III. Manipulating
Complex Expressions
A. Multiplying
and Dividing Algebraic Fractions
B. Addition of Algebraic Fractions
C. Equations
D. Equivalent Equations
E. Linear Equations
F. Inequalities
G. Simultaneous
Equations
A. Multiplying and Dividing Algebraic Fractions
When
multiplying algebraic fractions, we search for common factors:
factors that are common to both numerators and denominators.
These factors are canceled, and the resulting fraction simplified,
if possible. When dividing algebraic fractions, we invert the
divisor (the fraction we're dividing by) and then multiply, searching
for common factors.
Example 1
Perform the division.
3a - 4 divided by 3ab - 4b
2 - 3a 4
- 9a
Solution
First, invert the divisor
and express the division as the multiplication.
3a - 4 x 4 - 9a
2 - 3a 3ab - 4b
Next, factor any factorable expressions.
4 - 9a =
(2 - 3a)(2 + 3a)
3ab - 4b = b(3a - 4)
The given division is now written
and the common factors (in red) canceled.
3a -
4 x (2 - 3a)(2 + 3a) = 2 + 3a
2
- 3a b(3a - 4) b
B. Addition of
Algebraic Fractions
When adding (or subtracting)
algebraic fractions, we proceed as in arithmetic:
- Find the lowest common denominator
(LCD) of the fractions.
- Write each fraction using the
LCD of the fractions.
- Add (or subtract) the numerators.
The denominator will be the LCD. Simplify the resulting fraction.
Example 2
Express as a single fraction:
2x + 1 - x - 3
2a 3a
Solution
The lowest common denominator is 6a. Each fraction is written
with its denominator as 6a. This gives
2x + 1 - x - 3 =
3(2x+1) - 2(x-3)
2a 3a
6a
6a
=
6x + 3 - 2x - 6
6a
6a
=
6x + 3 - 2x + 6
6a
=
4x + 9
6a
C. Equations
An
equation is a mathematical statement that two algebraic expressions
are equal. The letters in the equation are the unknowns. An equation
may be thought of as asking the question, "What numerical
value of the unknown satisfies the equation?" (Satisfies
means to make both sides equal.) The value of the unknown is
called the solution of the equation, or root of the equation.
We solve an equation by finding the numerical value of all of
its roots.
There
are equations that do not have real roots. For example , the
equation x = -4 does not have a real root;
the square of any real number, whether positive or negative (e.g.,
+ 2 or -2) is positive. Hence, there is no real root to x = -4.
D. Equivalent Equations
Two equations
that have the same roots are said to be equivalent. To solve
an equation, we often put the equation in a form that is more
easily solved with one or both of the following operations:
- Add or subtract the same term
to or from each side of an equation.
- Multiply or divide each side
of an equation by the same number or algebraic expression.
E. Linear
Equations (linear equations do not contain squares
or square roots)
To
solve a linear equation that contains fractions, first remove
all fractions by multiplying both sides by the lowest common
denominator of all the fractions.
Example 3
x/3 + x/2 = 5
Solution
Multiply both sides by the LCD which is 6: (3 x 2)
2x + 3x = 30
5x = 30
x = 6
Then
move all terms containing the unknown to one side and all terms
that do not contain an unknown to the other side. Factor out
the unknown from all terms that contain the unknown; finally,
divide each side by the coefficient of the unknown.
Example 4
Solve a(a + t) = b
- bt for t
Solution
To solve for t, first
remove the parentheses:
a +
at = b - bt
Move "- bt" to the
left side and "a" to the right
side:
bt + at = b - a
We can factor t out of the two terms on the left side; we can
also factor the right side to obtain t(b + a) = (b + a)(b - a).
Divide both sides by (b + a) and find the solution to be
t = b - a.
F.
Inequalities
An
inequality is simply a comparison of two quantities or expressions.
The following symbols with their meanings are used:
> is greater than
< is less than
>= is greater than or equal
to
<= is less than or equal to
If we place all numbers on a
number line with negative numbers to the left of zero and positive
numbers to the right of zero, then if A is to the right of B
we state that A > B; if A is to the left of B then A <
B. Consequently, we conclude that -5 > -7 and -3 < 2.
-7 -5 -3 0 2
We also form inequalities using
algebraic symbols. The inequality 3x + 2 > x - 6 is solved
just like an algebraic equation is solved. We subtract 2 from
each side of the inequality so that 3x > x - 8. Then subtract
x from each side so that 2x > - 8. Divide by 2 and obtain
x > - 4. This is the solution. Any number greater than - 4
satisfies the inequality.
There are several rules that
we must follow when manipulating inequalities:
- The same number or algebraic
expression can be added or subtracted from each side of an inequality.
- The same positive number (or
positive algebraic expression) can multiply or divide each side
of an inequality.
- Both sides of the same type
of inequality can be added and the inequality remains. (If x
< y and w < z, then x + w < y + z.)
- If a negative number (or negative
algebraic expression) multiplies or divides each side of an inequality,
the inequality sign must be reversed. (Be sure to remember this;
it often leads to errors!)
Example 5
Solve the inequality
2x - 2 > x - 5.
Solution
The inequality is treated in much the same manner as an algebraic
equation. Add 2 to each side:
2x > x - 3.
Subtract x from each side and
the solution is x > -3
NOTE: Retain the same inequality
symbol throughout the solution unless an operation reverses the
symbol.
Example 6
Solve the inequality 3r + 5 > 6r - 7.
Solution
First, subtract 5 from each side:
3r
> 6r - 12
Next, subtract 6r from each side:
G. Simultaneous
Equations
Two
unknowns in two equations are solved by either of two methods.
1. Substitution
2. Addition or Subtraction.
1. The substitution method:
a. Solve the first equation for
one unknown in terms of the other unknown and substitute the
result into the second equation.
b. Solve the resulting equation
for the unknown.
c. Substitute the value of this
known unknown into either original equation and solve for the
second unknown.
Example 7
Solve for x and y.
x - y = 2
2x + y = -5
using
the substitution method.
Solution
Using the substitution
method, the first equation gives
x = 2 + y
Substitute this into the second equation and solve for y:
2(2 + y) + y = -5
4 + 2y + y = -5
3y = -9
y = -3
Substitute this value back into the first equation and solve
for x:
x + 3 = 2
x = -1
The solution is x = -1, y = -3.
2. The addition or subtraction
method:
a. Multiply one equation by a
properly chosen number so that one of the unknowns has the same
coefficient in both equations.
b. Add or subtract the equations
so that one of the unknowns is eliminated.
c. Solve the resulting equation
for the remaining unknown.
d. Substitute the value of this
known unknown into either original equation and solve for the
second unknown.
Example 8
Solve for x and y
x - y = 2
2x + y = -5
This time
use the addition or subtraction method.
Solution
We simply add the two equations since this will eliminate y:
x
- y = 2
+ 2x + y = - 5
3x =
-3
x =
-1
Substitute this value back into the first equation and solve
for y:
(-1) - y = 2
-y = 3
y = -3
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