Mathematics 96 (3581)
CA 5: Definition of Subtraction
Mt. San Jacinto College
Menifee Valley Campus
Spring 2013
__________________
Name
This class handout is worth a maximum of five (5) points. It is due no later than the end
of class on Friday, 12 April.
NOTE: You may need to study this entire handout carefully several times before you
begin the exercises it contains. You may need to study example exercises carefully
several times before you attempt the exercise sets that follow them. Also, the order in
which the exercises occur may not necessarily be the order in which you complete them.
If the solutions to a particular exercise set elude you, skip to another one. You are being
given two weeks to complete this handout because you’ll probably need to study it,
attempt some of the exercises and then take a break, continuing with it a day or two later.
From arithmetic of signed numbers, we’ve noticed that subtraction and addition of the
opposite yield the same result. For example, 7 – 5 (“seven subtract five”) and 7 + (-5)
(“seven plus the opposite of five”) both equal 2. This same relationship holds true for
algebra. In other words, every subtraction can be interpreted as addition of the opposite.
For example, 7x – 4y and 7x + (-4y) are equivalent. The following property makes
precise the connection between subtraction and addition of the opposite:
The Definition of Subtraction If x and y are real numbers, then
x - y = x + (-y)
(1).
The following four equations are expressions of the Definition of Subtraction:
-2 - 8 = -2 + (-8)
9c – (-7t) = 9c + 7t
6 – (x + 5) = 6 + [-(x + 5)]
(4 – z) – 9 = (4 – z) + (-9)
Example 1. Insert the missing symbol(s) (e.g. grouping symbol(s), a constant or a
variable expression) to create an equation that expresses the Definition of Subtraction.
( x 2  4x  4 ) - 1 =
x 2  4x  4
Solution: Notice that the given equation is a contradiction (the solution set is the empty
set). That is, there is no solution. Note that the equation is of the form
1
z–1 = z
where z corresponds to the expression x 2  4 x  4 . Not only must we insert symbols to
make the equation an identity (solution set R) but the equation must express the
Definition of Subtraction! We can create an identity by subtracting the constant one from
the right-hand side. This will create an equation of the form
z–1 = z–1
(an identity) where, as before, z corresponds to the expression x 2  4 x  4 . To finish
the exercise, we must express the Definition of Subtraction. We can do so by rewriting
one of the subtractions as addition of the opposite. That is, we must rewrite z – 1 = z – 1
as either
z – 1 = z + (-1)
(which takes the form of formula (1), x – y = x + (-y), where x corresponds to z and y
corresponds to 1)
or
z + (-1) = z – 1.
Choosing the first of these two options yields
( x 2  4x  4 ) - 1 =
x 2  4 x  4 + (-1)
Note that this equation has the same form as formula (1), where x in the formula
corresponds to the expression x 2  4 x  4 in the equation and y in formula corresponds
to the constant one. That is, the modified equation now expresses the Definition of
Subtraction.
Example 2. Insert the missing symbol(s) (e.g. grouping symbol(s), a constant or a
variable expression) to create an equation that expresses the Definition of Subtraction.
(4 – z) – 9 =
+ (-9)
Solution: The left-hand side of the equation is already of the form x – y, where x
corresponds to the expression 4 – z and y corresponds to 9. That is, the left-hand side of
the equation takes the form of the left-hand side of formula (1). Notice that the symbol
nine (9) appears on both sides of the equation. If we compare the equation with formula
(1), where x corresponds to 4 – z and y corresponds to 9, we can determine what to insert
on the right-hand side of equation so that the equation will express the Definition of
2
Subtraction. Creating a vertical alignment between the equation and formula (1), we
have
(4 – z) – 9 =
+ (-9)
x
- y =
x
+ (-y)
It appears that we need to insert the expression 4 – z in the gap directly right of the equal
sign. That way, all occurrences of x in formula (1) will correspond to the expression 4 –
z in the equation. Doing so, we have
(4 – z) – 9 = (4 – z) + (-9)
and the equation now expresses the Definition of Subtraction, as required.
Exercise 1. Insert the missing symbol(s) (e.g. parentheses, a constant or a variable
expression) to create an equation that expresses the Definition of Subtraction. (To
receive full credit (three points), you must complete at least seven of the following eight
parts correctly. If you complete five or six parts correctly, you’ll earn two points.
Completing three or four parts correctly equates to one point. Correctly completing two
parts (or fewer) yields zero points. PLEASE USE A PENCIL OR INK OTHER
THAN BLACK!
a.
b.
7j
7j
 18 =
9
9
y 2  4 y  15
( y 2  4 y  15)  34
=
c.
2 5
- (-9)
15
d.
 23 
( t 2  3t  6 ) +    =
 5
e.
17 –
f.
( z2  7z 
=
2 5
15
t 2  3t  6
= 17 + [-(c + 8)]
11
 3
) -   =
4
 4
3
+
3
4
g.
11z
7 yt
h.
+
=
-5
76
 76 
= ( p2  3 p  9 ) -   
7
 7 
While the existence of grouping symbols in an expression often implies the operation of
multiplication, this is not always the case. Grouping symbols, such as parentheses, are
typically utilized to express the Definition of Subtraction (see formula (1)). They
“surround” the opposite, separating the plus and minus signs. Utilized in this way,
parentheses help make the expression x + (-y) more “readable” than it would be without
them. However, these parentheses do not imply multiplication. The operation that
combines x and –y on the right-hand side of formula (1) is addition, not
multiplication.
In the following examples, expressions containing grouping symbols are simplified. In
each expression, some grouping symbols may imply multiplication while others may not.
In these examples, and particularly in the exercise that follows, pay close attention to the
operations adjacent to the grouping symbols. These must be performed as indicated. In
other words, don’t assume the operation is multiplication simply because grouping
symbols are nearby!
Example 3. Express each subtraction as addition of the opposite and simplify.
5 – 6(x – 7) -3 + (-2x)
Solution: Notice that multiplication is implied by the first set of parentheses present in
the expression. Notice also that the second set of parentheses is preceded by a plus sign,
so it is addition, not multiplication, that combines the term -3 to the term -2x.
As directed by the instructions, before doing anything else, we’ll first change each
subtraction into addition of the opposite:
= 5 + (-6)(x + [-7]) + (-3) + (-2x)
To simply, we’ll begin by performing the multiplication, distributing the factor -6 into the
binomial factor x – 7. This yields the expression
= 5 + (-6x) + 42 + (-3) + (-2x)
4
Adding like terms yields the simplified answer: -8x + 44, or 44 – 8x.
NOTE: There are two parts to the answer: the expression in which every subtraction has
been changed to addition of the opposite:
5 + (-6)(x + [-7]) + (-3) + (-2x)
and the final, simplified expression:
-8x + 44 (or 44 – 8x)
Example 4. Express each subtraction as addition of the opposite and simplify.
5x – 7(2x – 3y) -5
Solution: Before doing anything else, we’ll first change each subtraction into addition of
the opposite:
= 5x + (-7)(2x + [-3y]) + (-5)
We’ll perform the multiplication, distributing -7 into the binomial 2x – 3. This yields the
expression
= 5x + (-14x) + 21y + (-5)
Adding like terms yields the simplified answer: -9x + 21y - 5, or 21y – 9x - 5.
Exercise 2. Express each subtraction as addition of the opposite and simplify. To receive
full credit (two points), you must complete at least six of the following eight parts
correctly. If you complete between three and five parts correctly, you will receive one
point. Correctly completing two parts (or fewer) yields zero points.
NOTE: There are two parts to the answer: the expression in which every subtraction has
been changed to addition of the opposite (BEFORE DOING ANYTHING ELSE) and the
simplified final expression (see the NOTE following Example 3.)
a.
( t 2  3t  6 ) - 2( 4t 2  5t  2 ) - (-8t)
5
b.
4v – (5u + 2) + (-9v) – 8(7v –u)
c.
2r + 5(2s – 9) - 16
d.
11z + 3(8 – z) – (z - 7) – (13 – z)
e.
 4d 2  d  3 - ( 2d 2  5d  1) 2
6
f.
y – 4(5y + 6) – (-7y) – 8 + 9y
g.
8 – 2(3 – j)j + 4j – (5j – 2)(-3)
h.
r(2 –5r)2 – (-6j) + 3 – (8 + j)
7