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Carlisle Math Team
Meet #1 – Category 5
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Self-study Packet
1. Mystery: ?
2. Geometry: Angle measures in plane figures incuding
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
For current schedule or information,
See http://www.imlem.org
Meet #1, Algebra
1
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Meet #1 – Algebra
Words or ideas you should know for this topic:
The Variable or Unknown: A letter that “stands for” a number. For example,
X+2 = 5, what is X? X=3. We usually use X, Y, or Z, but we can use □, D,
or A5 too. A first step in many word problems is to pick something you don’t
know (“The Unknown”) and call it X and then write equations as if you knew
what it was. Example: “Bob is twice as old as Alice.” Ok, Alice’s age is the
unknown, so call it A, and call Bob’s age B. Now you can write B=2*A.
Implied multiplication:
3X means 3 times X, just like “3 Apples” means Apple+Apple+Apple.
2(X+1) means 2 times (X+1).
(1)(2)(3) means 1*2*3 = 6.
3AB means 3*A*B.
We also sometimes use · or * to mean times: 3·4 = 3*4 = 3(4) = 12.
Watch out for 2½ – This usually still means 2.5 and not 2 · ½ !
Operations on an Equation: Example: 2X = 12, divide each side by 2: X = 6.
You can always do the same thing to both sides of an equation and it’s still
an equation. This is a key idea for solving any equation. Example:
2X+7= 13
Subtract 7 from each side:
2X = 13-7 = 6, or 2X=6
Now divide each side by 2:
X
= 3.
Identity: Something that is always true. For example, 1=1 is true
regardless of what value for X you assume. However, 2X=6 is only true if
X=3 so it is not an identity. Example: What value of N makes this an
identity? 3(X+1) = 3X + N. If N=3, this is always true for any X, so N=3
makes it an identity.
Distribute is when you “multiply out” a value outside
parentheses. For example, 2(X+3) = 2X + 6. You distributed
the 2. Another example: 3 rooms, each containing an apple
and two parakeets is the same as 3 apples and 6 parakeets, or
3(A+2P) = 3A+6P.
Watch out on distributing a negative number! For example,
3G – 2(G – 1) … Here you must distribute a negative 2:
3G – 2G + 2 (-2 · -1 = +2)
Collect like terms or Combine like terms: Put your apples in one bag, and
your oranges in another. 3X+2+1X+3+1X=10 looks messy, but if you
combine all the X terms you get 3X+1X+1X+2+3=10 or 5X+5=10. Can you
solve it now? X=1.
Meet #1, Algebra
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Simplify the expression: (2X+4)/2 is simpler when written X+2. “Is there
any way to rearrange this so it’s not as messy?” Example:
Simplify 2(X+1)-2X.
First distribute the 2: 2X+2-2X. Now collect like terms: 2X-2X+2 or 2.
Solve for is when you put one variable on one side of an equation all by itself
and everything else is on the other side. You use the Operations on an
Equation to do this. You can often do this by 1) distribute; 2) combine like
terms; 3) use Operations on an Equation to isolate the unknown variable.
If a question just says “Solve this equation” then you should find the one
unknown (such as X) and solve for that.
Evaluate the expression if x=3: This means to replace every x with 3 and
see what you get. For example, evaluate 10X+7(2X+1) if X=3. You can
“plug in” or replace the X’s with 3: 10·3+7(2·3+1) = 30+7(7) = 30+49=79
Rodin’s The Thinker
“If A#B means …” problems
We’re used to plus, minus, times, and divided by, but what the heck is # or
@ or other funny symbols between two numbers? These problems invent an
“operation” just for the problem. For example, if A#B means 2A+B, then
you can do 3#1 by putting in 3 for A and 1 for B: 2(3)+1 = 7. As with
PEMDAS, you should do parentheses first and then left-to-right:
If A#B means (for today only) 2A+3B, what is 3#(1#2) ?
Answer: 1#2 = 2·1+3x2=8. 3#8 = 2
Adding/subtracting fractions
Many times the problem “looks” like algebra but is really mostly just a
fraction problem. You remember that to add two fractions you must put
each over a common denominator.
What is X/2+X/3 if X=1?
Answer: 1/2+1/3 = 3/6+2/6 = 5/6. We wrote 1/2 as 3/6 so we could add.
Meet #1, Algebra
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“Find the value of W that makes this an identity”
All of these problems can be solved by
1) Simplify (perhaps by distributing something first);
2) Solve for W.
Here’s an example from 1999’s Meet #1:
Find the value of F so that the following equation is an identity,
where any value of N produces a true statement:
-5 – 2(5 – 4N) = 5N + 3(N – F)
To solve this, distribute the -2 and +3 and collect like terms:
-5 – 10 + 8N = 5N + 3N -3F
-15
+ 8N = 8N – 3F
(and subtract 8N from each side)
-15
= -3F
Now divide each side of the equation by -3:
-15/-3
= -3F/-3
and we get 5 = F
So, the answer is 5.
Square Root
This is the opposite of square. For example, 32 means 3·3 or 9. So we can
say that √9 “the square root of 9” must be 3. What is √49? It’s 7, because
7·7=49. If you know your perfect squares, then square roots are easy: 1, 4,
9, 16, 25, 36, 49, 64, 81, 100, 121, … These have square roots of 1,2,3,…
The “team round” often has a big scary square root such as
A B 
C
E
D
if
A  11, B  30, C  20, D  4, E  3
The big square root will usually have a nice whole number answer since we’re not using a
calculator. Just put in the numbers and see what perfect square you get. Answer: √49 = 7.
Meet #1, Algebra
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Solutions are on the next page.
Hints were not part of the original meet.
Category 5
Algebra
Meet #1, October 2004
1. Find the value of C so that the equation below is an identity.
132x  5  7x  C  4 9x  8  3x
Hint: Distribute the multiplication into the parentheses, collect all the x terms, get C on one side by itself.

2. Evaluate the expression below for x 
5
7
and y  .
8
12
18x  9y  (4 x 6y)

Hint: First rewrite without parentheses (distribute the -1 carefully!) Then collect like terms and simplify. Then put
in the x and y values.

3. In a certain triathlon, the competitors run five times as far as they swim and
they bicycle four times as far as they run. If the total distance traveled in the race
is 52 miles, how many miles do the competitors run?
Answers
1. _______________
2. _______________
3. _______________
Meet #1, Algebra
Hint: Which event is the shortest ? Call that distance X. What is the distance of
each event as a multiple of X? Now do the total distance as a multiple of X.
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Solutions to Category 5
Algebra
Meet #1, October 2004
Answers
1. -97
1. First we distribute and combine like terms on both
sides of the equation as follows:
132x  5  7x  C  4 9x  8  3x
26x  65  7x  C  36x  32  3x
2. 0
3. 10
33x  65  C  33x  32
Notice that we have 33x on both sides of the equation.
We can leave it or we can subtract it from both sides. The
point is that this equation will an identity (true for all
values of x) if we findthe value of C that makes 65 + C
equal to –32. Solving for C, we get:
65  C  32
C  97
2. Substituting the values given into the equation, we get
18x  9y  4 x  6y  18x  9y  4 x 6y
 14 x 15 y
5
7
 14  15 
8
12
5
7
 7 5
4
4
35 35


4
4
0
3. Let x be the distance in miles that the competitors swim. If they run five times
 5x miles. If they bicycle four times as far as they
as far as they swim, then they run
run, then they bicycle 4(5x) = 20x miles. The total distance the competitors run,
swim, and bicycle is 52 miles, which gives us the equation: 5x + x + 20x = 52.
Combining like terms, we get 26x = 52, so x = 2. This means that the competitors
swim 2 miles, so they must run 2  5 = 10 miles.
Meet #1, Algebra
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Category 5
Algebra
Meet #1, October 2003
1. If AB means A 2  5B , then find the value of 11 79.



Hint: Remember to do inside the parentheses first, making A be 7 and 9 be B: 7 2 – 5(9)
2. Find the value of K so that the equation below is an identity. (An identity is an
equation that is true for all real values of x.)
3x  52x  7  K  12x  38  7x  2x

Hint: Distribute, collect like terms, simplify, solve for K.
3. If 37  4 x  3  4 3x  2, y  24  5x , and z  4y  8 , find the value of z.


Answers
1. _______________
2. _______________
3. _______________
Meet #1, Algebra

Hint: Solve the first equation for X. You’ll need to distribute and collect like
terms. You should have a nice simple answer for x.
Then you can do the second equation too and get y. Keep going.
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Solutions to Category 5
Algebra
Meet #1, October 2003
Answers
1. 101
2. 59
3. 44
1. Evaluating 79 first, according to the rule, we get
7 2  5 9  49  45  4 . Next we evaluate 114 and get
112  5  4  121  20  101


of
2. Simplifying and combining like terms on each side

the equation, we get
3x  52x  7  K  12x  38  7x  2x
3x 10x  35  K  12x  24  21x  2x
7x  35  K  7x  24
Adding 7x to both sides of the equation, we get:
35  K  24
Adding 35 to both sides, we find that K = 59.


3. First we must solve the first equation
for x.
37  4 x  3  4 3x  2
37  4 x  3  12x  8
37  16x 11
48  16x
x3
Substituting 3 for x in the second equation, we get:
y  24  5  3
y  24
15
y9
Finally, we substitute 9 for y in the third equation and
get:
z  4 9  8
z  36  8
z  44
Meet #1, Algebra
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
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Category 5
Algebra
Meet #1, October, 2002
1. Simplify the expression below.
5x  43x  5 73x  5 93x  5 4x 13
Hint: Rewrite 4(3X+5) by distributing the 4: 4*3X + 4*5 = 12X+20.
Be careful when distributing the -9, your neighbor may distribute +9 by mistake.
2. Evaluate the expression below for x 
3
5
and y  .
4
6
122x  y 32x  y
Hint: Rewrite it with the x and y values replaced by their fractions.
To do 2*3/4 + 5/6 you’ll need a common denominator such as 12ths.
3. Find the value of N that will make the equation below an identity.
(An identity is an equation for which all real numbers are solutions.)
9x  3x  2 23x  8 2N
Answers
1. _______________
2. _______________
3. N = ___________
Meet #1, Algebra
Hint: Do the distributed multiplications such as -3(X+2) = -3X-6
Then try to solve for N, which means get N on one side and everything else on
the other side.
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Solutions to Category 5
Algebra
Meet #1, October, 2002
Answers
1. 7x – 3
2. 26
3. N = 11
1. To simplify the expression, we could use the
distributive property on each set of parentheses as
follows:
5x  43x  5 73x  5 93x  5 4x 13
= 5x  12x  20 21x  35 27x  45 4x 13
Then we combine like terms and get the simplified
expression 7x – 3. Alternatively, we might notice that all
three sets of parenthese contain the same 3x + 5. Thus
there are 4 + 7 – 9 = 2 of these, so the original expression
can be rewritten as: 5x  23x  5 4x 13.
This simplifies to: 5x  6x  10 4x  13 = 7x – 3.
2. Replacing the x’s and the y’s in the expression with the values given, we get:
 3 5  3 5
 9 10  9 10




12
2


3
2


12
 4 6  4 6
2 12  12
  3
2  


 


  12 12
18 10 18 10
28  8 




 12


3


12
12 12 12 12
12
  3
 
  28 2 

 

  12
26.
3. First, we can distribute and combine like terms: 9x  3x  2 23x  8 2N
9x  3x  6  6x  16 2N
6x  6  6x 16 2N
With 6x on each side of the equation, we are guaranteed the same amount on each
side for all real values of x. We only need to find the value of N that will make
everything else equal so the equation will be an identity.
6  16 2N
2N  6  16
2N  22
N = 11
Meet #1, Algebra
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Category 5
Algebra
Meet #1, October, 2001
1. Simplify:
6 2 x  1  3x  5 x  4  43x  7  18
4
Hint: First get rid of the parentheses such as 6(2x+1) = 12x+6
Be very careful about -4(3x+7)… distribute the -4 and not +4.
2. Find the value of z that makes the following equation an identity:
14 x  8  23x  5  2 x  3   z  6 x 
Hint: Simplify by distributing the 2( … ) and then collecting terms with x.
What value of z makes this always true for any x you pick?
3. Cindy plans to sell her car and she wants to get $8000 for it. She knows that if a buyer makes
an offer of a and she makes a counter-offer of b and if they negotiate back and forth, going
exactly half way each time, the final price they settle on will be
a  2b
. If someone offers her
3
$6000, what should her counter-offer be, so that they settle on the $8000 that she wants?
Answers
1. _______________
2. _______________
3. _______________
Meet #1, Algebra
Hint: The final price is $8000, which is (a+2b)/3. The initial offer is $6000
which is a. Her first counter-offer is b. What is b?
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Solutions to Category 5
Algebra
Meet #1, October, 2001
Answers
1. 2x  6
2. z  5
3. $9000
1. To simplify the expression, we must first use the
distributive property to eliminate the parentheses as
follows:
6 2 x  1  3x  5 x  4  43x  7  18
4
12 x  6  3x  5x  20  12 x  28  18
4
Now, combining like terms, we get:
8 x  24 8 x 24


 2x  6
4
4
4
2. First we simplify each side of the equation:
14 x  8  23x  5  2 x  3   z  6 x 
14x  8  6x  10  2x  3  z  6x
8x  2  8x  3  z
Since 8x  8x for any value of x, this equation will be
an identity when  2  3  z , which happens only when
z  5.
3. We must solve the following equation for b, the
counter offer:
6000  2b
 8000
3
Multiplying both sides by 3, we get:
6000  2b  24,000
2b  18,000
b  9000
Cindy should make a counter offer of $9000.
Meet #1, Algebra
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Category 5
Algebra
Meet #1, October, 2000
3 A2  B 3
, then find the value of 5   7  3 . Express your answer as a
20
mixed number in simplest form.
1. If A  B means
Hint: Start inside the parentheses – do the 73 first. You should get a nice
whole number.
2. Find the value of C so that the equation below will be an identity.
(An identity is an equation in which any value of the variable will make the equation a true
statement.)
27x  83x  5  4 7x  C  55x  4
Hint: Multiply out and collect the X terms such as27x-24x…
And then solve for C.
3. Evaluate the expression if x 
2
.
7
513x  8  213x  8  413x  8  3x  32
Answers
1. _______________
2. _______________
3. _______________
Meet #1, Algebra
Hint: You can make it easier by noticing the repeating (13x-8) values – you have
5 of them, 2 of them, -4 of them. How many of them does that make?
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Solutions to Category 5
Algebra
Meet #1, October, 2000
Answers
1.  7
1
20
2.  5
3. 20
1. To find 5   7  3 , we must first compute 7  3 .
Substituting 7 for A and 3 for B, we obtain
3  7 2  33 3  49  27 147  27 120



 6.
20
20
20
20
Now we compute 5  6 :
3  52  63 3  25  216 75  216  141
1



 7
20
20
20
20
20
2. To solve this problem we first distribute (watching out
for negative signs), then combine like terms:
27x  83x  5  4 7x  C  55x  4
27 x  24 x  40  28 x  4C  25x  20
=
3x  40  3x  4C  20
=
Notice that we have 3x on both sides of the equation.
Now we have to make the rest of the equation balance;
we need to find C so that  40  4C  20 . Adding 20 to
both sides and dividing by 4, we find that C must be -5.
3. The observant algebra student will notice that the three
sets of parentheses contain the same 13x  8 . This
student would then simplify the expression before
substituting. There are 5  2  4 or 3 of those
13x  8 ’s.
513x  8  213x  8  413x  8  3x  32
=
313x  8  3x  32
Then we distribute and combine like terms.
39 x  24  3x  32
=
42 x  8
=
2
And finally we substitute
for x.
7
2
42   8  12  8  20
=
7
Meet #1, Algebra
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