Download Algebra Meeting (Multiple topics from pre-algebra and

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 www.mathcounts.org 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? 2006-2007 School Handbook Warm-Up 6-5
2. 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? 2005-2006 School
Handbook Warm-Up 2-3
3. 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. 2007-2008 School Handbook Workout 1-10
5. 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?
2005-2006 School Handbook Warm-Up 8-9
8. If y = 2x + 1, which of the following equations is true? (A) x = 2y + 1
(D) x = (1/2)y + (1/2) 2005-2006 School Handbook Warm-Up 7-9
(B) x = (1/2)y + 1
(C) x = (1/2)y – (1/2)
9. 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? 20062007 School Handbook Warm-Up 5-9
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.
2007-2008 School Handbook Warm-Up 4-9
2009–2010 MATHCOUNTS Club Resource Guide
45
11. According to the linear function represented in this table, what is the value of x when y = 8?
2004-2005 School Handbook Warm-Up 7-10
x
–4
1
6
y
23
20
17
12. 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? 2006-2007 School Handbook Warm-Up 12-2
13. What is the least positive integer value of x for which the inequality 3x > 2x + 1 is true? 2005-2006 School
Handbook Warm-Up 4-10
14. 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 2 + x + 36 = 0
that could be under the ink stain? 2004-2005 School Handbook Warm-Up 14-2
17. In the equation │x – 7│ – 3 = –2, what is the product of all possible values of x? 2006-2007 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.
46 2009–2010 MATHCOUNTS Club Resource Guide
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?
marbles Three bags each contain the
2. ___________
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. 2009. MATHCOUNTS Club Resource Guide Problem Set
meters This graph shows the linear relationship between the
9. ___________
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 xcoordinate 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
–4 23
1
20
6
17
units How far apart are the y-intercepts of the line with the equation
12. ___________
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. 2
x + 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. 2009. MATHCOUNTS Club Resource Guide Problem Set
Algebra Meeting Solutions
(2009-2010 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 36 – 28 = 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) = 42 – 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 + 36. 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 = 20 + 4x. 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. 2009. MATHCOUNTS Club Resource Guide Solution Set
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 x2 + 5x + 6 = x2 + 6x + 5. Subtracting x2
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 1 + 36 = 37, 2 + 18 = 20,
3 + 12 = 15 and 4 + 9 = 13. The possibility of 6 + 6 = 12 would not give us two distinct solutions.
The sum of these distinct possibilities is 37 + 20 + 15 + 13 = 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. 2009. MATHCOUNTS Club Resource Guide Solution Set