Module 1 Solutions and Notes for Select Practice Problems

This material was originally prepared by Mr. John Hemmerling with revisions by Mr. Louis Aquila.

Practice Exercise - Unit 5

Notes: The most important concept to learn and be able to use from this unit is how to handle a double negative. You have a double negative when there are two negative signs side-by-side with no number between them. A double negative always gives a plus sign.

Example:-(-4) + 5 = 9 / You can see that there are two negative signs in this example. This is a double negative because there is no number between them, only a parenthesis. The parenthesis is there to separate the two negatives so you can see that there are two distinct negative signs. Therefore, 4 + 5 = 9
Example:-4 - 7 = -11 / There are two negative signs in this example but the number 4 is between them. Therefore, you do not have a double negative, you are subtracting 7 from -4 and the result is -11.

Detailed Solutions:

10)-52 - (-38) = -14 / There are 3 negative signs in this problem and the second negative sign is the operation of subtraction. So use the rule a – b = a + (-b). Therefore, the problem can be rewritten as: -52 +(-(- 38))
= -52 + 38 = -14
16)-8 - 3 - 9 = -20 / There are three negative signs again but there is a number between each of them so there is no double negative in this problem.
You can solve this by using each operation working left to right:
-8 - 3 = -11 and -11 - 9 = -20
45)-4 [-(-2)](-3)(-2) = -48 / There are 5 negative signs in this problem. There is a double negative in front of the 2. You can simply note that the numbers are being multiplied and that 5 is odd so the answer must be negative. Once you know the sign of the answer, you can multiply the numbers in any order to complete the solution.

Practice Exercises - Unit 9

Notes:The Order of Operations is very important. Almost all calculators and programming languages use the same order of operations to evaluate expressions. This and the double negative are twovery important ideas covered in this module.

Detailed Solutions:

7.4 + 3[4 + 3(0 - 2)] = -2
The expression within the parentheses must be evaluated first: / (0 - 2 ) = (-2)
We now have: / 4 + 3[4 + 3(-2)]
The expression within the brackets must be evaluated next: / [4 + 3(-2)] = [4 + (-6)] = [-2]
We now have: / 4 + 3[-2]
Multiplication/division must be done before addition/subtraction: / 4 + 3[-2] = 4 + (-6) = -2
A common mistake students make is to add 4 + 3 first, rather than do the multiplication first.
24)-7 – 4[-5 + 5(-32 + 4)] = 173
The expression within the parentheses must be evaluated first.
We have -32 which is -33 = -9. / (-32 + 4) = (-9 + 4) = (-5)
We now have: / -7 – 4[-5 + 5(-5)]
The expression within the bracket must be evaluated next. We have multiplication and addition here. As you know, the multiplication goes first, but we need to find first. = 4 thus we have: / [-5 + 5(-5)] = [-5(4) + 5(-5)]
= [-20 + (-25)]
= [-45]
We now have: / -7 -4[-45]
Do the multiplication before the subtraction: / -7 - (-180) = -7 + 180 = 173

Practice Exercises - Unit 12, Set 4

Notes: These practice problems are mostly review, but there are 14 problems from unit 12, set 4 of the Course Guide that are a little tougher than the rest. Problems # 38 - #51 on this Set are fractions that have addition/subtraction and multiplication in the numerators and denominators. The fraction bar acts as a grouping symbol and divides these problems into two pieces. You should work above and below the fraction bar and do the division that the fraction bar represents last.

Detailed Solutions

Evaluate the numerator first: / -2(6) + 3(-6) = -12 + -18 = -30
Evaluate the denominator next: / (-3)(-12 + 10) = (-3)(-2) = 6
Put the fraction back together with the new values and divide. /
Evaluate the numerator first.
Find the value inside the [...] first. / -2[(-3)7 + 8(-5)] = -2[(-21) + (-40)]
= -2[-61]
= 122
Evaluate the denominator next.
Notice that there is a double negative there. / -12 - (-18) = -12 + 18 = 6
Put the fraction back together with the new values and simplify the fraction. /