Download Shady Side Academy Middle School Review Packet for students

Shady Side Academy
Middle School
Review Packet
for students entering Geometry
* Some work taken from Geometry, by Jurgensen, Brown & Jurgensen, Houghton Mifflin Company, 1997
DIRECTIONS: Read through the information and study the given examples.
Complete all boxed exercises on notebook paper using proper form.
SECTION I: LINEAR EQUATIONS & PROPERTIES OF EQUALITY
A. Solving Linear Equations
To solve a linear equation, use the operations that are the opposite of the ones in the equation (inverse
operations).
Examples
1.
Solve to find the value of the variable. Show proper algebraic transformations.
x + 7 = 12
2.
3m + (m – 2) = 10
3.
4t + 23 = 9t – 7
x + 7 – 7 = 12 – 7
3m + m + (–2) = 10
4t + 23 + 7 = 9t – 7 + 7
x=5
4m + (–2) = 10
4t + 30 = 9t
4m + (–2) + 2 = 10 + 2
4t + (–4t) + 30 = 9t + (–4t)
4m = 12
30 = 5t
4m 12

4
4
30 5t

5
5
m=3
6=t
Exercises – Find the value of the variable in the given equation.
1.
(n – 20) + 5n = 28
3
a =9
4
2.
3.
2(h + 5) = 3(h – 2)
B. Properties of Equality for all real numbers
Addition Property
If a = b and c = d, than a + c = b + d
Multiplication Property
If a = b, then ca = cb
Substitution Property
If a = b, then either a or b may be substituted for the other in any equation.
Reflexive Property
a=a
Symmetric Property
If a = b, then b = a
Transitive Property
If a = b and b = c, then a = c
Exercises – Name the property that justifies the algebraic transformation shown.
1.
4b – 5 = –2
2.
15f + 7 = 12 – 20 f
4b = 3____________
35f + 7 = 12____________
b = 0.75__________
35f = 5 ________________
f=
1
_______________
7
3.
If k  5 = 10, then k = 50
_____________________
C. Systems of Linear Equations
Example 1
Solve the given system by the Substitution Method:
y = 5 – 2x and 5x – 6y = 21
1.
Substitute 5 – 2x for y in the second equation,
2.
Now substitute 3 for x
then solve for x:
in the other equation:
5x – 6(5 – 2x) = 21
5(3) – 6y = 21
5x  6(5) – 6(–2x) = 21
15 – 6y = 21
5x –30 + 12x = 21
–6y = 6
17x – 30 = 21
y = –1
 S = (3, –1)
17x = 51
x=3
Example 2
Solve the given system by the Linear Combination Method:
6x – 2y = 26 6x – 2y = 26
4x + y = 22
Multiply by 2 8x + 2y = 44
14x
= 70
x=5
Substitute 5 for x in one of the original equations:
4(5) + y = 22
20 + y = 22
y=2
 S = (5, 2)
Exercises – Solve each system of linear equations.
1. y = 3x
5x + y = 24
2. 3x + y = 19
2x – 5y = -10
3. 5x + 2y = 19
3x – 4y = 1
4. 3x + 2y = 71
2x – y = –4
SECTION II: OPERATIONS WITH NUMBERS
To show work in proper form, you must do the following:
1. Write the given expression.
2. Complete the operations within any symbols of inclusion – parentheses, brackets, vinculums (aka
fraction bars). Do the innermost operation and write the result. Use an = sign to connect the new
expression to the original one.
3. Keep doing the operations until you reduce the expression to a single number. Use = signs
to connect each expression to the one before (see examples below).
Example 1
Evaluate 24  8  2  2
24  8  2  2 = 24   4  2
= 24  2
= 48
Example 2
Evaluate  24  8  2  2
 24  8  2  2 =  24  8  2  2
= 192  2  2
= 96  2
= 94
Example 3
Evaluate
68
4
28
68
48
4= 4
28
10
 4 108 + 4
= 4 54  4
= 8 54
Exercises – Evaluate the following expressions showing work in proper form
.
1.
2.
14  6   9  2
14  6   9  2
4.
800  50  2  7
5.
19 
25  7
54
3.
800   50  2  7
6.
28 
8 6
57
SECTION III – VARIABLES AND SUBSTITUTION
A variable is a letter that represents a number. To evaluate an expression containing a variable, your must be
told what value to use for the variable, then you substitute that number for the variable and evaluate the
expression as before.
Example 1
Evaluate the expression by substituting the given values of the variable. x +7, if :
a) x is 8;
b) x is 93.
a) x + 7 = 8 + 7
b) x + 7 = 93 + 7
= 15
= 100
Example 2
Evaluate the expression by substituting the given values of the variable.
a)
40 40
=
x
5
b)
=8
40 40
=
x
80
40  40
=
80  40
1
=
2
40
, if:
x
a) x is 5;
b) x is 80;
c) x is 60.
40 40
=
x 60
40  20
=
60  20
2
=
3
c)
Exercises – Evaluate the expressions by substituting in the given values of the variable.
1.
 72  n   5 , if:
2.
a) n is 6;
b) n is 9;
c) n is 0.
 y  31  4 , if:
3.
7x, if:
a) y is 80;
b) y is 57;
c) y is 31.
4.
a) x is 8;
b) x is 11.
k7
, if:
k 3
a) k is 5;
b) k is 3;
c) k is -7.
SECTION IV – POWERS AND EXPONENTS
In the expression 45, the 4 is called the base, the 5 is called the exponent, and the whole expression is called a
power. To evaluate a power, you simply do the multiplication.
Example 1
Evaluate 53
3
Evaluate  
4
Example 2
5 = 555
 3   3
  =
2
 4   4
= 125
=
2
2
3
2
9
16
***IT IS IMPORTANT TO KNOW THE DIFFERENCE BETWEEN  3 AND 32 !!!
2
 3
2
means multiply –3 and –3 together to get +9.
32 means “the opposite of 32 ” which is –9.
 3
2
=  3  3 = +9
32 =   3 =   3  3 = –9
2
SECTION V – ORDER OF OPERATIONS
In order to evaluate a given expression, you must perform the given mathematical operations following the order
of operations. The order is:
1. Complete the operations within any symbols of inclusion – parentheses, brackets, vinculums (aka
‘fraction bars’).
2. Evaluate any exponents.
3. Divide and/or multiply, in order, from left to right.
4. Subtract and/or add, in order, from left to right.
Example 1
Evaluate 60  7  5  6  2   24
60  7  5  6  2   24 = 60  7  5  3  24 Inside the ( ), divide before adding.
Example 2
= 60  7 8  24
Do what is inside the ( ) first.
= 60  7  8  16
= 60  56  16
= 4  16
= 20
Evaluate the exponent.
Multiply before adding or subtracting.
Add & subtract in order from left to right.
Answer
Evaluate 3x 2  5x if x is 4.
3x 2  5x = 3   4   5  4
= 3  16  5  4
= 48  20
= 28
2
Substitute 4 for x in both places. Use ( ) with the exponent.
Raise to powers before multiplying.
Multiply before adding or subtracting.
Answer
Exercises – Evaluate the given expressions using order of operations.
1.
a) 5  2  4  3
b)
c)
4.
2.
5  2  4  3
5  2   4  3
30  10  5  3
a)
7  32
3.
b) 7  32
c)
5.
 7  3
8  5  43
x 2  2x  1 , if :
a) x is 6;
2
b) x is 4.
6.
74  4 3   9  4  52
SECTION VI – QUADRATIC EQUATIONS
All quadratic equations in standard form can be solved using the Quadratic Formula.
In other words, if ax2 + bx + c = 0, with a  0, then x 
b  b 2  4ac
.
2a
If the quadratic equation can be factored, it can be solved quickly that way.
Example 1
Example 2
3x2 + 14x + 8 = 0
3x2 + 14x + 8 = 0
(3x + 2)(x + 4) = 0
x
b  b 2  4ac
2a
 3x + 2 = 0 or x + 4 = 0
x
14  142  4(3)(8)
2(3)
x
14  196  96
14  10

6
6
x
2
or x = -4
3
x
a = 3, b = 14, c = 8
2
or x = -4
3
Exercises – Solve each quadratic equation. Factor where you can if you like.
1. 2m2 + m = 0
2. y2 + 8y + 12 = 0
3. 3x2 + 3x = 4
4. x(x + 5) = 14
SECTION VII – RATIONAL ALGEBRAIC EXPRESSIONS
The rules for fractions apply to all rational algebraic expressions:
1.
a c a+c
+ =
,b0
b b
b
Example 1
8w
=2w
4
2.
a c ac
• = , b  0, d  0
b d bd
Example 2
5t - 10 5(t - 2)

15
15
t-2

3
3.
a c a d
÷ = • , b  0, c  0, d  0
b d b c
Example 3
x+6
x+6

2
36 - x
(6 - x)(6+ x)
1
=
6-x
Exercises – Simplify the rational algebraic expressions below.
1.
-18r 3 t
=
12rt
2.
33ab - 22b

11b
5a + 5b
=
a 2 - b2
3.
4.
3x 2 - 6x - 24
=
3x 2 + 2x - 8
SECTION VIII – RADICAL ALGEBRAIC EXPRESSIONS
The symbol
always indicates the positive square root of a number. The radical
Example 1
Example 2
16
=
3
56 = 4 14
16
3
4
3

3 3
4 3

3
 4  14

 2 14
64 can be simplified.
Example 3
(3 7)2 = (3 7)(3 7)
 3 3 7  7
 9 7
 63
Exercises – Simplify the rational algebraic expressions below.
1.
98 
2.
250

48
3.
(9 2) 2 
4.
300  30 