meet #3 algebra · 12 28 a b 7 + 3(2y - 5 ) < 52 category 5 algebra meet #3 - january, 2014 50th...
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Park Forest Math Team
Meet #3
AlgebraAlgebra
Self-study Packet
Problem Categories for this Meet: 1. Mystery: Problem solving
2. Geometry: Angle measures in plane figures including supplements and complements
3. Number Theory: Divisibility rules, factors, primes, composites
4. Arithmetic: Order of operations; mean, median, mode; rounding; statistics
5. Algebra: Simplifying and evaluating expressions; solving equations with 1 unknown including identities
Important information you need to know regarding ALGEBRA Absolute value; inequalities in one variable including interpreting line graphs
Absolute Value is the distance a number is from zero. Absolute value is never negative. The symbol for absolute value is
and
Inequalities
• To solve an inequality, solve as if it were a regular equation. Remember to switch the direction of the inequality sign only if you multiply or divide by a negative!
≥- 3 X + 6 -17 - 86 A B
Category 5 Algebra Meet #3 - January, 2016
1) What are the two values of N that make this absolute valueequation true?
2) The graph represents the solution set of the inequality
What is the value of A + B ?
3) Solve the following inequality for C:
5(C + 2) - 4(2C - 3) + 7(3C - 8) < - 6(0.5C - 9) - 64
Express your answer as a common fraction.
ANSWERS
1) ___ and ___
2) A + B = ____
3) C < ______
Solutions to Category 5 Algebra Meet #3 - January, 2016 Answers
1) - 5 ; 19(any order)
2) - 12
3)
1) Either N - 7 = 12 or N - 7 = - 12. So, either N = 19 or X = - 5.
2) original inequality having added 17 to both members having divided both members by - 3
So, A + B = - 29 + 17 = - 12
3) 5(C + 2) - 4(2C - 3) + 7(3C - 8) < -6(0.5C - 9) - 64 original inequality 5C + 10 - 8C + 12 + 21C - 56 < -3C + 54 - 64 distribute
18C - 34 < -3C - 10 combine terms 21C < 24 add 3C; add 34
C < 8/7 divide 21
12 28
A B
7 + 3(2Y - 5 ) < 527 + 3(2Y - 5 ) < 527 + 3(2Y - 5 ) < 527 + 3(2Y - 5 ) < 52
Category 5 Algebra
Meet #3 - January, 2014 50th anniversary edition50th anniversary edition50th anniversary edition50th anniversary edition
1) If C = 5 , then what is the sum of all values of C that make this
sentence true?
2) The graph below represents the set of all values of X that make
X - NX - NX - NX - N ≤ 8≤ 8≤ 8≤ 8 true. What is the value of N ?
X
3) If and
then the set of all possible values of Y is represented in the graph below:
What is the value of A + B ?
ANSWERS
1) ______
2) ______
3) ______
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1)
Answers
7 + 3(2Y - 5 ) < 527 + 3(2Y - 5 ) < 527 + 3(2Y - 5 ) < 527 + 3(2Y - 5 ) < 52
Solutions to Category 5 Algebra Meet #3 - January, 2014
Answers 1) The two solutions are 5 and -5.
Their sum is zero.1) 0
2) The absolute value sentence can be2) 20 translated as, "The distance between a
number, N, and all solutions is at most3) 15 8 units." N can be found by locating the
midpoint of 12 and 28, which is 20.
3) 7 + 6Y - 15 < 527 + 6Y - 15 < 527 + 6Y - 15 < 527 + 6Y - 15 < 52
6Y - 8 < 526Y - 8 < 526Y - 8 < 526Y - 8 < 526Y < 606Y < 606Y < 606Y < 60
Y < 10Y < 10Y < 10Y < 10
also, 7 - 3(2Y - 5 ) < -87 - 3(2Y - 5 ) < -87 - 3(2Y - 5 ) < -87 - 3(2Y - 5 ) < -87 - 6Y + 15 < -87 - 6Y + 15 < -87 - 6Y + 15 < -87 - 6Y + 15 < -8
22 - 6Y < -822 - 6Y < -822 - 6Y < -822 - 6Y < -8 - 6Y < -30- 6Y < -30- 6Y < -30- 6Y < -30
Y > 5Y > 5Y > 5Y > 5
Therefore, A = 5 and B = 10 and their sum, A + B, is 5 + 10 = 15.
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Meet #3 January 2012
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Category 5 – Algebra
1. How many integers do not satisfy the inequality below?
| |
2. Find the positive difference between the two solutions to the equation:
|
|
3. The graph below describes the solution to the inequality: | |
Find the value of
Answers
1. _______________
2. _______________
3. _______________
-2 0 2 4 6
Meet #3 January 2012
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Solutions to Category 5 – Algebra
1. Let’s solve the inequality: | |
If the argument is positive we get:
If the argument is negative we get: , so the solution to the
inequality is { } What integers do not fall in this range?
{ } - a total of integers.
2. In the positive case:
we get
or
In the negative case:
we get
or
The differnce between the two solutions is .
3. The graph depicts all the points on the number line whose distance from is no
more than . In other words, it is the visualization of: | | , which makes
.
The abosolute value function measures the distance between points on the line.
Answers
1.
2.
3.
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Category 5 - Algebra
Meet #3, January 2010
1. What is the positive difference between the least and largest integers that satisfy the
inequality below?
𝑥
5− 1 ≤ 2
2. What value of the parameter M in the inequality 𝑀 ∙ 𝑥 − 1 ≥ 3 will make the
solution agree with the line graph below?
Express your answer as a decimal.
3. For how many integers N (excluding zero) does the inequality below hold true?
30
𝑁 > 2 ∙ 𝑁
2 0 1 3 -1
Answers
1. _______________
2. _______________
3. _______________
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Solutions to Category 5 - Algebra
Meet #3, January 2010
1. 𝑥
5− 1 ≤ 2 can be written (multiply both sides by 5) as 𝑥 − 5 ≤ 10.
If the argument on the left is positive we get 𝑥 − 5 ≤ 10 𝑠𝑜 𝑥 ≤ 15. If the argument
is negative we get 𝑥 − 5 ≥ −10 𝑠𝑜 𝑥 ≥ −5. Taken together the range of values that
make the inequality true are: −5 ≤ 𝑥 ≤ 15.
The largest integer solution is 15, the least is −5, and the difference is therefore 20.
2. If the argument in the inequality 𝑀 ∙ 𝑥 − 1 ≥ 3 is positive then we can write:
𝑀 ∙ 𝑥 − 𝑀 ≥ 3 𝑂𝑟 𝑥 ≥3+𝑀
𝑀=
3
𝑀+ 1 and if the argument is negative we can write:
𝑀 − 𝑀 ∙ 𝑥 ≤ 3 𝑂𝑟 𝑥 ≤3−𝑀
−𝑀=
𝑀−3
𝑀= 1 −
3
𝑀
Comparing these to the graph we require that 1 +3
M= 3 AND 1 −
3
M= −1
The solution to both is 3
M= 2 Or M =
3
2= 1.5.
Alternatively, by looking at the graph solution we can ask ourselves: What inequality
is described by the graph? And the answer is that the graph depicts points outside the
region (-1, 3), or in other words points whose distance from the middle of the region is
at least 2 (half the size of the region). So we can conclude that the graph depicts the
inequality x − 1 ≥ 2 (1 being the midpoint of the region (-1, 3)).
Comparing this inequality to the one given we can again calculate that 𝑀 = 1.5 is the
correct value.
3. Since both sides of the inequality are positive let’s see what happens when N is
positive: 30
𝑁> 2 ∙ 𝑁 means 𝑁2 < 15 and so 𝑁 < 4 or in other words N can take on
the values {1, 2, 3}. If we assume N is negative, the same will apply (no need to
reverse the inequality, as both sides are positive due to the absolute value), so N can
take on the values {−1, −2, −3}. Overall there are 6 possible values for N.
Answers
1. 20
2. 1.5
3. 6
Category 5
Algebra
Meet #3, January 2008
1. What is the positive difference between the two solutions to this equation?
6 3 12x− =
2. For what value of K is the solution set of the inequality below given by the
graph below?
32�34 3 7� 5 6 3 24 6
- 4 0
3. For how many integer values of y is 18
2 y− a positive integer?
Answers
1. _______________
2. _______________
3. _______________
Solutions to Category 5
Algebra
Meet #3, January 2008
1. |8 3 9:| � �1
6 3 34 � 12 or 3 12
334 � 6 or 3 18
4 � 32 or 6
The positive difference between – 2 and 6 is 6 – (– 2) = 8
2. – 2�34 3 7� 5 > 3 24 6 364 14 5 > 3 24 6 344 8 5 >
This inequality becomes an equality at the border point 4 � 34
34�34� 8 � > � 16 8 � 1�
3. Since ? is an integer, 2 3 ? will be an integer. In order for H ����IH to be a
positive integer, 18 must be divisible by 2 3 ? but you don’t need to worry about
the positive part since the absolute value bars will take care of that. Basically we
want to know how many values of y make 2 3 ? a factor of 18 remembering that 2 3 ? J 0 KL ? J 2. So 2 3 ? � 18, 9, 6, 3, 2, 1, 31,32,33,36,39,318 which
means that ? � 316,37,34,31, 0, 1, 3, 4, 5, 8, 11, 20 for 12 integer values.
Answers
1. 8
2. 24
3. 12
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Category 5 Algebra Meet #3, January 2006 1. How many integer values of n satisfy the inequality below?
18n + 1
> 3
2. Find the sum of the two solutions to the equation below. Express your answer as a mixed number with the fraction part in lowest terms.
3x − 7 + 9 = 42 3. For what value of B does the solution to the inequality below match the graph below?
3 5x − 4( )− 3x + B > −21
Answers 1. _______________ 2. _______________ 3. _______________
-6 -4 -2 0 2 4 6 8 10 12 14
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Solutions to Category 5 Algebra Meet #3, January 2006
1. There are ten (10) integer values of n that satisfy the inequality. They are -6, -5, -4, -3, -2, (not –1), 0, 1, 2, 3, and 4.
For example, when n = -6, we get 18
−6 + 1= 18
−5= 18
5= 3
35
,
which is greater than 3.
2. First let’s subtract 9 from both sides of the equation.
3x − 7 = 33
The expression inside the absolute value bars could be either negative or positive, so we have to solve the two separate equations below.
3x − 7 = 333x = 40
x = 403
or 3x − 7 = −33
3x = −26
x = − 263
The sum of these two solutions is 403
+ − 263
� � �
� � �
= 143
= 4 23
.
3. Let’s simplify the inequality first. 3 5x − 4( )− 3x + B > −21
15x −12 − 3x + B > −2112x −12 + B > −21
12x > −21+ 12 − B
12x > −9 − B
x > −9 − B12
We know from the graph that x > 3. This means that −9 − B
12
must equal 3, or equivalently –9 – B = 36. Solving for B, we get B = –9 – 36 = –45. Alternatively, one could substitute 3 for x from the beginning and then solve for B.
Answers 1. 10
2. 423
3. –45