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MAT 102 Practice Final
(1) Solve: 3(x - 4) = 4(x + 2) + 7
(2) Solve: 2 | x  1| 3  7
(3) Solve the system by graphing (graph the lines and find the point of interesection)
x  2y  9
2 x  y  2
(4) Solve: 2 x3  10 x 2  28 x.
(6) Solve the rational equation:
2
3
5

 2
x  1 x  4 x  3x  4
(7) Solve the linear system:
y = 3 + 4x,
x - 2y = -13
(8) Solve the linear system:
3x + 2y = 14,
x - 3y = -10
(9) Nate and Honey work in the college sports ticket office. Tickets to a baseball game
are $3 for students, and $6 for the general public. If the total take for the game was $330,
on sales of 70 tickets, how many students and how many members of the public attended
the game?
(10) Graph the system of inequalities:
x  2 y  4,
2x  3y  6
(11) Simplify the radical expression:
3
27x9
1
(12) Rewrite as a radical expression, and simplify if possible: ( x6 y8 ) 2
(13) Simplify the radical expression:
98w9 z 6
(14) Multiply: (4  3) 2
(15) Rationalize the denominator:
(16) Solve the radical equation:
52
7 9
y2  y  4
(18) Add the complex numbers: (3  4i)  (5  7i)
(19) Multiply the complex numbers: (3  4i )(5  7i )
(20) Solve by completing the square: x 2  6 x  34  0
(21) Solve using the quadratic formula: 3x 2  4 x  8
(22) Joe can finish a job in 6 hours, while it takes his sister Jane 9 hours to finish the
same job. How long will it take the two of them working together?
(23) Solve using u-substitution:
x4 - 8x2 + 15 = 0.
(24) Evaluate the polynomial f ( x)  2 x 2  5x  2, x  3
(25) Evaluate the function f ( x)  100  4 3 x at x = 125
(26) Find the vertex of the graph of:
f(x) = x2 + 4x + 5
(28) For the given functions, find (f + g)(x), and (f - g)(x):
f ( x)  4 x 2  7 x  3, g( x)  4 x 2  6x  2
(29) Graph the parabola: f(x) = x 2  7 x  6
(30) Find the maximum value of the function f ( x)  2 x 2  8x  25
(31) If P( x)  2.3x 2  50 x  11, C ( x)  3x  5 , then find C ( x)  P( x) .
Solutions:
(1) ) 3(x - 4) = 4(x + 2) + 7,
3x - 12 = 4x + 8 + 7
3x - 12 = 4x + 15,
-3x
-3x
-12 = x + 15
-15
-15
-27 = x
(2) 2 | x  1| 3  7
+3 +3
2 | x  1| 10
(parentheses first - distribute)
(combine like terms 8 and 7)
(move variable to one side)
(get rid of the 15 - subract)
(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:)
 9
(3) The intercepts of the first line are (9,0) and  0,  .
 2
The intercepts of the second line are (0,2), and (-1,0). Graph the intercepts and then draw
the lines between them:
The solution is (1,4)
(4) 2 x3  10 x 2  28 x.
-28x -28x
3
2
2 x  10 x  28 x  0
2 x( x 2  5x  14)  0,
(get LHS = 0)
(gcf = 2x)
(inside pharentheses - trinomial, lead coefficient = 1)
-1 14
-2 7
(-2 + 7 = 5)
2 x( x  7)( x  2)  0,
(factors = 0)
2 x  0, x  7  0, x  2  0,
x= 0 -7 -7 +2 +2,
x = 0, x = -7, x = 7.
x = {0,7}
Note: you could also use the quadratic formula.
(5)
(6)

3  
5
 2
( x  1)( x  4) 

 ( x  1)( x  4),

 x  1 x  4   ( x  4)( x  1) 
2( x  4)  3( x  1)  5,
5 x  5  5,
5 x  10,
x  2.
(7) Note: on the exam, you will probably be asked to solve a number of linear systems.
They will specify the method, but as it's a multiple-choice exam, there's no way of telling
which method you used. Therefore, use the method that you think is best.
y = 3 + 4x,
x - 2y = -13
Since this system is already solved for a variable (y in the first equation), it is probably
best to go ahead and solve by substitution.
x - 2(3 + 4x) = -13,
x - 6 - 8x = -13,
+6
+6
-7x = -7,
x=1
(back-solve for y)
y = 3 + 4(1) = 7. The solution is (1,7).
(8) I cold do this one by substitution or elimination:
Elimination:
Substitution:
3x + 2y = 14,
x = -10 + 3y
(solve for a variable)
-3(x - 3y) = -10,
3(-10 + 3y) + 2y = 14,
-30 + 9y + 2y = 14
3x + 2y = 14,
+30
+30
-3x + 9y = 30
11y = 44,
11y = 44,
y = 4, x = -10 + 3(4) = 2,
y = 4, x - 3(4) = -10,
+12
+12
The solution is (2,4)
x=2
(9) First, identify the unknowns with variable names:
x = number of student tickets,
y = number of general public tickets
x + y = 70
(total number of tickets)
3x + 6y = 330 (take = amount from students + amount from public)
Solve this:
-3(x + y) = 70
-3x - 3y = -210
3x + 6y = 330
3x + 6y = 330
3y = 120,
y = 40, and x + 40 = 70, so x = 30.
(10) Graph the borders, pick a test point:
(11) It's a cube root, so you're looking for cubes – 27 is a cube, and x9 can also be thought
of as one:
3
27 x9  3 33 ( x3 )3  3 x3
1
(12) ( x6 y8 ) 2  x
(13)
6*
1
2
y
8*
1
2
 x3 y 4
98w9 z 6  7w4 z3 2w
49*2 w8 *w
z6
F O
I
L
(14) (4  3)  (4  3)(4  3)  16  4 3  4 3  3 3  16  8 3  3  19  8 3
2
(15) You have to multiply on top and bottom by the conjugate of the denominator:
52
5  2 ( 7  9)
5 7  9 5  2 7  18
35  9 5  2 7  18

*


2
2
74
7 9
7  9 ( 7  9)
( 7)  9
(16) First, Isolate the radical, then square both sides:
y2  y  4
-y -y
( y  2) 2  (4  y ) 2
(quadratic equation – move everything to one side:)
y  2  16  8 y  y 2
-y -2 -2 -y
(solve it: )
0  y 2  9 y  14
0  ( y  7)( y  2)
y  7  0, y  2  0,
y  7, y  2
Check the solutions: y = 7 doesn't work, but y = 2 does.
(18) Just combine the real parts and imaginary parts as like terms (here it is in slow
(3  4i)  (5  7i) 
motion:) (3  5)  (4  7)i 
8  3i
F O
I
L
(19) (3  4i)(5  7i)  15  21i  20i  28i 2 = 15  i  28  43  i
-28(-1)
(20) First, move the 34 to the right:
2
x 2  6 x  34  0
 6 
(then add   = 9 to both sides:)
 34  34
 3 
x 2  6 x  34
9
9
(now, take the square root of both sides:)
( x  3)  25
2
x  3   25,
x  3  5i,
x  3  5i
(21) Get it in standard form (right hand side = 0), and identify a, b, and c:
3x 2  4 x  8,
3x 2  4 x  8  0,
a  3, b  4, c  8
Then put these in the formula:
x
4  42  4(3)(8) 4  112 4  4 7


2(3)
6
6
(22) Let x = number of hours for the two to complete the job. Note that Joe can do
the job per hour, and Jane can do
1
1
x  x 1
6
9
1
of the job per hour. The equation is
9
(multiply both sides by 18 to clear the fraction:)
part of job
part of job
joe does
Jane does
1
1
18( x  x)  (1)18
6
9
3 x  2 x  18,
5 x  18,
18
x ,
5
x  3.6 hours
1
of
6
(23)
x 4  8 x 2  15  0,
u  x 2 , u 2  ( x 2 )2  x 4 ,
u 2  8u  15  0,
(u  5)(u  3)  0
u  5  0, u  3  0,
u  5, u  3.
x 2  5, x 2  3,
x   5,  3
(24)
Evaluate means "plug in". f (3)  2(3)2  5(3)  2  2(9)  15  2  35
(25) f ( x)  100  4 3 125  100  4(5)  80

 b
 b   4
(26) Vertex =   , f       
, f (2)   2, (2) 2  4(2)  5  (2,1).
 2a  2a    2(1)



(28) (a) ( f  g )( x)  f ( x)  g ( x)  4 x 2  7 x  3  (4 x 2  6 x  2)  8 x 2  x  5.
(b) ( f  g )( x)  f ( x)  g ( x)  4 x 2  7 x  3  (4 x 2  6 x  2)  13x  1.
(29)
 7
vertex =   ,
 2
 7 
f      (3.5, (3.5) 2  7(3.5)  6)  (3.5  6.25)
 2 
y-intercept = (0,6)
x 2  7 x  6  0,
x-ints: ( x  6)( x  1)  0,
x  6, x  1
So graph these features and the parabola through them.
(30) The graph of the function will be a parabola pointing down, so the maximum is the

 b
8
 b  
, f (4)   4, 42  8(4)  25  (4, 41).
vertex   , f       
 2a  2a    2(2)



(31) C ( x)  P( x)  (2.3x 2  50 x 11)  (3x  5)  2.3x 2  53x  6