Study Guide MAT 102 Final
We're going to go topic-by-topic.
Rational Expressions
You're going to be performing some operations on rational expressions, which are
'fractions with polynomials'.
x2  3
x2  2 x  8
'Domain' means all of the x-values that you can plug into the function. What we have to
do is figure out all of the values that we can't plug in first. That's accomplished by setting
the denominator equal to 0, and solving:
x 2  2 x  8  0,
Ex (1): Find the domain of the rational function. f ( x) 
( x  4)( x  2)  0,
x  4  0, x  2  0,
x  4, x  2
So the domain is the set of all real numbers except x  4, x  2 .
Note that factoring is a significant skill, as it always is with rational expressions.
x2  5x  6 x2  x  2
 2
x2  4
x  4x  3
We need to factor all of they polynomials involved, and we also need to 'invert and
multiply by the second RE, then cancel all like factors:
x 2  5x  6 x 2  4 x  3 ( x  3)( x  2) ( x  2)( x  1)
=1
 2

*
x2  4
x  x  2 ( x  2)( x  2) ( x  3)( x  1)
Ex (2) : Divide:
x
3
 2
x  2 x  8 x  7 x  12
Find the common denominator by factoring the denominators:
( x  4)( x  2)
( x  4)( x  3) ,
LCD = (x – 4)(x + 2)(x – 3), convert to LCD:
( x  3)
x
3
( x  2)


( x  3) ( x  4)( x  2) ( x  4)( x  3) ( x  2)
Ex (3): Subtract:
2
x 2  3x  6
( x  3)( x  4)( x  2)
1 4
 3
x
x
Ex (4): Simplify the complex rational expression:
3 2

x2 x
2
2 x2
3   2  3x 2 
2
x x 
x  3x  2
Multiply by the LCD x2 on top and bottom:
2
1  2
4 x2  1
4  2  x  4x2  x2
x
x
2
3
5

 2
x  1 x  4 x  3x  4
We have to clear the fractions by multiplying through by the LCD:

3  
5
 2
( x  1)( x  4) 

 ( x  1)( x  4),

 x  1 x  4   ( x  4)( x  1) 
Ex (5): Solve the rational equation:
2( x  4)  3( x  1)  5,
5 x  5  5,
5 x  10,
x  2.
Ex(6):
Solving Linear Systems:
These have two equations with two variables. The main solution methods are substitution
and elimination('addition'):
Ex (7): Solve the system by substitution:
x  2y  7
2 x  5 y  13
x  2y  7
2 x  5 y  13
x  2 y  7,
(The x in the first equation has coefficient 1 – solve for it)
 2y  2y
x = 2y + 7
(substitute this into the other equation)
2(2y + 7) + 5y = -13,
4y + 14 + 5y = -13,
-14
-13
9y = -27,
27
y
, y  3
(back-solve for x: )
9
x – 2(-3) = 7,
-6
-6
x = 1.
The solution is the ordered pair (1,-3).
Ex (8): Solve the system by addition:
3 x  5 y  6,
5 x  4 y  27
You can eliminate either variable here. Either way, you’re going to have to multiply both
equations. I picked y to eliminate, as the coefficients have opposite signs already. The
coefficients of y both ‘go into’ 20, so make one coefficient -20, and the other 20:
4(3 x  5 y  6),
5(5 x  4 y  27)
12 x  20 y  24,
25 x  20 y  135
37 x
 111,
111
 3,
37
3(3)  5 y  6
9
9
x
(backsolve for y )
15
 3,
5
The solution is (3,3).
5 y  15, y 
Ex (9):
Solution:
Solving Quadratic Equations: We have different methods that we can use here –
factoring, the square root principle, completing the square, factoring, the quadratic
formula. Let's review these:
Ex (10): Solve by extracting the square roots: x 2  25
Take the square root of both sides of the equation:
x   25,
x   25i
x  5i
Ex(11): Solve by factoring: x 2  48  16 x
First, we need to get standard form by subtracting 16x from both sides:
x 2  16 x  48  0, (factor:)
( x  12)( x  4)  0, (factors = 0:)
x  12  0, x  4,
x  12, x  4, x  {4,12}
Ex(12): Solve using the quadratic formula: x 2  2 x  6  0,
a = 1, b = 2, c = 6, and
2  (2)2  4(1)(6) 2  20 2  5 4i
x


2(1)
2
2
2  2 5i
2
Ex(13): Solve the quadratic equation:
You could use the quadratic formula (feel free) or factoring, starting with the gcf = 4:
4 x 2  12 x  40 x  0

4( x 2  3 x  10)  0
4( x  4)( x  2)  0,
x  4  0, x  2  0,
x  {4, 2}
Radicals Let's look at some typical problems:
Ex(14):
Solution:
3
Ex(15): Simplify: (196 x12 y 6 ) 2
Everything in the parentheses goes to the power outside:
3
3
12*
(196 x12 y 6 ) 2  196 2 x
3
2
y
6*
3
2
 ( 196)3 x18 y 9  143 x18 y 9  2744 x18 y 9
m
n
note that with fractional exponents, x  n x m , the top m is the power, the bottom n is
the root.
Ex(16): Simplify: 162x 7 y10 .
7
10
3
1
Solution: 162 x7 y10  162 x 2 y 2  81 2 x 2 y 5  9 x3 y 5 2 x
Note that the whole number part of the exponent comes outside, the numerator stays
inside.
Ex(17): Simplify 3 x 2 y 7 3 x 6 y 6
First, multiply the radicands, then simplify using the fractional exponent:
3
2
x y
7 3
8
13

1
8 13 3
x y  x y  x y
6
6
2
2
 x3 y 3  x 3 y
3
4
1
3
8 13

 x2 y 4 3 x2 y
Ex(18): Simplify, then rationalize the denominator:
49
12
a
a
We can use the radical rule:
,

b
b
49
49
7
7
. Now rationalize the denominator:



12
12
4 3 2 3
7
3 7 3 7 3


6
2 3 3 2(3)
Ex(19): Solve the rational equation: 5 x  4  6  3
First, isolate the radical by adding 6 to both sides of the equation:
5x  4  9
then square both sides to eliminate the radical:
( 5 x  4) 2  92 ,
5 x  4  81,
4 4
5 x  85,
x  17
Ex(20): Perform the radical operation: ( 5  6)( 7  4) .
Use FOIL: ( 5  6)( 7  4)  7 *5  4 5  6 7  24  35  4 5  6 7  24
Ex(21): Perform the radical operation. 3  4 27  3 12
Pull out the roots first:
3  4 27  3 12  3  4 9 3  3 4 3
 3  4(3) 3  (3)(2) 3  3  12 3  6 3
 19 3
Absolute Value: You should be able to solve both absolute value equations, as well as
absolute inequalities.
Ex(22): Solve the equation: 2 | x  1| 3  7
2 | x  1| 3  7
+3 +3
2 | x  1| 10
(first, isolate the absolute value:)
2
10
| x  1| ,
2
2
| x  1| 5
x - 1 = 5, x - 1 = -5
+1 +1
+1 +1
x = 6, x = -4.
(now use the absolute value property:)
Ex(23): Solve the absolute inequality, and give the solution in interval notation:
x 5  3
Here the operation is greater than (>), and therefore we have to 'split' the inequality into
separate ones:
(> - 'split' the inequality like: )
x 5  3
x  5  3 or x  5  3,
5 5
5 5
x  2 or x  8,
Interval notation: (, 2) U (8, )
Ex(24): 2 x  1  3  4
We have to keep the special rules for absolute inequalities in mind:
2 x  1  3  4 (first, isolate the absolute value:)
2x 1  3  4
(now, with operation < we 'wedge' the inequality between 3, -3:)
3 3
| 2 x  1| 7,
7  2 x  1  7
(solve the compound inequality:)
1
1 1
6  2 x  8
6 2x 8
 
 ,
2 2 2
3  x  4.
Interval notation [-3,4]
The Vertex of the Quadratic Equation/Function You will, on the exam, be provided with
 b
 b 
the formula v    , y     . You may be asked to use it in a variety of contexts.
 2a  2a  
Some of you have struggle with this mightily, but let's take another crack ('K?)
Ex(25): Find the vertex of the parabola y  3x 2  12 x  16

 b
 b    12
v    , y     
, y (2)   2,3(2) 2  12(2)  16   2, 4 
 2a  2a    2(3)



Ex (26):
Solution:
Ex(27): Is the vertex of this function the maximum or minimum value? Find the vertex.
y  ( x  3)2  4
First of all, the negative lead coefficient means the parabola points down, which means
the vertex is a maximum. When the function is written in this form, it is easy to detect the
vertex, using the 'translates' like we did early in the course. The horizontal translate is the
number we are subtracting from x inside of the parentheses, 3. The vertical translate is the
number we are adding outside of the parentheses, 4. The vertex is (3,4).
Complex Number Operations: 'treat i as a variable':
Ex(28): Multiply: (3  4i )(5  7i)
Use foil:
(3  4i )(5  7i)
15  21i  20i  28i 2
 28(1)
43  i
Ex(29): Divide:
3  4i
5  7i
Multiply by the conjugate of the denominator on top and bottom:
3  4i (5  7i) 15  21i  20i  28i 2 15  41i  28 13  41i
13 41



  i
2
2
5  7i (5  7i)
5 7
25  49
74
14 74