Warm-Up If y = 2x + 1, which of the following equations is true? (A) x = 2y + 1 (B) x = (1/2)y + 1 (C) x = (1/2)y (1/2) (D) x = (1/2)y + (1/2) School Handbook Warm-Up This graph shows the linear relationship between the time in seconds, x, for Caroline s walk and the distance in meters, y, Caroline is from her starting point. The graph passes through the point (20, 30). According to the graph, how many meters will Caroline walk in exactly one hour? School Handbook Warm-Up Several points are plotted on a graph. For each point, the x-coordinate is the length of a side of a square while the y-coordinate is the perimeter of that same square. One such point is (2, 8) since a square with side length 2 units has a perimeter of 8 units. What is the slope of the line connecting the points? Express your answer in simplest form School Handbook Warm-Up MATHCOUNTS Club Resource Guide 45

2 11. According to the linear function represented in this table, what is the value of x when y = 8? School Handbook Warm-Up 7-10 x y How far apart are the y-intercepts of the line with the equation y = 2x + 3 and the line that goes through the point (4, 2) and has a slope of 1? School Handbook Warm-Up What is the least positive integer value of x for which the inequality 3x > 2x + 1 is true? School Handbook Warm-Up If x y = 6 and x + y = 12, what is the value of y? 2006 School Competition Countdown Round #6 15. For what value of x will (3 + x) (5 + x) = (1 + x) (2 + x) be true? 2007 School Competition Target Round #4 16. Milton spilled some ink on his homework paper. He can t read the coefficient of x, but he knows that the equation has two distinct negative, integer solutions. What is the sum of all of the distinct possible integers x that could be under the ink stain? School Handbook Warm-Up x + 36 = In the equation x 7 3 = 2, what is the product of all possible values of x? School Handbook Warm-Up 6-8 Answers: 8; 4 marbles; 12; 4/3; 4; $35; 15; C; 5400 meters; 4; 21; 3 units; 2; 3; 1; 85; 48 Possible Next Steps Problem #10 has students examine the relationship between the side length of a square and its perimeter. Based on their solution to this problem, can students predict what the slope of the line would be if they were to graph the relationship between the side length of an equilateral triangle and its perimeter? Similarly, students can graph the relationship between the side length of a square and its corresponding area to see that these relationships are not always linear MATHCOUNTS Club Resource Guide

3 Algebra Student Sheet 1. Maya initially thought that 8 was a solution to 2x + 20 = 4x 4, but when she checked her work, she realized that her answer was incorrect. After substituting 8 for x, how much greater is the expression on the left side of the equals sign than the expression on the right side? 2. marbles Three bags each contain the same number of marbles. (The weight of an empty bag would not affect the weight on the scale.) If three bags of marbles plus two more marbles balance 14 marbles, as shown, how many marbles are in each bag? 3. The function f (x) is defined by f (x) = x 2 x. What is the value of f (4)? 4. What is the value of x such that (x, 0) is a solution of the equation y = 3x 4? Express your answer as a common fraction. 5. What integer is tripled when nine is added to three-fourths of it? 6. $ A fitness center charges a membership fee plus a fixed amount per day for each day the center is used. If Claire paid $68 for 22 days of use and Adrian paid $71 for 24 days of use, how many dollars is the membership fee? 7. What number should be added to both the numerator and the dominator of 1/5 to get a fraction equivalent to 4/5? 8. If y = 2x + 1, which of the following equations is true? (A) x = 2y + 1 (B) x = (1/2)y + 1 (C) x = (1/2)y (1/2) (D) x = (1/2)y + (1/2) Copyright MATHCOUNTS, Inc MATHCOUNTS Club Resource Guide Problem Set

4 9. meters This graph shows the linear relationship between the time in seconds, x, for Caroline s walk and the distance in meters, y, Caroline is from her starting point. The graph passes through the point (20, 30). According to the graph, how many meters will Caroline walk in exactly one hour? 10. Several points are plotted on a graph. For each point, the x- coordinate is the length of a side of a square while the y-coordinate is the perimeter of that same square. One such point is (2, 8) since a square with side length 2 units has a perimeter of 8 units. What is the slope of the line connecting the points? Express your answer in simplest form. 11. According to the linear function represented in this table, what is the value of x when y = 8? x y units How far apart are the y-intercepts of the line with the equation y = 2x + 3 and the line that goes through the point (4, 2) and has a slope of 1? 13. What is the least positive integer value of x for which the inequality 3x > 2x + 1 is true? 14. If x y = 6 and x + y = 12, what is the value of y? 15. For what value of x will (3 + x) (5 + x) = (1 + x) (2 + x) be true? 16. Milton spilled some ink on his homework paper. x 2 + x + 36 = 0 He can t read the coefficient of x, but he knows that the equation has two distinct negative, integer solutions. What is the sum of all of the distinct possible integers that could be under the ink stain? 17. In the equation x 7 3 = 2, what is the product of all possible values of x? Copyright MATHCOUNTS, Inc MATHCOUNTS Club Resource Guide Problem Set

5 Algebra Meeting Solutions ( MCP Club Resource Guide) Problem 1. When Maya substitutes her 8 in for x, she finds that the left side of the equation has a value of 2(8) + 20 = 36 and the right side of the equation has a value of 4(8) 4 = 28. The expression on the left of her original equation is then = 8 greater than the expression on the right side. Problem 2. Since the 2 loose marbles on the left balance with 2 marbles on the right, we know that the contents of the three plastic bags must balance with the other 12 marbles on the right. If three equal bags have a total of 12 marbles, then each bag must have 12 3 = 4 marbles. Problem 3. When 4 is plugged in for x, we see f(4) = Simplifying the expression on the right gives us 16 4 = 12. Problem 4. If (x, 0) is a solution of the equation y = 3x 4, then we can solve the equation 0 = 3x 4 to find the value of x. Adding 4 to both sides of the equation, we get 4 = 3x. Dividing both sides by 3, we get 4/3 = x, or x = 4/3. Since the point (x, 0) is on the x-axis, we also could use a graphing calculator to see where the graph of y = 3x + 4 intersects the x-axis or where the lines y = 3x + 4 and y = 0 intersect. Problem 5. Translating the English to algebra and letting our unknown integer be x, we have 3x = (3/4)x + 9. Multiplying both sides of the equation by 4 gives us 12x = 3x Subtracting 3x from both sides (9x = 36) and then dividing by 9 results in x = 4. Problem 6. Adrian s total fee was $71 $68 = $3 more than Claire s and was for 2 additional days. This tells us each day s fee is $3 2 = $1.50. For Claire s 22 days, she would have paid 22 $1.50 = $33 for her daily fees and $68 $33 = $35 for the membership fee. Problem 7. Translating from English to algebra and letting our unknown number be x, we have (1 + x)/(5 + x) = 4/5. When cross products are set equal, this simplifies to 5(1 + x) = 4(5 + x) or 5 + 5x = x. Subtracting 4x and 5 from both sides gives us x = 15. Problem 8. The equation y = 2x + 1 is to be solved for x in terms of y. Subtracting 1 from both sides of the equation, we get y 1 = 2x. Now we can divide both sides of the equation by 2, which gives (y 1)/2 = x. Distributing this division by 2, we can rewrite this as x = (1/2)y 1/2, which is choice C. Problem 9. The graph shows a linear relationship, which tells us we can use proportional reasoning for this problem. We want to know how far Caroline can walk in one hour. The glitch is that we are given information in seconds and looking for information in hours. If Caroline walks 30 meters in 20 seconds she will walk 30 meters 3 = 90 meters in 20 seconds 3 = 60 seconds or 1 minute. One hour is 60 minutes, so Caroline will walk 90 meters 60 = 5400 meters in 1 minute 60 = 60 minutes or 1 hour. If we want to use the graph, we could use the fact that 1 hour is 3600 seconds and y = 1.5(3600) = 5400 meters. Problem 10. The perimeter of a square is always four times its side length, so the equation of the line is y = 4x. If we add one unit to the side length of a square, then the perimeter will increase by four units. The slope of the line is 4. We also could choose two points on the line and apply the slope formula. We know (2, 8) is on the line, and because a square with a side Copyright MATHCOUNTS, Inc MATHCOUNTS Club Resource Guide Solution Set

6 length of 1 unit has a perimeter of 4 units, the point (1, 4) is on the line. Using the slope formula, we get (8 4)/(2 1) = 4/1 = 4. Problem 11. The x-values increase by 5 while the y-values decrease by 3. If we write the next three ordered pairs in the table, we get (11, 14), then (16, 11), then (21, 8). We have stumbled onto the desired y-value of 8, and the corresponding x-value is 21. Problem 12. The y-intercept of the line with the equation y = 2x + 3 is said to be 3, which means more specifically the point (0, 3). The slope of the other line is 1, which means that as we move 1 unit to the right the line drops down 1 unit. If we start at the point (4, 2) and move to the left 1 units, the line goes up 1 unit. If we go four units left, the line goes four units up to the point (0, 6), which means that the y-intercept of this line is 6. These two y-intercepts are 6 3 = 3 units apart. Problem 13. We start with the inequality 3x > 2x + 1. If we subtract 2x from both sides of the inequality, we have x > 1. The least integer that is greater than 1 is 2. Problem 14. If we add the expressions on the left sides of the two equations together, they will equal the sum of the expressions on the right sides of the two equations. This gets us to 2x = 18. Now we see x = 9. Substituting into the first equation, we have 9 y = 6, so y = 3. Problem 15. The first step is to set the cross products of the equation equal to each other. This gives us (3 + x)(2 + x) = (1 + x)(5 + x) and simplifies to x 2 + 5x + 6 = x 2 + 6x + 5. Subtracting x 2 from both sides gives us 5x + 6 = 6x + 5. Now we subtract 5x and 5 from each side to see x = 1. Problem 16. The fact that the equation has two distinct negative, integer solutions means that it can be factored into the product of two binomials of the form (x + a)(x + b), with integers a and b. This in turn means that the coefficient under the ink stain must be a sum of two integers whose product is equal to the 36 that we can see. The possibilities are = 37, = 20, = 15 and = 13. The possibility of = 12 would not give us two distinct solutions. The sum of these distinct possibilities is = 85. Problem 17. Adding 3 to both sides of the equation, we get the simpler x 7 = 1. The value of x 7 could be either 1 or 1, so x = 8 or 6. The product of these two values is 48. Copyright MATHCOUNTS, Inc MATHCOUNTS Club Resource Guide Solution Set