```Module 6: Notes and Solutions to Select Practice Problems.
The examples in the Course Guide are very much like the problems in the homework so
you should look at those as well as the solutions here. These are solutions for Practice
Problems-Unit 1.
(3)
-a - 3 = -10
You should perform opposite/inverse operations 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 not alone yet. There is a minus sign next to it so we are not finished. You
can get rid of 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 it's
opposite of division is multiplication so
reciprocal (Check out Module 5 in the
multiply everything by 5.
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
Don't be fooled by the appearance of the problem. This one 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
Here's another one that looks backwards.
3x
2
4
3x
72
22
4
3x
9
4
4
 3x  4 
 9     
3
 4  3 
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 one has one variable in two different places. 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
You need to combine like terms so you have to remove the parenthesis. Remember to
distribute the negative sign.
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
I don't really like working with fractions so I usually get rid of them when I can so
multiply everything by 2 first.
2  3( x  4)    92
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.
 b  b 2  4ac
where ax2 + bx + c = 0 and a  0 .
x
2a
It is just another formula that you have used in the calculator exercise from Module 2. 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) so each
solution below will use both methods (when applicable) and check the answers.
4.
x2 + 5x + 6 = 0
You can also use the Quadratic Formula:
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) must equal zero so x = -2
(x + 3) must equal zero so x = -3
4 + (-10) + 6 = 0 √
9 + (-15) + 6 = 0 √
11.
16t2 - 25 = 0
16t2 - 25 = 0
(4t - 5)(4t + 5) = 0
You can also use the Quadratic Formula:
In order for this product to be equal to zero:
 0  0 2  4(16)( 25)  1600
x

2(16)
32
40 5
x

32 4
 40
5
x

32
4
a = 16, b = 0, c = -25
(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 √
2x2 - x = 0
18.
You can also use the Quadratic Formula:
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
Here's one that cannot be factored (try it and you'll see) 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
 3  393
 3  393
OR
16
16
Check: I'm only going to do one of these since the math follows the same steps for both.
8x2 + 3x - 12 = 0 so
Thus, x 
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
You are not required to do each problem using both methods but you should be able to
use each method to solve a quadratic equation.
```