Download Pre-Calculus Algebra Summary 2 One

Pre-Calculus Algebra
Summary 2
One-dimensional mathematics: equations
Equations in one variable: a statement in which two algebraic expressions containing one
variable are equal. The two expressions are called the sides of the equations.
Linear equations in one variable. A linear equation in one variable is equivalent to on equation
of the form ax + b =0 where a and b are numbers and a  0. A linear equation is also called a
first-degree equation because the variable x is of degree 1.
Domain of a variable.
Sometimes the admissible values of a variable are limited by some restrictions. The set of all
admissible values of a variable is called the domain of the variable.
Examples
1. Find the solution of the
equation 3x + 5 = -7
2. Find the positive numbers
for which 4x +7 =2x+3
3. Find the real solution for
Solve: x = -4
The solution set is {-4}
Solve: x = -2
No solution because the domain limits
the solution to positive numbers
The solutions set is empty, that is
{ }
The solution set is {-4, 4}
Solve: x  4
2
the equation x  16
4. Find the real solution for
the equation x 2  9  5
5. Find the negative numbers
Solve: x 2  4 , no
solution.. There is no
real number that
satisfies the equation
Solve: x   5
The solutions set is empty, that is
{ }
The solution set is {  5 }
2
for which x  5
Some restrictions on the possible values of a variable.
1. The denominator of a fraction can not be zero.
Example:
6
14
25
For the equation


, x  2, x  3, x  3, x  1
2
x  2 x  9 x 1
2. When looking for solutions within the domain of real numbers, the radicand of an even index
radical can not be negative.
Example:
For the equation x  3  3  x  2, x  3  0, that is , x  3
Note: a radical is a number of the form n x , where n is called the index and x is called the
radicand.
3. The input to a logarithm can not be zero or negative.
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Solutions or roots of an equation in one variable.
Since an equation is a statement, it may be true or false depending on the value given to the
variable.
The admissible values of the variable that result in a true statement are called the solutions or
roots of the equation.
Solution set: the set of all admissible values that result in a true statement is called the solution
set of the equation. The set of solutions of an equation is called the solution set of the equation.
An equation can have one solution, a finite number of solutions or infinitely many solutions.
Equivalent equations: two or more equations that have the same solutions set are called equivalent
equations.
Procedures that result in equivalent equations. To indicate that the equations are similar we use
the symbol 
1. Interchange the two sides of the equation (this is called the symmetric property of equality)
Example: x + 5 = 3x -4  3x – 4 = x + 5
2. Simplification of the two sides of the equation by combining similar terms, eliminating
parenthesis, and so on.
Example: 3(x – 2) +4(3 – 2x) = 5 – (9x – 6)  3x - 6 +12 - 8x = 5 – 9x +6
 -5x + 6 = 11 - 9x
3. Add or subtract the same expressions on both sides of the equation.
Example: -5x + 6 = 11 - 9x  (-5x + 6) + 9x = (11 – 9x) + 9x  4x + 6 = 11
 (4x + 6 ) - 6 = (11) – 6  4x = 5
4. Multiply or divide both sides of the equation by the same nonzero number or expression.
4x 5
5
1
Example: 4x = 5 
  x   1  1.25
4 4
4
4
5. If one side of the equation is 0 and the other side can be factored, then by the Zero-Product
Property, set each factor to zero.
Example: (x - 2)(x + 3) = 0  x – 2 = 0 or x+ 3 = 0  x = 2 or x = -3
6. Careful!
Multiplication of both members of an equation by an expression containing the variable could add
extra solutions, therefore the equation that results is not equivalent to the original equation. Those
solutions are called extraneous solutions.
When you need to multiply both members by an expression containing the variable, the solutions
should be checked on the original equation to detect the possibility of extraneous solutions. The
extraneous solutions should be eliminated from the solution set. We use the symbol  to indicate that
the left side of the equation implies the right side, but the right side does not necessarily implies the left
side. Therefore the equations are not equivalent (  ).
Example: Solve the equation
5
 1
x3
5
5
 1 
(x  3)  1(x  3), to eliminate the denominator;
x3
x3
 5   x  3  x  5  3  x  2
x= -2 checks in the original equation, therefore the solution set is {-2}
6
15
Solve the equation

3
2
x 1 x 1
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6
15
6
15

3

3
x  1 x2  1
x  1 ( x  1)( x  1)
6
15

( x  1)( x  1) 
( x  1)( x  1)  3( x  1)( x  1)  6x  6  15  3x 2  3
x 1
( x  1)( x  1)
 0  3x 2  6x  24  x 2  2x  8  0  ( x  4)( x  2)  0  x  4  0 or x  2  0
 x  4 or x  2.
They both check on the original equation, the solution set is {-2, 4}
Solving linear equations.
Problem 1. Solve the equation 5 + 3x = 9x – 11
Solution: 5 + 3x = 9x – 11  9x – 11 = 5 + 3x  (9x – 11) -3x = (5 + 3x) – 3x
6x 16
8
1

x 2
 6x -11 = 5  (6x – 11) + 11 = 5 + 11  6x = 16 
6
6
3
3
Mentally: 5 + 3x = 9x – 11
 9x -11 = 5 + 3x symmetric property of equality
 6x – 11 = 5 subtracting 3x from both sides of the equation
6x = 16 adding 11 to both sides of the equation

16 8
2
Dividing by 6 both sides of the equation.

x
 2
6 3
3
Problem 2. Solve the equation -3x + (12 – 2x) = 4x – (8 – x)
Solution: -3x + (12 – 2x) = 4x – (8 – x)
 -3x + 12 -2x = 4x – 8 +x  -5x + 12 = 5x - 8 Simplification of both sides
Subtracting 5x from both sides
 -10x +12 = -8
 -10x = -20 Subtracting 12 from both sides.
Solution set is { 2 }
 x =2 Dividing both sides by -10
4
5
Problem 3. Solve the equation
5
y
2y
4

 5 
4
5
5
   5  2y    2y multiplication by 2y if y  0 .
y
2y
y

 2y 
 8 – 10y = 5 Simplification of both sides of the equation
 -10y = -3 Subtracting 8 from both sides of the equation
3
,dividing both sides by 10.
y

10
2x
6
Problem 4. Solve the equation

2
x3 x3
2x
6
2x
 6


2 
( x  3)  
 2 ( x  3) if x  3  0
Solution: x  3 x  3
x3
x3

Solution:
 2x  6  2( x  3)  2x  6  2x  6  4x  12  x  3
The solution appears to be x = -3. But x = -3 is not possible because it will make the
denominator of the fractions in the original equation equal to 0.
Therefore the equation has no solution.
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Problem 5. Solve the equation
x
x3
3


2
2
2
x 1 x  x x  x
Solution:
x
x3
3
x
x3
3





2
2
2
( x  1)( x  1) x( x  1) x( x  1)
x 1 x  x x  x
x
x3
3

x( x  1)( x  1) 
x( x  1)( x  1) 
x( x  1)( x  1), x  0, 1,  1
( x  1)( x  1)
x( x  1)
x( x  1)
 x 2  ( x  3)( x  1)  3( x  1)
 x 2  x 2  4x  3  3x  3   x  6
Problem 6. Solve
 x  6
The solution set is {6}
a b
  c, c  0 for x
x x
Solution:
a b
 a b
  c, c  0     x  cx, c  0  a  b  cx , c  0  cx  a  b, c  0
x x
 x x
ab
x
c
7. Careful! Squaring both sides of an equation does not necessarily lead to an equivalent
equation. The process of squaring both sides could result in extraneous solutions. The solutions
should be checked on the original equation to detect for the possibility of extraneous solutions.
Examples:
The solution of 2x = 4 is x = 2.
Squaring both sides leads to 4x 2  16  x 2  4  x  2
-2 is an extraneous solution, it does not check in the original equation.
Solve the equation 3  3x  1  x
Solution:
3  3x  1  x  3x  1  x  3 

3x  1
2  x  32
 3x  1  x 2  6x  9
 0  x 2  9x  8  0  0  ( x  8)( x  1)  0  x  8  0 or x  1  0  x  8 or x  1
The solution x =1 does not check, it is a extraneous solution.
Answer: the solution set is { 8 }
8. Careful! Dividing both sides of an equation by an expression containing the variable may result in
loosing some of the solutions because the divisor could be zero. Do not divide by a expressions
containing variables!
2
Example: Find the solution of the equation x  5x
Incorrect: x 2  5x 
x 2 5x

x5
x
x
Correct: x 2  5x  x 2  5x  0  x( x  5)  0  x  0 or x  5  0  x  0 or x  5
Note. In the process of dividing by x the root x = 0 was lost.
Answer: The solution set of x 2  5x is {0, 5}
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Solving linear and quadratic equations with absolute value.
Case 1. If a is a positive real number and u is an algebraic expression, then
|u| = a is equivalent to the two equations u = a or u = -a
Case 2. If a =0 and u is an algebraic expression, then
|u| = 0 is equivalent to the two equations u = 0
Case 3. If a is a negative real number and u is an algebraic expression, then
|u| = a has no solution.
Example 1. Solve the equation |3x| = 12
| 3x | 12  3x  12 or 3x  12  x  4 or x  4
Solution set: {-4, 4}
Example 2. Solve the equation |1 – 2x| + 6 = 9
| 1  2x | 6  9 | 1  2x | 3  1  2x  3 or 1  2x  3  2x  2 or  2x  4
 x  1 or x  2
Solution set: {-1, 2}
x 1
 1
2 3
x 1
x 1
x 1
  1 |   1 or   1  3x  2  6 or 3x  2  6
2 3
2 3
2 3
Example 3. Solve
 3x  8 or 3x  4
x
8
4
or x  
3
3
4
8
Solution set :  or  
3
3
Example 4. Solve |3x – 5| = -4
This equation does not have a solution because a = -4 (a is negative)
Example 5. Solve |6 – 3x | = 0
| 6  3x | 0  6  3x  0   3x  6  x  2,
The solution is {2}
Example 6. Solve the equation |3x - |2x+1|| = 4
Solution: |3x - |2x+1|| = 4  3x - |2x+1| = 4 or 3x - |2x+1| = - 4
 - |2x+1| = 4 – 3x or - |2x+1| = -4 -3x
 |2x+1| = 3x – 4 or |2x+1| = 3x +4
 2x + 1 = 3x – 4 or 2x + 1 = -(3x – 4) or 2x + 1 = 3x +4 or 2x + 1 = -(3x + 4)
 -x = -5 or 5x = 3 or -x = 3 or 5x = -5
3
or x= -3 or x = -1
 x = 5 or x 
5
3
is not a solution, x= -3 is not a solution, x = -1 is a
 x = 5 is a solution, x 
5
solution.
The solution set is {-1, 5}
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