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