```Module 6: Notes and Solutions to Select Practice Problems.
These solutions were prepared by Mr. John Hemmerling with revisions by Mr. Louis
Aquila
Practice Exercises-Unit 1.
(3)
-a - 3 = -10
Perform the opposite/inverse operation until you get the variable alone on one side of the
equal sign. The opposite of -3 is +3 so:
-a - 3 +3 = -10 +3
-a = -7
The variable is still not isolated. There is a minus sign next to it, so we are not finished.
You can remove the minus sign by multiplying both sides of the equation by -1.
(-1)(-a) =( -7)(-1)
Thus, a = 7
Check: -a - 3 = -10 so -7 - 3 = -10 must be true.√
(14)
 3e
 9
5
There are two ways to solve this equation.
1) There are two operations here. The most 2) There is a fraction in front of the
obvious is probably the division by 5. The
variable. Multiply a fraction by its
opposite of division is multiplication so
reciprocal (Check out Module 5 in the
multiply everything by 5.
Course Guide if you want to review this
topic) and you get 1. Multiplying by 1
gives you the variable by itself. Since this
5  3e    95
is an equation, you must multiply
 5 
everything by the reciprocal.
-3e = -45
 5   3e 
 5 


   9 

The only operation left is multiplication by
  3  5 
  3
(-3). The opposite of multiplication is
division so divide everything by (-3)
As you can see, the variable is alone now
and all that's left is the simplification of the
right-hand side (rhs) of the equation.
 3e  45

3
3
 9  5  45
e

 15
Thus e = 15
3
3
Check:
 3e
 3(15)  45
 9 so

 9
5
5
5
(39)
-14 = 5p + 18
This problem could just as easily been written as 5p + 18 = -14. Anyway, the solution of
the original problem follows.
-14 - 18 = 5p + 18 -18
- 32 5p

so p = -32/5
5
5
 32
 14  5 
 18
Check: -14 = 5p + 18 so
must be true. √
5
 14  32  18
(47)
7
3x
2
4
Observe that it makes no difference if the variable term is on the right or the left.
3x
2
4
3x
72
22
4
3x
9
4
4
 3 x  4 
   9     
3
 4  3 
12  xor x  12
7
Check: 7 
3x
312 
 2 so 7 
 2 must be true.
4
4
36
7
2
√
4
7 92
(60)
2b - 9 = 5b
This problem has the variable b on both sides of the equal sign. You have to get them
together to solve the equation. Adding the opposite of either term to both sides of the
equation will do it.
2b - 2b - 9 = 5b - 2b
-9 = 3b
 9 3b

so b = -3
3
3
Check: 2b - 9 = 5b so 2(-3) - 9 = 5(-3) must be true.
-6 -9 = -15 √
(68)
y - (2 + 2y) = 6
First remove the grouping symbols. Remember to distribute the negative sign over every
term within the parentheses. Then you need to combine like terms.
y - (2 + 2y) = 6
y - 2 - 2y = 6
-2 - y = 6
-2 + 2 - y = 6 + 2
-y = 8 so y = -8
Check: y - (2 + 2y) = 6 so -8 - (2 + 2[-8]) = 6 must be true.
-8 - (2 + [-16]) = 6
-8 - (-14) = 6
-8 + 14 = 6 √
(74)
 3( x  4)
 9
2
First, multiply everything by 2 to remove the denominator.
3( x  4) 
 2  
  (2)  9 
2


We now have the equation:
-3(x - 4) = -18
-3x + 12 = -18
-3x + 12 - 12 = -18 - 12
-3x = -30
x = 10
Check:
 3( x  4)
 3(10  4)
 9 so
 9 must be true.
2
2
 3(10  4)
 9
2
 30  12
 9 √
2
 18
 9
2
Notes for Unit 2.
For the quadratic equation ax2 + bx + c = 0 (in standard form) and a  0 , the Quadratic
Formula is.
 b  b 2  4ac
2a
To use this formula, put the numbers in for a, b and c and follow the order of operations.
The Quadratic Formula can be used to find the solutions to all quadratic equations (in
standard form). Each problem below will use the quadratic formula and where applicable
we will also solve the equation by factoring. You do not have to solve a quadratic
equation both ways. First try to factor the expression ax2 + bx + c. If you can do this,
then you will be able to find the solutions without using the quadratic formula. If you
cannot factor the expression ax2 + bx + c, then you have no recourse but to use the
quadratic formula. Only use the quadratic formula if you cannot factor. Below where the
expression ax2 + bx + c can be factored, we still solve the problem using the formula
simply to convince you that both approaches give the same answers.
x
4.
x2 + 5x + 6 = 0
x2 + 5x + 6 = 0
a = 1, b = 5, c = 6
(x + 2)(x + 3) = 0
x
In order for this product to be equal to zero:
 5  5 2  4(1)(6)  5  25  24

2(1)
2
 5  1  5 1

2
2
 5 1  4
x

 2
2
2
 5 1  6
x

 3
2
2
Check: x2 + 5x + 6 = 0 so (-2)2 + 5(-2) + 6 = 0 and (-3)2 + 5(-3) + 6 = 0.
x
(x + 2) = 0, so x = -2
(x + 3) = 0, so x = -3
4 + (-10) + 6 = 0 √
9 + (-15) + 6 = 0 √
11.
16t2 - 25 = 0
16t2 - 25 = 0
(4t - 5)(4t + 5) = 0
In order for this product to be equal to zero:
a = 16, b = 0, c = -25
 0  0 2  4(16)( 25)  1600
x

2(16)
32
40 5
x

32 4
 40
5
x

32
4
(4t - 5) = 0
4t - 5 + 5 = 0 + 5
4t = 5 so t = 5/4
OR
(4t + 5) = 0
4t + 5 - 5 = 0 - 5
4t = -5 so t = -5/4
Check: 16t2 - 25 = 0 so 16(5/4)2 - 25 = 0 (Check is the same for x = -5/4 )
16(25/16) - 25 = 0
25 - 25 = 0 √
18.
2x2 - x = 0
2x2 - x = 0 so x(2x - 1) = 0
a = 2, b = -1 , c = 0
In order for this product to be equal to zero:
 (1)  (1) 2  4(2)(0) 1  1

2
(
2
)
4
x=0
11
OR
x
0
2x - 1 = 0
4
2x - 1 + 1 = 0 + 1
11 1
x

2x = 1 so x = 1/2
4
2
2
2
2
Check: 2x - x = 0 so 2(0) - 0 = 0 √ and 2(1/2) - (1/2) = 0
x
1 1
2    0
4 2
√
1 1
 0
2 2
26.
8x2 + 3x - 12 = 0
The expression 8x2 + 3x - 12 cannot be factored (try it), so you must use the Quadratic
Formula: a = 8, b = 3, c = -12
x
 3  32  4(8)( 12)  3  9  384  3  393


2(8)
16
16
Thus, x 
 3  393
 3  393
OR
16
16
Check: I'm only going to do one of these since the math is the essentially the same for
both.
8x2 + 3x - 12 = 0 so
2
  3  393 
  3  393 
  3
  12  0
8



16
16




  3  393   3  393    3  393 


  3
  12  0


 

2
16
16


 

 9  3 393  3 393  393  9  3 393


 12 32   032 

32
16




402  6 393  2  9  3 393  12(32)  0
402  6 393  (18)  6 393  384  0
402  402  0
Once again, if you can factor, then solve the quadratic equation by factoring. Otherwise,
you must use the quadratic formula.
```