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1.04 Equations and Properties of Equations Linear, Absolute Value, and Literalfrom Intermediate Algebra: One Step at a Time © 1998 Pages 39-50 Dr. Robert J. Rapalje Seminole State College of Florida PROPERTIES OF EQUATIONS Methods of solving equations are as varied as the types of equations to be solved. However varied the strategies may be, all must be executed according to and without violating several properties of equations: 1. REFLEXIVE PROPERTY a = a. Any number is equal to itself.
2. SYMMETRIC PROPERTY If a = b, then b = a.The order in which the equality is given does not matter. For example, you can say "X=4" or "4=X", the meaning is the same--the value of X is 4.
3. TRANSITIVE PROPERTY If a = b and b = c, then a = c.The word "trans" means "across." If you can get from point "a" to "b", and then from "b" to "c", then you can get from "a" across "b" to "c."
4. ADDITION PROPERTY If a = b, then a + c = b + cIf a = b, then a - c = b - c. The same number may be added (or subtracted) from both sides of an equation.
5. MULTIPLICATION PROPERTY If a=b, then ac = bc If a=b and c≠ 0, then a/c = b/c. Both sides of an equation may be multiplied or divided by the same non-zero number. LINEAR EQUATIONS A linear equation in one variable is an equation in which the highest degree of the variable is one (no variable squared, cubed, or higher terms). We usually think of an equation being "linear" as opposed to being "quadratic". If it is a linear conditional equation, there will be only one solution. This section provides an opportunity to distinguish between conditional equations, identities, and contradictions. EXAMPLE 1. Solve for x: 5(3x - 4) - x (x - 5) = x (5 - x) Solution: 15x – 20 – x2 + 5x = 5x - x2 - x2 + 20x - 20 = 5x - x2 + x2 - 20x -20x + x2 - 20 = -15x
EXAMPLE 2. Solve for x: 5(3 x - 4) - x (x -5) = 4 x (5- x ) + 3 x 2 Solution: 15x – 20 – x2 + 5x = 20x - 4x2 + 3 x 2 - x2 + 20x - 20 = 20x - x2
- 20 = 0 NO WAY! CONTRADICTION—NO SOLUTION! EXAMPLE 3. Solve for x: 5(3 x -4) - x (x -5) = x (20- x) - 20 Solution: 15x – 20 – x2 + 5x = 20x - x2 - 20 - x2 + 20x - 20 = - x2 + 20x - 20 + x2 - 20x + 20 + x2 - 20x + 20 0 = 0 IDENTITY—True for all values of x! EXAMPLE 4. Solve for x: 5(3 x - 4) - x (x - 5) = x (15 - x) - 20 Solution: 15x – 20 – x2 + 5x = 15x - x2 - 20 - x2 + 20x - 20 = - x2 + 15x - 20 + x2 - 15x + 20 + x2 - 15x + 20 5x = 0 x = 0 EXERCISES: Solve the equations for x. Identify which are contradictions, identities, or conditional equations. 1. 4(x+3) = 6(2 x -5) - 2 x 2. 6(x +3) = 3(2 x -3) + 27
3. 6(x +3) = 3(6-2 x) + 4 x 4. 6(x +3) - 3(5-2 x) = 12 x
5. 6(x +3) - 3(6-2 x) = 12 x 6. x (x -6) = 4 - x (2- x)
7. x (x -2) = 4 - x (2- x) 8. x (3 x -8) = 12 x - 3 x (4- x)
ABSOLUTE VALUE EQUATIONS The absolute value of a number refers to the "size" of a number or the "magnitude" of a number without regard to whether the number is positive or negative. You remember that the absolute value of a number cannot be negative. The following formal definition of absolute value may at first appear to contradict this last statement.
DEFINITION: | x | =
x if x
≥ 0
= -x if
x < 0 Does
it appear that in the second part of the definition | x | = - x,
that the absolute value of x equals a "negative"?
What you must remember is that in the second part of the
definition, x is itself negative, that the absolute value of x
is actually the negative of the negative, which is positive!
This formal definition of absolute value of x confirms the
fact that there are generally two cases to consider--there are two
solutions to be found. Consider
the simple example, |
x | = 3. Obviously,
the solutions are x = 3
and x = -3.
Likewise |Junk| = 3 has
two solutions: Junk = 3 and Junk = -3.
The next exercises illustrate a concept in which the variable,
instead of being "
x " or "Junk",
is "
x - 2” or “3 x + 2”:
EXERCISES.
Solve the absolute value equations: 1. | x |
= 4
2.
| x | = 6
3. | x |
= 10
x
= ____ or
x = ____ 4. | y | = 4
5. |$| = 6
6. |Junk| = 10 y
=____ or y =____ 7. | x
-2| = 4
8. |
x -2| = 6
9. | x
-2| = 10 x
-2 =
or x -2 =
x
=
or x =
10.
|2 x -2| = 4
11. |2
x -2| = 6
12. |2
x -2| = 10 2
x -2= _____ or
2 x -2= _____
2 x
=
or 2 x =
x
=
or x = 13.
|3 x +2| = 4
14. |3
x +2| = 6
15.
|3 x +2| =
10 3
x +2= _____ or 3 x
+2= _____
3 x
=
or 3 x =
x
=
or x = 16.
|2 x -7| = 5
17. |2
x -7| = 7
18. |3 x +6|
= 6
What
if you have |
x | = -4 or |3
x +6| = -4? Notice
that an absolute value cannot equal a negative. Therefore there is no solution.
The method used in the explanation so far is not valid for absolute
value equal to negatives. In
such cases there is no solution. Study
the following examples. EXAMPLE
1. Solve for x: EXAMPLE
2. Solve for x:
|2
x - 5| = 3
|2
x - 5| = -3 Solution:
Solution:
No Solution, since
absolute
2
x - 5 = 3 or
2
x - 5 = -3
value cannot equal a negative.
2
x = 8
2 x = 2
x
= 4 x
= 1 Check:
(Just for fun)
x
=4
x =1
|2(4)-5|
= 3 |2(1)-5|
= 3
|3|
= 3
|-3| = 3 Notice
that in Example 2, because the absolute value equals a negative number,
there are not two solutions to solve.
In fact, if you try to solve two cases as in Example 2, you missed
the problem completely. Whenever
an absolute value of any variable equals a negative number, there is no
solution! EXAMPLE
3. Solve for x:
|2
x
- 5| = | x
- 40|
Since
there are two absolute values in each of these examples, it might appear
that there should be two cases for each for a total of 2 times 2 = 4 cases
to solve. The four cases are
as follows:
Case
I: Positive
= Positive
Case II:
Positive = Negative
(2
x - 5) = (x
- 40)
(2
x - 5) = - (x
- 40)
Case
III: Negative = Negative Case
IV: Negative = Positive
-
(2 x - 5) = - (x
- 40)
- (2 x - 5)
= (x - 40) However,
before solving all four cases, notice that Case III is actually Case I,
where both sides of the equation were multiplied by -1.
Also, Case IV is the same as Case II, with both sides multiplied by
-1. Therefore, you need only
solve Cases I and II. Solution:
Case
I: 2
x - 5 = x
- 40
Case II: 2
x - 5 = -( x
- 40)
x
= -35
2 x - 5 = -
x + 40
3
x = 45
x
= 15
Check:
x = -35
x = 15
|2(-35)-5|
= |(-35)-40|
|2(5)-5| = |5-10|
|-70-5|
= |-75|
|10-5| = |-5|
|-75|
= |-75|
| 5| = |-5|
The
solution for Example 3 is x
= -35, 15. EXAMPLE
4. Solve
for x:
|2
x - 5| = |2
x - 15| Solution:
Case I:
2
x - 5 = 2
x - 15
Case II:
2
x - 5 = -(2
x - 15)
-5
= -15
2
x - 5 = -2
x + 15
No
Solution for Case I
4
x =
20
x
= 5
The only solution is
x
= 5.
SUMMARY
I.
For
c ł
0, |a
x + b| = c has two cases to solve:
a
x + b = c
or
a
x + b = -c
[Note:
If c=0, the two cases are the same!]
II.
For c<0, |a
x + b| = c has No
Solution!
III.
|a
x + b| = |c
x + d| has two cases to solve:
a
x + b = c
x + d
or
a x + b = -(c x
+ d)
EXERCISES: 1.
|2 x - 7| = 5
2.
|2 x - 7| = -5 3.
|3 x + 6| = -18
4.
|3 x + 6| = 18 5.
|3 x - 5| = 5
6.
|3 x - 5| = 10
7.
|4 x - 12| = 0
8.
|4 x + 12| = 0 9.
|2 x - 3| = | x + 6|
10.
|2 x + 3| = |4 x - 9|
11.
|3 x - 4| = |12 - x |
12.
|3 x + 4| = |12 - x | 13.
|2 x + 4| = |12 - 2 x |
14.
|3 x - 5| = |5 + 3 x | 15.
|2 x - 3| = |3 - 2 x |
16.
|8 - x | = |8 + x | LITERAL EQUATIONS Frequently
equations (formulas) to be solved are expressed in terms of several
different letters.
You will probably recognize some of the formulas that we use here,
since many of them come from science, business, geometry, and other areas
of life.
Other formulas have been made up especially for practice in this
section. These
are also called literal equations,
because there are so many different "litters" (joke) in them.
In these equations, you will be solving for one variable (letter)
in terms of all the other variables (letters) in the equation.
The following steps will be helpful in solving literal equations.
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