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: Pre-Calculus - Chapter 2A
Chapter 2A - Solving Equations
Introduction and Review of Linear Equations
An equation is a statement which relates two or more numbers or algebraic expressions. For example,
the equation
24  6
is a statement relating the equality of the expression 2  4 with the number 6. This statement is always
true. The equation
x  3  12
is a statement relating the equality of the expression x  3 with the number 12. This statement may or
may not be true depending on what numerical value the variable x is assigned.
A solution is a value of the variable which makes the equation true. It can be seen by inspection that
x  9 is a solution of the above equation since
9  3  12
To verify that a number is a solution to an equation, replace the variable with that value. If the
resulting statement is true, the value is a solution, otherwise, it is not a solution.
We will explore methods for solving many different kinds of equations in this section. The simplest
kind is the linear equation, which is an equation that can be written in the form
ax  b  0
This kind of equation is solved by isolating the variable, namely, undoing what has already been done
to it.
Example:
Solve the equation 6  3x  x − 4 for x.
Solution:
Begin by moving all the variable expressions to one side of the equation and all the
numbers to the other side:
6  3x  x − 4
subtract 6 from both sides
3x  6 − 6  x − 4 − 6 Note the commutative property
3x  x − 10
now subtract x from both sides
3x − x  x − x − 10
x − x cancels to 0
2x  −10
divide both sides by 2
x  −5
Final solution
We can check our solution by replacing x with −5 in the original equation:
6  3−5  −5 − 4 ?
6  −15  −9 ?
− 9  −9 
Therefore, our solution is correct.
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: Pre-Calculus
©
: Pre-Calculus - Chapter 2A
Question:
Why does the strategy in the previous example work?
Answer:
It is easy to verify the strategy for an equation of the form ax  b  0 using the
properties of real numbers:
ax  b  −b  0  −b
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ax  0  0  −b
Additive Inverse Property
ax  −b
1  ax  1  −b
a
a
b
1x  −a
x  − ba
Identity for Addition
Multiplying by 1
a is the same as dividing by a
: Pre-Calculus
Multiplicative Inverse
Identity for Multiplication
©
: Pre-Calculus - Chapter 2A
Solving Quadratic Equations
A quadratic equation is an equation that can be written in the form
ax 2  bx  c  0
This is referred to as the standard form of a quadratic equation. We will discuss three methods for
solving quadratic equations: factoring, completing the square, and using the quadratic formula.
I. Solving by Factoring
The following theorem is the basis for solving equations by factoring:
Theorem:
If ab  0, then a  0 or b  0 or both.
Therefore, the following strategy can be applied to solve equations by factoring:
1. Write the equation in standard form (if it is not already in standard form)
2. Factor
3. Set each factor equal to zero and solve.
Example 1:
Solve x 2  5x − 6 for x
Solution:
We first write the equation in standard form:
x 2  5x − 6
subtract 5x from both sides
x − 5x  5x − 5x − 6
2
x 2 − 5x  −6
add 6 to both sides
x − 5x  6  −6  6
2
x 2 − 5x  6  0
now factor the left
x − 2x − 3  0
set each factor equal to 0 and solve
x − 2  0 or x − 3  0
x  2 or x  3
We can check both of our solutions:
2 2  52 − 6 ?
3 2  53 − 6 ?
4  10 − 6 ?
9  15 − 6 ?
4  4
9  9
Both 2 and 3 are solutions.
Example 2:
Solution:
©
Solve 2a 2  3a − 14  0 for a
The equation is already in standard form, so factor and solve:
: Pre-Calculus
©
: Pre-Calculus - Chapter 2A
2a 2  3a − 14  0
2a  7a − 2  0
2a  7  0 or a − 2  0
2a  −7 or a  2
a  − 7 or a  2
2
We can check both of our solutions:
22 2  32 − 14  0 ?
2− 7  2  3 − 7 − 14  0 ?
2
2
2 49   3− 7  − 14  0 ?
24  32 − 14  0 ?
4
2
49 − 21 − 14  0 ?
8  6 − 14  0 ?
2
2
49 − 21 − 28  0 ?
0  0
2
2
2
0  0
Therefore, both − 7 and 2 are solutions to the equation.
2
II. Solving by Completing the Square
The equation x 2  9 can easily be solved by knowing that 3 2  9 (so x must be 3). However, there is
another number whose square is 9. Recall that the product of two negative numbers is a positive
number. Thus x  −3 is also a solution to the equation since −3 2  −3−3  9.
In general, the solution to the equation x 2  c (where c is positive) is x  c or x  − c . This can be
written more concisely as x   c . This method is often referred to as the square root extration
method or simply the square root method.
Example 3:
Solve each of the following equations.
a)
x 2  40
b)
x − 2 2  17
Solution:
a)
The solution is given by (remember to simplify your radicals!)
x   40
x  2 10
b)
Note that if
x − 2 2  17
then
x − 2   17
or (adding 2 to both sides)
x  2  17
Given a quadratic equation ax 2  bx  c  0 (especially one which is not easily factored), we can
rewrite the equation so that the left-hand side is a perfect square (like the second example above). The
process, which works on every quadratic expression, is described below:
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: Pre-Calculus
©
Example 4:
Solution 1:
: Pre-Calculus - Chapter 2A
Solve y 2  7y  3  0 by completing the square
We begin by arranging the left-hand side to look like y 2  ky.
y 2  7y  3  0
subtract 3 from both sides
y  7y  −3
2
Now we half and square the linear coefficient (the number in front of y) and add this number to BOTH
sides of the equation:
y 2  7y  −3
y 2  7y  49  −3  49
4
4
49
−12
2
y  7y 

 49
4
4
4
y 2  7y  49  37
4
4
2
7
2
add
or 49 to both sides
4
get a common denominator on right
add the numerators on right
Why did we do this? Because the left-hand side is now a perfect square. Recall the special product
x  a 2  x 2  2ax  a 2
Notice that if we take one-half the coefficient of x and square it, we obtain
2a 2  a 2
2
This will work no matter what a is.
Since the left hand side is a perfect square we can ”undo” the square by taking the square root of both
sides.
y 2  7y  49  37
4
4
2
y 7
 37
2
4
y  7   37
2
4
“factor” the left
now solve as example 2 above
simplify the radical
37
y 7  
2
2
37
−7  37

y  −7 
2
2
2
Solution 2:
follows:
subtract 7 from both sides
2
final solution
Later it will be more useful when completing the square to organize our numbers as
y 2  7y  3  0
y 2  7y  3  0
y 2  7y 
7
2
2
7
2
3−
y 7
2
2
From here, we add 37 to both sides and continue as before.
4
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: Pre-Calculus
2
0
− 37  0
4
©
: Pre-Calculus - Chapter 2A
Solve 9x 2  10x  1  0 by completing the square.
Example 5:
Solution 1:
Since the coefficient of x 2 is not 1, we have to be more careful when writing the
2
left-hand side as x  kx:
9x 2  10x  1  0
subtract 1 from both sides
9x 2  10x  −1
divide both sides by 9
x 2  10 x  − 1
9
9
x 2  10 x  25  − 1  25
9
81
81
9
10
25
−9
25
x


 16
x2 
9
81
81
81
81
2
x 5
 16
9
81
x  5   16   4
81
9
9
x  −5  4
9
9
half and square 10 
9
5
9
2
or 25
81
common denominator on right
factor left side (perfect square)
solve using square roots
subtract 5 from both sides
9
x  − 5  4  − 1 or x  − 5 − 4  −1
9
9
9
9
9
Solution 2:
square here:
You must also be extremely careful if you use the alternate method of completing the
9 x 2  10 x 
9
9x 2  10x  1
9 x 2  10 x  1
9
2
2
5
1−9 5
9
9
2
− 16
9 x 5
9
9
2
5
9 x
9
2
x 5
9
0
0
0
0
 16
9
 16
81
Then proceed as above.
2
Question:
Why −9 5
9
Answer:
We added the 5 inside the parentheses, so the 9 distributes over it:
9
2
2
9 x 2  10 x  5
 9x 2  10x  9 5
9
9
9
in step 3?
2
so we have actually added 9 5
to our expression. To balance this, we must subtract 9 5
9
9
the same side or add it to the other side.
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: Pre-Calculus
2
from
©
: Pre-Calculus - Chapter 2A
III. Solving with the Quadratic Formula
We can use the technique of completing the square on the general quadratic equation ax 2  bx  c  0
to derive a general formula for the solutions to a quadratic equation:
ax 2  bx  c  0
subtract c from both sides
ax  bx  −c
divide both sides by a
2
c
x2  b
ax  −a
half and square b
a 
b2
c
b2
x2  b
a x  4a 2  − a  4a 2
2
2
−c4a
 b2
x 2  ba x  b 2 
a4a
4a
4a
2
2
x 2  ba x  b 2  b − 4ac
4a 2
4a
2
2
x b
 b − 4ac
2a
4a 2
b
2a
2
2
or b 2
4a
get a common denominator on right
add numerators
factor left side
use square root method
b 2 − 4ac
4a 2
simplify radical ( 4a 2  2a)
 b 2 − 4ac
x b 
2a
2a
2
b − 4ac
x− b 
2a
2a
2
−b  b − 4ac
x
2a
subtract b from both sides
2a
x b  
2a
combine common denominator fractions
The final step gives us the Quadratic Formula.
The solution to any quadratic equation of the form
ax 2  bx  c  0
−b  b 2 − 4ac
x
2a
is given by
This formula must be memorized.
Remember: a is the coefficient of x 2 , b is the coefficient of x, and c is the constant term in this formula.
Example 6:
Solution:
©
Solve x 2  7x  3  0 using the Quadratic Formula
For the quadratic formula a  1, b  7, and c  3.
: Pre-Calculus
©
: Pre-Calculus - Chapter 2A
x
−7  7 2 − 413
substitute in formula
21
−7  49 − 12
2
−7  37
x
. Thus, we have
2
−7  37
x
≈ −0. 458 62 or
2
−7 − 37
x
≈ −6. 541 4
2
x
Example 7:
Solve 5x 2  x − 1 using any method.
Solution:
First write the equation in standard form. The quadratic formula is generally the
easiest method to use when the quadratic does not easily factor.
5x 2  4x − 1
5x 2 − 4x  1  0
For the quadratic formula, a  5, b  −4 (be careful with signs), and c  1.
x
−−4  −4 2 − 451
25
4  −20
simplify the complex radical (*See Note Below)
10
4  2i 5
x
now factor and cancel the common 2
10
2 2i 5
x
10
2i 5
x
5
Note that the solutions above are complex numbers. Be sure to simplify any radicals in your final
solutions!
*Note:
In simplifying the complex radical, recall that
x
i
−1
So
−20 

4  5  −1
4  5  −1
 2 5i
 2i 5
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: Pre-Calculus
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: Pre-Calculus - Chapter 2A
Equations in Quadratic Form
An equation is in quadratic form when one power of x is twice as large as the (only) other power of x.
More formally, an equation in quadratic form can be written as ax 2n  bx n  c  0. To solve an
equation of this type, we first “change” the variable by renaming things. For example, if we define
u  x n in the equation above, then u 2  x n  2  x 2n . Therefore, our equation above becomes
au 2  bu  c  0 which is a quadratic equation we can solve for u. Once we solve for u, we find x by
raising our solution(s) to the 1n th power. Remember the 1/n th power of a number a, a 1/n , is a number
b such that b n  a. That is,
a 1/n  b  b n  a .
Example 1:
Solve b 4 − 9b 2 − 112  0
Solution:
The equation above is in quadratic form, so let u be the smaller power of b; that is,
2
2
u  b . Then u  b 2  2  b 4 , so our equation becomes u 2 − 9u − 112  0. Thus,
9  −9 2 − 41−112
u
21
9  529
2
9

23
u
2
u  16 or u  −7
u
Since u  b 2 , we have
b 2  16 or b 2  −7
b  4 or b  i 7
2
1
Solve x 3 − 4x 3  −3
1
2
Write in standard form, and then let u  x 3 . Then u 2  x 3 :
Example 2:
Solution:
2
3
x
x
2
3
− 4x
− 4x
1
3
1
3
 −3
3  0
u − 4u  3  0
2
u − 3u − 1  0
u  3 or u  1
Since u  x
1
3
x
x
1
3
1
3
3
 3 or x
1
3
 3 3 or x
1
1
3
3
 13
x  27 or x  1
Question:
Answer:
©
In the equation 2x 10 − 7x 5 − 8  0, what should u equal?
Set u  x 5 . The equation is then transformied into 2u 2 − 7u − 8 .
: Pre-Calculus
©
: Pre-Calculus - Chapter 2A
Rational Equations
A rational equation is an equation which involves rational expressions, or fractions.
The easiest way to solve a rational equation is to eliminate the fractions by multiplying both sides of the
equation by the least common denominator.
Example 1:
Solve 2x − 3  −1
x−2
Solution:
We begin as if we were going to add the rational expression on the left-hand side. The
LCD is xx − 2. However, instead of getting a common denominator, we multiply both sides of the
equation by this LCD:
2 − 3
xx − 2  −1xx − 2
x
x−2
2 xx − 2 − 3 xx − 2  −x 2 − 2x
x
x−2
2x − 2 − 3x  −x 2  2x
2x − 4 − 3x  −x 2  2x
− x − 4  −x 2  2x
x 2 − 3x − 4  0
x − 4x  1  0
x  4 or x  −1
Let’s check both solutions:
2 − 3  −1?
4−2
4
1 − 3  −1 ?
2
2
−1  −1 
3
2 −
 −1?
−1
−1 − 2
−2 − 3  −1 ?
−3
−2 − −1  −1 ?
−1  −1 
This next example is very important as it demonstrates how easy it is to get a wrong answer when
solving an equation involving rational expressions in x.
5  x1
x−4
x−4
Solution:
The LCD is x − 4 (remember the most it occurs in any one fraction is once). Multiply
both sides of the equation by x − 4:
Example 2:
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Solve 2 
: Pre-Calculus
©
: Pre-Calculus - Chapter 2A
5
x − 4  x  1 x − 4
x−4
x−4
5
2x − 4 
x − 4  x  1
x−4
2x − 8  5  x  1
2
2x − 3  x  1
x4
But watch what happens when we check our solution:
2 5  41
4−4
4−4
5
2
 5 UNDEFINED!!!
0
0
Therefore, our ”solution” is not really a solution at all (this is called an extraneous solution).
How did we get x  4 as a solution?
The reason is based on restrictions on our variable. In the first equation
2 5  x1
x−4
x−4
x cannot be 4, since division by 0 is not allowed. However, when we eliminate the fractions, this
restriction is eliminated:
2x − 4  5  x  1
Question:
Answer:
24 − 4  5  4  1
55
Not only is 4 a valid number for this equation, it is a solution! This is why you must check solutions to
rational equations. We always need to be on the lookout for the introduction of false solutions to the
original equation.
So if x  4 is not a solution, what is a solution to this equation? It turns out that this particular equation
does not have any solutions. The easiest way to see that there are no real solutions is to look at the
graphs of 2  5 and x  1 . There is a real solution if and only if their graphs intersect.
x−4
x−4
5
2 + x - 4 in red
4
x + 1 in blue
x-4
Notice that for any x the graph of 2  5 always lies above the graph of x  1 . Thus, the equation
x−4
x−4
has no real solution. You will be asked in one of the exercises to show algebraically that this equation
has no solution.
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: Pre-Calculus
©
: Pre-Calculus - Chapter 2A
1  1 8 .
x−2
x−2
Solution:
As in the previous examples we multiply by the least common denominator, which is
x − 2. Remember, if we get x  2 as a solution, it is extraneous and not valid.
Example 3:
Find all solutions of the equation 2 
2
2x − 2 
1  1 8
x−2
x−2
multiply by x − 2
1 x − 2  1x − 2  8 x − 2
x−2
x−2
2x − 4  1  x − 2  8
2x − 3  x  6
subtract x from both sides
x−3  6
add 3 to both sides
x9
Let’s check that x  9 is indeed a solution.
2
©
: Pre-Calculus
cancel denominators
1
9 − 2
2 1
7
14  1
7
7
15
7
8
9 − 2
 1 8
7
 7  8
7
7
 15
7
 1
©
: Pre-Calculus - Chapter 2A
Radical Equations
When an equation contains one or more radicals, the following strategy can be used:
1. Isolate one radical
2. Raise both sides of the equation to the appropriate power to remove the radical (for example,
if the radical is a square root, square both sides of the equation)
3. If necessary, repeat the process until all radicals have been removed, then solve the resulting
equation.
Example 1:
Solution:
Solve 3 x  7 − 4  8
Isolate the radical x  7 , then square both sides.
3 x7 −4  8
add 4 to both sides
3 x  7  12
divide both sides by 3
x7  4
x7
2
square both sides
 4 2 square and square root cancel
x  7  16
subtract 7 from both sides
x9
Let’s check our answer:
3 97 −4  8?
3 16 − 4  8 ?
34 − 4  8 ?
12 − 4  8 ?
8  8
As with rational equations, checking our solution will be very important in finding any extraneous
solutions.
Example 2:
Solution:
Solve x  5  1  x
x5 1  x
subtract 1 from both sides
x5  x−1
square both sides
2
x5
 x − 1 2
cancel on left, FOIL on right
x  5  x 2 − 2x  1
subtract x from both sides
5  x − 3x  1
subtract 5 from both sides
0  x 2 − 3x − 4
factor right side
0  x − 4x  1
set each factor equal to 0 and solve
2
x  4 or x  −1
But watch what happens when we check our solutions:
©
: Pre-Calculus
©
: Pre-Calculus - Chapter 2A
45 1  4?
−1  5  1  −1 ?
9 1  4?
4  1  −1 ?
31  4?
2  1  −1 ?
4  4
3  −1 X
Therefore, our only solution is x  4. The second ”solution”, x  −1, is an extraneous solution which
arose when we squared both sides of the equation.
As with rational equations, it is very important that we check solutions to make sure they satisfy the
original equation.
Sometimes it is necessary to isolate and eliminate more than one radical in an equation. In many cases,
the only way to do this is one at a time!
Example 3:
Solve 2 x − x − 3  3
Solution:
Although it does not matter which radical is eliminated first, it is often best to save the
”simplest” one for last. Therefore, we will eliminate x − 3 first.
2 x − x−3  3
add x − 3 to both sides
2 x  3 x−3
subtract 3 from both sides
2 x −3 
2
2 x − 3 
x−3
square both sides
2
x−3
FOIL left, cancel on right
4x − 12 x  9  x − 3
subtract x from both sides
3x − 12 x  9  −3
add 12 x and 3 to both sides
3x  12  12 x
3x  4
 x
12
x4  x
4
x  4 2   x 2
4
x 2  8x  16  x
16
2
x  8x  16  16x
divide by 12
factor and cancel numerator
square both sides again
FOIL numerator on left, cancel on right
multiply by 16 to eliminate fraction
subtract 16x from both sides
x − 8x  16  0
factor left side
x − 4  0
set factor equal to 0 and solve
2
2
x4
Let’s check our answer:
2 4 − 4−3  3?
22 − 1  3 ?
4−1  3?
3  3
Thus, our solution is valid.
©
: Pre-Calculus
©
: Pre-Calculus - Chapter 2A
Absolute Value Equations
The method for solving equations which contain the absolute value function comes from the definition
of absolute value. Recall that
x
|x| 
if x ≥ 0
−x if x  0
Therefore, for the absolute value of an expression to equal a number, either the expression or the
negative of the expression must equal that number. To solve such equations we rewrite every absolute
value equation as two separate equations.
Example 1:
Solution:
then
Solve |x|  5.
There are two cases to consider the first case is x ≥ 0 and the second is x  0. If x ≥ 0,
|x|  x  5
So x  5 is a solution, but we need to be careful here. Is 5 ≥ 0? Remember we got the solution by
assuming that x ≥ 0 so we need to check that our solution satisfies this condition.
For the other case, if x  0
|x|  −x  5
x  −5 .
Notice that in this case the solution x  −5 satisfies the assumption that x  0. Thus, we have two
solutions x  5.
Example 2:
Solve |y − 7|  3
Solution:
We again consider the cases where the argument of the absolute value function is
greater or equal than zero, or less than zero. That is y − 7 ≥ 0 or y − 7  0.We rewrite the equation into
two equations: y − 7  3 and −y − 7  3 (Remember that we must change all or none of the signs of
the expression! |y − 7| is NOT the same as y  7!!!)
case 1: y − 7 ≥ 0 |y − 7|  y − 7  3 case 2: y − 7  0 |y − 7|  −y − 7  3
y  10
−y  7  3
−y  −4
y4
Notice: in both cases the solution minus 7 satisfies the appropriate assumption. That is, 10 − 7 ≥ 0 and
4 − 7  0.
Let’s check our solutions:
|10 − 7|  3 ? |4 − 7|  3 ?
Both solutions are valid.
©
: Pre-Calculus
|3|  3 ?
|−3|  3 ?
3  3
3  3
©
: Pre-Calculus - Chapter 2A
A geometrical interpretation of absolute value inequalities.
Recall that |a − b| refers to the distance between the points a and b on the number line. Therefore, the
equation
|x − 7|  3
can be thought of as, ”the distance between x and 7 is 3?”, or more appropriately, ”what number or
numbers are 3 units from 7 on the number line?” See the illustration below:
3 units each
4
Example 3:
Solution:
7
10
Solve the equation |xx − 1|  2.
As in the previous examples we break the problem into two cases.
Case 1: xx − 1 ≥ 0
|xx − 1|  xx − 1  2
x2 − x − 2  0
x  1x − 2  0
Thus, we have two solutions x  −1 and x  2.
Case 2: xx − 1  0
|xx − 1|  −xx − 1  2
x2 − x  2  0
1 1−8
2
7
x 1 
i
2
2
In this case there are no real solutions. Thus, the final answer is x  −1 and x  2.
x
©
: Pre-Calculus
©
: Pre-Calculus - Chapter 2A
Equations in Several Variables
Many equations or formulas involve several variables. It is often useful to ”solve” these types of
equations for one particular variable in terms of all the others. All of the techniques discussed in this
section can be used to solve such equations; the key is to determine what kind of equation you have in a
given variable.
Example 1:
The formula for the surface area of a cylinder is given by S  2r 2  2rh where r is
the radius of the cylinder and h is the height. Solve this equation for h.
Solution:
Although there is a square in the equation, h is not squared, so this is a linear equation
in h. We solve by isolating h. As we do, we treat every other variable as if it were just a number
S  2r 2  2rh
subtract 2r 2 from both sides
S − 2r 2  2r 2 − 2r 2  2rh
2rh  S − 2r 2
2rh  S − 2r 2
2r
2r
2
S
−
2r
h
2r
divide both sides by 2r
note we cannot cancel the 2r!!!
Example 2:
Solve the equation in Example 1 for r.
Solution:
This time the desired variable r is squared, so the equation is a quadratic equation. We
use the quadratic formula to solve for r.
S  2r 2  2rh
0  2r 2  2rh − S
Here a  2, b  2h, and c  −S:
−2h  2h 2 − 42−S
r
substitute into formula
22
−2h  4 2 h 2  8S
4
−2h  4 2 h 2  2S

simplify radical
4
−2h  2  2 h 2  2S
factor 2 in numerator

4
2 −h   2 h 2  2S
reduce

4
−h   2 h 2  2S
r
2
It appears that we have two solutions. However, for this particular equation the variable r represents a
length so we must have r ≥ 0. Thus, we can discard the solution with the minus sign in front of the
radical, and the radius of a cylinder in terms of its height and surface area equals

−h   2 h 2  2S
2
As you can see, solving an equation for one variable can be very different from solving the same
equation for a different variable.
r
©
: Pre-Calculus
©
: Pre-Calculus - Chapter 2A
Exercises for Chapter 2A - Solving Equations
1.
2.
3.
4.
5.
6.
For problems 1-3, show the following are solutions to the given equation.
x−1  x1;x  0
x−1
x1
−12  12y 2  10y; y  3
2
x − x  20; x  25
For problems 4-6, solve the equations for the given variable.
−12  12y  10y
2x  1  −2x  1
y
y
5y
1
−
 y−
3
4
6
7.
8.
9.
10.
11.
For problems 7-11, solve by factoring
x 2 − 7x − 8  0
x 2  −3x
7x 2  16x  −9
16x 2 − 18  2x
x 3 − x 2 − 30x  0
12.
13.
14.
15.
16.
17.
18.
For problems 12-15, solve by completing the square
x 2  6x − 10  0
3y 2  12y  8  0
a 2 − 3a  9
2x 2  5  x
Solve equation 12 by using the quadratic formula
Solve equation 13 by using the quadratic formula
Solve equation 15 by using the quadratic formula.
19.
20.
21.
22.
23.
For problems 19-23, solve the equation by the method of your choice.
x 2 − 13x  30  0
3c 2 − 21  0
−6t 2  5t  3
y 2 − 4y  1  0
3x 2 − 4x − 15  0
24. Brady throws a ball into the air. The equation s  −16t 2  30t  4 gives the height of the ball s (in
feet) t seconds after throwing it. When will the ball hit the ground after he throws it?
For problems 25-29, solve the equations for the given variable.
25. x 4 − 10x 2  9  0
2
1
1
1
26. x 3 − 4x 3  −3
27. b 4 − 3b 2 − 28  0
28. y 2  4y 4 − 12  0
29. −16x −2 − 6x −1  1  0
For problems 30-36, solve for the given variable.
30. 1 − 5x  3
4
©
: Pre-Calculus
©
31.
32.
33.
34.
35.
36.
: Pre-Calculus - Chapter 2A
2  1
y−2
y1
x−1  x1
x−1
x1
a − 1  a  11
a−1
2
y−3
 22
y − 16
y 2 − 4y
2  3x  0
x5
x−4
x 2 − 4x  4  x 2  5x  6  0
x2
x−2
5  x  1 has no real solutions. Show,
x−4
x−4
algebraically that this equation has no solutions.
37. We saw in an example that the equation 2 
For problems 38-49, solve for the given variable.
38. 3  2 x  1  7
39. 3 x  7 − 4  8
40. y − 4 − y  −4
41. x  5  x − 1
42. z − z  20
43. y − 5  y  4 − 1
44. |x − 9|  3
45. |6b − 3| − 12  0
46. 2 x − 5  3
5
47. |y  6|  |2y − 2|
48. 3|2x  1| − 4  6
49. 5|x − 1| − 6  −9
50.
51.
52.
53.
54.
55.
56.
©
For problems 50-56, solve the equation for the indicated variable.
V  r 2 h for h
V  r 2 h for r
V 1  P 2 for V
1
V2
P1
V 1  P 2 for P
1
V2
P1
A  1 hb 1  b 2  for b 1
2
F for F
P
FU
v  2as for s
: Pre-Calculus
©
: Pre-Calculus - Chapter 2A
Answers to Exercises for Chapter 2A - Solving Equations
1.
Substitute the value into the equation:
0−1  01 ?
01
0−1
−1  1 ?
−1
1
− 1  −1
2.
2
− 12  12 3
 10 3
2
2
9
− 12  12
 10 3
4
2
− 12  27  15 ?
?
?
15  15
3.
25 − 25  20 ?
25 − 5  20 ?
20  20
4.
− 12  12y  10y
− 12  12y − 12y  10y − 12y
− 12  −2y
−12  −2y
−2
−2
y6
5.
2x  1  −2x  1
2x  1 − 1  −2x  1 − 1
2x  −2x
2x  2x  −2x  2x
4x  0
4x  0
4
4
x0
6.
Although this does not look like a linear equation at first, it will become one when we multiply both
sides by the lowest common denominator:
y
y
5y
−
 12 y −
12 1 
3
4
6
12  4y − 3y  12y − 10y
y  12  2y
y  12 − y  2y − y
12  y
©
: Pre-Calculus
©
: Pre-Calculus - Chapter 2A
7.
x 2 − 7x − 8  0
x − 8x  1  0
x − 8  0 or x  1  0; x  8 or x  −1
8.
Remember to put all terms on one side of the equation:
x 2  −3x
x 2  3x  0
xx  3  0
x  0 or x  3  0; x  −3
NOTICE that if we had divided both sides by x, we would have lost the solution x  0:
x 2  −3x
x  −3 ONLY
x
x
9.
7x 2  16x  −9
7x 2  16x  9  0
7x  9x  1  0
7x  9  0 or x  1  0
7x  −9 or x  −1
x  − 9 or x  −1
7
10.
16x 2 − 18  −2x
16x 2  2x − 18  0
28x 2  x − 9  0
28x  9x − 1  0
8x  9  0 or x − 1  0
8x  −9 or x  1; x  − 9 or x  1
8
11. Although this equation is not quadratic, it can still be factored:
x 3 − x 2 − 30x  0
xx 2 − x − 30  0
xx − 6x  5  0
x  0 or x − 6  0 or x  5  0
x  0 or x  6 or x  −5
12.
x 2  6x − 10  0
x 2  6x  9 − 10 − 9  0
x 2  6x  9  19
x  3 2  19
x  3   19
x  −3  19
©
: Pre-Calculus
©
: Pre-Calculus - Chapter 2A
13.
3y 2  12y  8  0
3y 2  4y  8  0
3y 2  4y  4  8 − 12  0 (don’t forget to distribute the 3!)
3y  2 2  4
y  2 2  4
3
4
3
2
y2  
(now rationalize)
3
y2  
2 3
3
2 3
y  −2 
(common denominator)
3
−6  2 3
y
3
y2  
14.
a 2 − 3a  9
3
2
3
a−
2
a− 3
2
a 2 − 3a 
2
 9
3
2
2
2
 9 9
4
2
 45
4
45
a− 3  
2
2
3  45
33 5
a

(simplify radical)
2
2
15.
2x 2  5  x
2 x2 − 1 x 
2
1
4
2x 2 − x  5  0
2 x2 − 1 x  5  0
2
2
2
5−2 1
0
4
2 x− 1
4
x− 1
4
2
2
 − 39
8
 − 39
16
x − 1   − 39
4
16
39
x− 1  
i
4
4
39 i
x 1 
4
4
1  39 i
x
4
©
: Pre-Calculus
©
: Pre-Calculus - Chapter 2A
16. Let a  1, b  6, and c  −10 in the quadratic formula:
x
−6  6 2 − 41−10
21
−6  36  40
2
−6  76

2
−6  2 19

2
2 −3  19

 −3  19
2

17.
Let a  3, b  12, and c  8 in the quadratic formula:
y
−12  12 2 − 438
23
−12  144 − 96
6
−12  48

6
−12  4 3

6
2 −6  2 3
−6  2 3


6
3

18.
2x 2 − x  5  0
Let a  2, b  −1, and c  5 in the quadratic formula:
x
−−1  −1 2 − 425
22
1  1 − 40
4
1  −39

4
1  39 i

4

19. This equation factors easily, although other methods could be used:
x 2 − 13x  30  0
x − 3x − 10  0
x − 3  0 or x − 10  0
x  3 or x  10
20. We can easily solve this equation by isolating the variable, although other methods also work:
3c 2 − 21  0
3c 2  21
c2  7
c 7
©
: Pre-Calculus
©
: Pre-Calculus - Chapter 2A
21. The quadratic formula is probably best for this (here a  −6, b  5, and c  −3):
− 6t 2  5t − 3  0
t
−5  5 2 − 4−6−3
2−6
−5  −47
−12
−5  47 i

−12

22. Again, we use quadratic formula with a  1, b  −4, and c  1:
y
4  −4 2 − 411
21
4  12
2
42 3

2
2 2 3

2
 2 3

23. We will use the quadratic formula to make an observation later. Here a  3, b  −4, and c  −15:
x
−−4  −4 2 − 43−15
23
4  196
6
4

14

6
18
x
or x  − 10
6
6
5
x  3 or x  −
3

24. Set s  0 and solve:
− 16t 2  30t  4  0
t
−30  30 2 − 4−164
2−16
−30  1156
−32
−30

34

−32
t  −64 or t  4
−32
−32
1
t  2 or t  − (not valid)
8

2 seconds after releasing the ball.
25. Let u  x 2 . Then u 2  x 4 , so our equation becomes
©
: Pre-Calculus
©
: Pre-Calculus - Chapter 2A
u 2 − 10u  9  0
u − 9u − 1  0
u  9 or u  1
Therefore,
x 2  9 or x 2  1
x  3 or x  1
1
3
2
3
26. Let u  x . Then u 2  x , so our equation becomes
u 2 − 4u  −3
u 2 − 4u  3  0
u − 3u − 1  0
u  3 or u  1
x
1
3
 3 or x
1
3
1
x  3 3 or x  1 3
x  27 or x  1
27. Let u  b 2 . Then u 2  b 4 , so our equation becomes
u 2 − 3u − 28  0
u − 7u  4  0
u  7 or u  −4
b 2  7 or b 2  −4
b   7 or b  2i
1
4
1
2
28. Let u  y . Then u 2  y , so our equation becomes
u 2  4u − 12  0
u  6u − 2  0
u  −6 or u  2
y
1
4
 −6(no solution) or y
y
24
 16
29. First, rewrite the equation in fraction form:
−16 − 6  1  0
x
x2
2
Now eliminate the fractions by multiplying by x :
− 16 − 6x  x 2  0
x 2 − 6x − 16  0
x − 8x  2  0
x  8 or x  −2
30. LCD is 4x:
©
: Pre-Calculus
1
4
2
©
: Pre-Calculus - Chapter 2A
4x1 − 4x 5x
 4x 3
4
4x − 20  3x
x − 20  0
x  20
Check:
1− 5 
20
1− 1 
4
3 ?
4
3 
4
x  20 is the solution to the equation.
31. LCD is y  1y − 2 (or simply cross-multiply since the fractions are isolated on both sides):
y  1y − 2
2
y1
 y  1y − 2
1
y−2
2y − 2  1y  1
2y − 4  y  1
y5
Check:
2  1
?
51
5−2
2  1 
3
6
Therefore, x  5 is the solution.
32. LCD is x  1x − 1:
x  1x − 1 x − 1  x  1x − 1 x  1
x1
x−1
x − 1x − 1  x  1x  1 remember to FOIL
x 2 − 2x  1  x 2  2x  1
− 4x  0
x0
Check:
0−1  01 ?
01
0−1
− 1  −1 
Therefore, x  0 is the solution
33. LCD is 2a − 1:
2a − 1 a − 1  2a − 1 a  11
2
a−1
a − 1a − 1  2a  11
a 2 − 2a  1  2a  22 quadratic equation
a 2 − 4a − 21  0
a − 7a  3  0
a  7 or a  −3
Check:
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7 − 1  7  11 ?
7−1
2
6  18 ?
6
2
33 
−3 − 1 
2
−4 
2
−2 
−3  11 ?
−3 − 1
8 ?
−4
−2 
Therefore, x  7 and x  −3 are the solutions
34. Factor first:
y−3
 22
y 2 − 4y
y − 16
y−3
2

y  4y − 4
yy − 4
The LCD is yy − 4y  4:
yy − 4y  4
y−3
2
 yy − 4y  4
yy − 4
y  4y − 4
y  4y − 3  2y
y 2  y − 12  2y
y 2 − y − 12  0
y − 4y  3  0
y  4 or y  −3
Check:
−3 − 3
2

−3 2 − 16
−3 2 − 4−3
−6 
2
9 − 16
9  12
− 6  −2 
7
21
4−3
2

4 2 − 16
4 2 − 44
2
1

16 − 16
16 − 16
1  2 Undefined
0
0
Therefore, the only solution is y  −3.
35. The LCD is x − 4x  5:
2
 x − 4x  5 3x
 0x − 4x  5
x − 4x  5
x5
x−4
2x  5  3xx − 4  0
2x  10  3x 2 − 12x  0
3x 2 − 10x  10  0 use quadratic formula
x
10  −10 2 − 4310
23
10  −20
6
10  2i 5

6
2 5i 5

6
5i 5

3

Since this is a complex solution, it will not be an extraneous solution.
36. First, see if factoring helps. Note that since this is multiplication, the LCD is not necessary!
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x 2 − 4x  4  x 2  5x  6  0
x2
x−2
2
x − 2
x  2x  3

0
x2
x−2
x − 2x  3  0
x  2 or x  −3
Check:
2 2 − 42  4 2 2  52  6

0 ?
2  2
2 − 2
0  20  0 UNDEFINED
0
4
−3 2 − 4−3  4 −3 2  5−3  6

0
−3  2
−3 − 2
25  0  0 ?
−1 −5
00 
Therefore, the only solution is x  −3.
37. Suppose 2  5  x  1 has a solution. Then it must be true that
x−4
x−4
2 5  x1
x−4
x−4
2x − 4
 5  x1
x−4
x−4
x−4
2x − 8  5  x  1
x−4
x−4
The only way two fractions with the same denominator can be equal is if the numerators are equal.
Thus, we must have
2x − 8  5  x  1
x−3  1
x  4.
Thus, if we have a solution it must equal 4, but 4 is not acceptible, since we cannot divide by 0.
38. First, isolate the radical, then square both sides:
32 x1  7
2 x1  4
x1  2
x1  4
x3
Check:
32 31  7 ?
32 4  7 ?
34  7 
Therefore, x  3 is the solution.
39. Again, isolate the radical and square both sides:
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: Pre-Calculus
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: Pre-Calculus - Chapter 2A
3 x7 −4  8
3 x  7  12
x7  4
x  7  16
x9
Check:
3 97 −4  8 ?
3 16 − 4  8 ?
12 − 4  8 
Therefore, x  9 is the solution.
40. Isolate the radical and square both sides. This time, you have a quadratic equation to solve after
squaring:
y − 4 − y  −4
y−4  y−4
y − 4  y 2 − 8y  16 (remember FOIL)
0  y 2 − 9y  20
0  y − 5y − 4
y  5 or y  4
Check:
5 − 4 − 5  −4 ?
4 − 4 − 4  −4 ?
1 − 5  −4 ?
0 − 4  −4 ?
1 − 5  −4 
0 − 4  −4 
Therefore, the solutions are x  5 and x  4.
41. As in the previous problem, isolate (already done), square and solve the resulting quadratic equation:
x5  x−1
x  5  x 2 − 2x  1
0  x 2 − 3x − 4
0  x − 4x  1
x  4 or x  −1
Check:
45  4−1 ?
−1  5  −1 − 1 ?
9 3 
4  −2 X
Therefore, the only solution is x  4.
42. Isolate the radical, square, and solve the resulting quadratic equation:
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: Pre-Calculus
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: Pre-Calculus - Chapter 2A
z − z  20
z − 20 
z
z2
− 40z  400  z
z2
− 41z  400  0
41  −41 2 − 41400
2
41

9

2
41
 9  25 or z  41 − 9  16
z
2
2
z
Check:
25 − 25  20 ?
16 − 16  20 ?
25 − 5  20 
16 − 4  20 X
Therefore, the only solution is x  25.
43. This time we must isolate one radical at a time:
y−5 
y4 −1
y − 5  y  4 − 2 y  4  1 (FOIL!!!)
y−5  y5−2 y4
− 10  −2 y  4
5
y4
25  y  4
y  21
Check:
21 − 5 
16 
21  4 − 1 ?
25 − 1 ?
4  5−1 
Therefore, the solution is x  21.
44. The absolute value leads to 2 equations:
|x − 9|  3
x − 9  3 −x − 9  3
x  12
−x  9  3
−x  −6
x6
Therefore, the solutions are x  12 and x  6. It is advisable to check that both solutions are valid.
45. Again write 2 equations (note the −12 does not change sign since it is not in the absolute value):
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: Pre-Calculus
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: Pre-Calculus - Chapter 2A
|6b − 3| − 12  0
6b − 3 − 12  0
− 6b − 3 − 12  0
6b − 15  0
− 6b  3 − 12  0
6b  15
b 5
2
− 6b − 9  0
− 6b  9
b  −3
2
Therefore, the solutions are b  5 and b  − 3 . It is advisable to check that both solutions are valid.
2
2
46. Write the 2 equations:
2x−5  3
5
2x−5  3 − 2x−5  3
5
5
2x  8
2
− x5  3
5
5
x  20
− 2 x  −2
5
x5
Therefore, the solutions are x  20 and x  5. It is advisable to check that both solutions are valid.
47. Although both absolute values change sign, you only have to change one side to solve (the other leads
to the same equations):
|y  6|  |2y − 2|
y  6  2y − 2 y  6  −2y − 2
−y  −8
y  6  −2y  2
y8
3y  −4
y  −4
3
Therefore, the solutions are y  8 and y  − 4 . It is advisable to check that both solutions are valid.
3
48. It is recommended to isolate the absolute value before writing the 2 equations to avoid sign errors:
3|2x  1| − 4  6
3|2x  1|  10
|2x  1|  10
3
2x  1  10
3
7
2x 
3
7
x
6
−2x  1  10
3
10
−2x − 1 
3
13
−2x 
3
13
x−
6
Therefore the solutions are x  7 and x  − 13 . It is advisable to check that both solutions are
6
6
valid.
49. Again isolate the absolute value:
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: Pre-Calculus
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: Pre-Calculus - Chapter 2A
5|x − 1| − 6  −9
5|x − 1|  −3
|x − 1|  − 3
5
Recall that absolute value represents the distance from zero on the number line. Therefore, it cannot
be negative, so this equation has no solutions.
50. The equation is linear in h, so we solve by isolating the variable:
V  r 2 h divide by r 2
V h
r 2
51. The equation is now quadratic in r. This quadratic can be solved by isolating r 2 and taking the square
root:
V  r 2 h divide by h
V  r2
h
r V
h
(In general, r represents the radius of a cylinder, so only the positive square root would make practical
sense).
52. The LCD is V 2 P 1 , so multiply both sides:
V 2 P 1  V 1  P 2 V 2 P 1 
V2
P1
P1V1  P2V2
V1  P2V2
P1
53. As in the previous problem, multiply by the LCD V 2 P 1 :
V 2 P 1  V 1  P 2 V 2 P 1 
V2
P1
P1V1  P2V2
P1  P2V2
V1
54. The equation is linear in b 2 , so solve by isolating the variable:
A  1 hb 1  b 2 
2
A  1 hb 1  1 hb 2
2
2
1
1
A − hb 2  hb 1
2
2
2A − hb 2  hb 1
2A − hb 2  b
1
h
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: Pre-Calculus - Chapter 2A
Other methods of solution are possible.
55. This equation is rational in F, so multiply by the LCD F  U:
F F  U
F  UP 
FU
PF  PU  F isolate F terms
PF − F  −PU
FP − 1  −PU
F  PU
1−P
56. Square both sides to eliminate the radical:
v
2as
 2as
2
s v
2a
v2
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: Pre-Calculus - Chapter 2B
Chapter 2B - Solving Inequalities
Introduction
An inequality is a statement comparing two quantities which may not be the same. The symbols , ,
≤, or ≥ are used in place of  to indicate which quantity is larger. The symbols are read as follows:
ab
a is less than b
a≤b
a is less than or equal to b
ab
a is greater than b
a ≥ b a is greater than or equal to b
Question:
a
a
a
a
a
Example 1:
If a  5 and b  8, which of the following statements are true?
b
(Answer:
b
(Answer:
≥b
(Answer:
b
(Answer:
≤b
(Answer:
False)
False)
False)
True)
True)
We learned earlier that
|3 − 5| ≠ 3  5
By evaluating each side, we find that
|3 − 5|  |−2|  2
35  8
Therefore, |3 − 5|  3  5
Example 2:
Find values of x which satisfy the inequality
x≥4
Solution:
Note that, if x  4, then the inequality is true since 4 ≥ 4 (This may not make sense at
first, but remember that the symbol means ”greater than or equal to”). Since 6 ≥ 4, then 6 also satisfies
the inequality. So does 4 12 . In fact, any value 4 or larger will make this inequality true.
Each of these numbers (like 4, 6, and 4 12 ) is a solution to the inequality. To solve an inequality means
to find all possible solutions of the inequality. Notice that this inequality has an infinite number of
solutions. We will find that this is true for most inequalities. Therefore, it is not feasible to list all the
solutions to an inequality. In general, there are three ways to describe the solutions of an inequality:
x≥4
In a statement or algebraic form
In graphical form on a number line
4
Using interval notation
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: Pre-Calculus
4, 
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: Pre-Calculus - Chapter 2B
Interval notation is used to describe a set of real numbers between 2 given numbers. The set a, b
represents the set of all real numbers x such that a  x  b. (All numbers between a and b). The set
a, b represents the set of all real numbers x such that a ≤ x ≤ b. In this latter case we include the two
endpoints of the interval. The symbols − and  are used to describe all numbers less than a given and
greater than a given number, respectively. The set −,  is used to represent all real numbers.
In summary open parentheses mean exclude the endpoint and square brackets mean include the
endpoint.
Question:
Which of the following numbers (−7, −5, −3, −1, 0) are in the interval −5, −1 ?
Answer:
−7 is not in the interval, −5 is in the interval, −3 is in the interval, −1 is not in the
interval, 0 is not in the interval
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: Pre-Calculus
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: Pre-Calculus - Chapter 2B
Linear Inequalities
To solve a linear inequality, isolate the variable in the same fashion as solving a linear equation.
However, there is one major difference: Whenever you multiply or divide an inequality by a negative
number, you must switch the inequality symbol (from  to  or vice versa).
Why?
Consider the following example:
57
If we multiply both sides of the inequality by −1, the result is
− 5  −7
This is due to the fact that, on the number line, the negative numbers are written ”backwards”, namely,
the smaller the absolute value, the larger the number.
Example 1:
Solution:
Solve for x: 7 − 2x  11
Solve for x the same way you would solve the equation 7 − 2x  11:
7 − 2x  11
subtract 7 from both sides
−2x  4
−2x  4
−2
−2
x  −2
divide by −2
remember to switch inequality!
The solution can be written algebraically (x  −2), using interval notation −2,  or a number line
which is shown below.
Example
Example 2:
Solution:
2
y−1
3y
Solve for y:
−
 −1
5
3
It is best to multiply by the LCD (least common denominator) first, which is 15:
y−1
3y
15
− 15
≥ 15−1
5
3
5y − 1 − 33y ≥ −15 distribute
5y − 5 − 9y ≥ −15
− 4y − 5 ≥ −15 add 5
− 4y ≥ −10 divide by − 4
y ≤ −10  5
−4
2
5
−, 
2
5
2
Notice how we indicate that 5/2 is included in the solution set by using a solid dot. If we want to
indicate that 5/2 is not included an unfilled circle will be used.
Sometimes two inequalities are combined in the same statement, such as a  x  b. As stated earlier,
this means the number x is between a and b. More formally, we say that
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: Pre-Calculus - Chapter 2B
a  x and x  b
Note: in order for this to be true, a must be less than b to start with. A statement such as 4  x  −2
cannot be true; there are no numbers greater than 4 and simutaneously less than −2.
Linear inequalities of this type may be solved by simply splitting them into two inequalities, solving
each separately, and finding all values common to both solutions.
Question:
Answer:
Example 3:
Solution 1:
Is there anything wrong with the statement 5  x  5?
Yes, there is no number which is both less than 5 and greater than 5.
Solve for x: −10  3x  5 ≤ 17
The combined inequality can be rewritten as
3x  5  −10 and 3x  5 ≤ 17 subtract 5 from both sides
3x  −15
3x ≤ 12
x  −5
x≤4
divide both sides by 3
The solution is all real numbers greater than −5 and less than or equal to 4. This can be written as
−5  x ≤ 4, or −5, 4, or we can use a number line as shown below.
5
4
As usual the red line is used to indicate the solution set, and filled and unfilled circles indicated whether
a point is in the solution set or not.
Solution 2:
Since the variable appears only in the middle, we can isolate it as we do for a single
inequality. Remember to do the same operations to all 3 parts and switch the inequality when
necessary!
− 10  3x  5 ≤ 17
− 15  3x ≤ 12
−5  x ≤ 4
As found earlier.
Example 4:
Solution:
Solve for x: x  4  2x − 7  3 − 3x
The combined inequality can be rewritten as
2x − 7  x  4
2x  x  11
2x − 7  3 − 3x add 7 to both sides
subtract x from both sides 2x  10 − 3x
x  11
5x  10
add 3x to both sides
divide both sides by 5
x2
Note that the solution is all real numbers greater than 11 and less than 2. Since no real numbers satisfy
both statements, there is no solution to the inequality. Note also that these two inequalities cannot be
solved together since there is no way to isolate x in the center.
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: Pre-Calculus - Chapter 2B
Absolute Value Inequalities
Inequalities involving absolute value can be rewritten using the concept of distance. For example, the
inequality |x|  3 can be thought of as ”what numbers are less than 3 units from 0 on the number line?”
3 units right
3 units left
3
0
3
From the diagram above it can be seen that any number between −3 and 3 satisfies the inequality
|x|  3. Similarly, the inequality |x|  3 can be thought of as ”what numbers are greater than 3 units
from 0 on the number line?” In this case, x can be greater than (to the right of) 3 or less than (to the left
of) −3.
The following generalizes this concept:
|x|  a is equivalent to − a  x  a
|x|  a is equivalent to x  a OR x  −a
The number a in the above is assumed to be positive. We have a similar statement for ≥ and ≤.
|x| ≤ a if and only if − a ≤ x ≤ a
|x| ≥ a if and only if x ≥ a OR x ≤ −a
Example 1:
Solution:
Solve for x: |x − 2|  1.
From the statement above, |x − 2|  1 can be rewritten as
− 1  x − 2  1 (add 2 to all sides)
1x3
which can also be written in interval notation as 1, 3. Notice that the numbers 1 and 3 are both 1 unit
away from 2 on the number line.
Example 2:
Solution:
Solve for x: |3x − 4|  6 ≤ 3.
We first isolate the absolute value term before trying to solve the inequality.
|3x − 4|  6 − 6 ≤ 3 − 6
|3x − 4| ≤ −3
Recall that absolute value represents distance; therefore, it cannot be negative; hence, it cannot be less
than −3. Therefore, this inequality has no solution.
Question:
Answer:
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Does x  5 satisfy the inequality |x − 4|  1?
No.
: Pre-Calculus
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: Pre-Calculus - Chapter 2B
Example 3:
Solve for x: |2x  4|  2 ≥ 7
Solution:
Before applying our rule for rewriting absolute value inequalities, we must first isolate
the absolute value term on one side of the inequality (Be careful not to write the left-hand side as
2x  6)
|2x  4|  2 ≥ 7 subtract 2 from each side
|2x  4|  2 − 2 ≥ 7 − 2
|2x  4| ≥ 5
This must be rewritten as two inequalities:
2x  4 ≥ 5 or 2x  4 ≤ −5 subtract 4 from both sides
2x ≥ 1
x≥ 1
2
2x ≤ −9
divide both sides by 2
9
or
x≤−
2
Therefore, our solution is x ≤ − 9 or x ≥ 1 . This can be written in interval notation one of two ways.
2
2
Either use the word ’or’ between the intervals: −, − 9  or  1 , , or we can use the union symbol,
2
2
which means either/or: −, − 9    1 , . A number line description of the solution set is shown
2
2
below.
or
9
2
2
1
2
Notice that the points − 9 and 1 are each 5 units away from the point −2. The following gives an
2
2
2
algebraic explanation of this.
|2x  4|  2 ≥ 7 factor a 2 from the absolute value term
2|x  2|  2 ≥ 7 divide by 2
|x  2|  1 ≥ 7 subtract 1
2
|x  2| ≥ 5
2
|x − −2| ≥ 5
2
Thus, the solution set is the set of points which are 5 or further from the point −2.
2
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: Pre-Calculus - Chapter 2B
Nonlinear Inequalities
If an inequality is not linear, we can use the following method:
1) Move every term of the inequality to one side.
2) Find the critical values- That is those values for which the expression is equal to zero or is
undefined.
3) The critical values will divide the real number line into intervals. Do not include the critical points
in the intervals.. Each interval is entirely in the solution set or entirely out of the solution set. To
determine which, test a number in the interval by substituting it into the inequality. If the chosen
number satisfies the inequality, then so does every other number in the interval, and if the number
does not satisfy the inequality, then no other number from that interval will satisfy the inequality.
4) Do this for each interval of the number line.
Example 1:
Solve for x: x  3x − 1 ≤ 0
Solution:
Since every term of the inequality is on one side, we find the critical values (Where
x  3x − 1  0 ) Recall that this equation is solved by setting each factor equal to 0. Therefore, the critical
values are
x  −3 or x  1
These critical values divide the number line into 3 intervals. −, −3, −3, 1, and 1, . We choose a number in
each interval, and test it by substituting into the original inequality as follows:
Therefore, the only interval which is in the solution set is between −3 and 1. Since the inequality is greater than
or equal to 0, the critical values (where the expression is equal to zero) are also part of the solution. Therefore,
our solution set is −3 ≤ x ≤ 1, or the interval −3, 1
Question:
Answer:
What are the critical values of the expression x 2 − 3x  2?
x 2 − 3x  2  x − 1x − 2
Thus, the critical values are 1 and 2.
Here we present a more detailed explanation of why the method described above works.
Example 2:
Solve for x: x  3x − 1 ≤ 0
Solution:
Before we begin, let’s recall the rules for signed numbers. When mulitplying or dividing 2
numbers, if the signs are the same the result is positive. If the signs are different, the result is negative.
Here we can view the left-hand side as a product of 2 factors, x  3 and x − 1. The inequality states that the
product is less than 0 (negative) or equal to 0. Based on the rules for signed numbers, this will occur when one
factor is positive and the other is negative.
Rather than deal with the different possibilities, we can test numbers between the critical values to determine
whether the expression is positive or negative. Recall that the critical values occur at those values of x in which
the expression is zero or undefined. We look at the critical values because those are the only places in which a
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polynomial expression could possibly change from positive to negative or negative to positive.
This can be illustrated on the number line (as the previous solution showed) or using a sign chart to analyze each
interval (NOTE: remember that the critical values are part of the solution since the expression is equal to 0. You
may wish to indicate this in your chart to remind you):
x3
Interval
x  −3 or −, −3
x−1
x  3x − 1
negative negative
positive
−3 ≤ x ≤ 1 or −3, 1 positive negative
negative
x  1 or 1, 
positive positive
positive
Since the expression is to be less than zero (negative) or equal to zero, the solution set contains only the interval
−3, 1 or −3 ≤ x ≤ 1. Since 0 is an allowed value, the finite endpoints of the intervals are also members of the
solution set.
Either the sign chart or the number line can be used to test each interval.
Example 3:
Solve for x: x 2 − 4x  12
Solution:
Remember, for nonlinear inequalities we first move all terms to one side of the inequality so that
we have 0 on the other side.
x 2 − 4x − 12  0
To find the critical values, we solve the equation
x 2 − 4x − 12  0
x − 6x  2  0
x  6 or x  −2
These critical values divide the number line into 3 intervals: x  −2, −2  x  6, and x  6. Notice that since the
inequality is strictly greater than 0, the critical values will not be part of the solution. The sign chart is given
below:
x−6
Interval
x  −2 or −, −2
x2
x − 6x  2
negative negative
positive
−2  x  6 or 2, 6 negative positive
negative
x  6 or 6, 
positive positive
positive
Since the inequality is greater than zero (positive), the solution is x  −2 or x  6
In interval notation we have
−, −2  6, 
You may have noticed in both of these examples that the overall sign alternates from positive to negative to
positive. In the next example, we show that this is not always the case.
Example 4:
Solution:
Solve for x: x 4  3x 3 − 4x 2 ≥ 0
Since all terms are on the same side, we find the critical values by solving the equation
x 4  3x 3 − 4x 2  0
x 2 x 2  3x − 4  0 remember common factors first!
x 2 x  4x − 1  0
x  0 or x  −4 or x  1
These critical values divide the number line into 4 intervals: −, −4, −4, 0, 0, 1, and 1, . Notice that we
are including the critical values since the inequality is greater than or equal to 0. The sign chart is given below
(Note that if you are choosing a specific number in each interval you must choose a fraction for the interval 0, 1
)!
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: Pre-Calculus - Chapter 2B
Interval
−, −4 or x  −4
x2
x4
x−1
x 2 x  4x − 1
positive negative negative
positive
−4, 0 or −4  x  0 positive positive negative
negative
0, 1 or 0  x  1
positive positive negative
negative
1,  or x  1
positive positive positive
positive
Since the inequality is greater than (positive) or equal to zero, the solution set is −, −4 or 1,  .
Question:
Answer:
Notice in this example that there is no sign change at x  0. Why isn’t there?
Notice that when we factored our expression, we obtained
x 2 x  4x − 1
The critical value x  0 came from the factor x, which is squared. Whenever you square a real number, the result
is always positive (notice the second column of the sign chart). This is why the sign of the expression does not
change at x  0.
Here we present examples of inequalities involving rational expressions. The important difference between
rational equations and inequalities is this: You cannot get rid of fractions by mulitplying the Least Common
Denominator in an inequality. The reason for this is that the expression we multiply by will be positive for some
values of x and negative for other values. This doesn’t matter in equations, but it does in inequalities as the
direction of an inequality sign changes when we multiply by a negative number. We must use the same strategy as
before. That is, locate the critical values, and then determine what happens in the intervals defined by the critical
values.
Solve for x: x − 1 ≥ 2
x−3
We begin by moving all terms to one side, in this case by subtracting 2:
x − 1 − 2 ≥ 0 get a common denominator
x−3
x − 1 − 2x − 3 ≥ 0
x−3
x−3
x − 1 − 2x − 6
≥ 0 distribute the negative sign
x−3
x − 1 − 2x  6 ≥ 0
x−3
−x  5 ≥ 0
x−3
To find the critical values, we must find where this expression is zero or undefined. The expression is zero when
the numerator is zero:
−x5  0
Example 5:
Solution:
− x  −5
x5
and the expression is undefined when the denominator is zero:
x−3  0
x3
The critical values 3 and 5 separate the number line into 3 intervals: −, 3, 3, 5, and 5, . Note that we are
including the critical value 5 since the expression is equal to zero, but we are not including the critical value 3
since this makes the expression undefined. In general, a critical value where the expression is undefined will not
be included in the solution. The sign chart is given below:
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: Pre-Calculus - Chapter 2B
−x  5
x−3
−, 3 positive negative negative
Interval
−x  5
x−3
3, 5
positive positive positive
5, 
negative positive negative
Since the expression must be greater than (positive) or equal to 0, the solution is
3, 5  x : 3  x ≤ 5 
Question:
Answer:
inequality
Example:
Why can’t you multiply by the LCD?
Recall that when you multiply or divide an inequality by a negative number, you must switch the
− 4x  −16
x4
However, to remove the fractions in our example, we would have to multiply our inequality by x − 3, which can
be positive or negative depending on x. Therefore, we do not know whether to switch the inequality or not. You
can analyze the different possibilities, but the method we used does so automatically.
Example 6:
Solve for x: x  4  2
x − 14
x−6
Solution:
Remember that you cannot get rid of the fractions in the inequality-you must move everything to
one side, and get a common denominator:
x4  2
x − 14
x−6
x  4 − 2  0 get a common denominator
x−6
x − 14
2x − 14
x  4x − 6
−
 0 FOIL/distribute in the numerators
x − 14x − 6
x − 6x − 14
x 2 − 2x − 24 − 2x − 28
 0 distribute the negative sign
x − 6x − 14
x 2 − 2x − 24 − 2x  28  0 combine like terms
x − 6x − 14
x 2 − 4x  4  0 factor the numerator
x − 6x − 14
x − 2 2
0
x − 6x − 14
The critical values are x  2 (where the expression is 0) , x  6, and x  14 (where the expression is undefined).
Since the inequality is strictly less than 0, none of the critical values are in the solution. The critical values divide
the number line into 4 intervals: −, 2, 2, 6, 6, 14, and 14, . The sign chart is given below:
Interval x − 2 2
x − 6
x − 14
x − 2 2
x − 6x − 14
−, 2 positive negative negative
2, 6
positive negative negative
positive
6, 14
positive positive negative
negative
14,  positive positive positive
positive
Since the expression must be less than 0 (negative), the solution is
6, 14  x : 6  x  14 
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: Pre-Calculus
positive
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: Pre-Calculus - Chapter 2B
Exercises for Chapter 2B - Solving Inequalities
1.
2.
3.
4.
5.
6.
7.
8.
9.
10.
11.
12.
13.
14.
15.
16.
17.
18.
19.
20.
21.
22.
23.
24.
25.
26.
27.
28.
29.
30.
31.
32.
33.
34.
©
For problems 1-3, determine whether the given values are solutions to the given inequality.
3x − 4  0;
x  10, x  −3, x  1
6x  13 ≥ 13x − 2;
x  −1; x  −5; x  8
2
x  −1; x  5; x  12
x − 10x  21  0;
For problems 4-9, write the following in interval notation, if possible. If not possible, state
why.
−12  x  7
9 ≤ x ≤ 10
−4 ≤ x  12
8x≥5
x  3 or x ≥ 11
5x7
For problems 10-37, solve each of the following, if possible.
2  3
4x − 12 ≥ 0
35.
x−1
x1
1
1
5x  7  2
36. x 
2
x1
−5x − 9  6
37. x − 1  3 ≤ x  2
x5
7 x− 3 ≤ 1
10
4
2x − 11 ≤ 12x  13
−1 ≤ 2x − 9  11
9  −2x − 2  8
−3x  10 ≥ x ≥ −5x − 12
−4x  11  6x − 10  9x − 7
|4x − 1|  8
|9 − x| ≤ 6
|5x| − 12  8
|x − 2|  8
|2x − 5| − 9 ≥ 12
|3x − 7|  12 ≤ 9
11  |2x  9|  10
x2 − x − 6  0
x 2 − 4x ≤ 45
2x 2  x − 3 ≥ 0
4x − 12  x 2
x 3 − 2x 2 − 24x ≤ 0
x 4 − 2x 3 − 24x 2 ≤ 0
1 −4  0
x
2x − 7 ≤ 3
x−5
3x  6 ≥ 2
x1
: Pre-Calculus
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: Pre-Calculus - Chapter 2B
Answers to Exercises for Chapter 2B - Solving Inequalities
1.
Test each value by substituting into the inequality:
310 − 4  0 ?
26  0 X
x  10 is not a solution to the inequality.
3−3 − 4  0 ?
− 13  0 
x  −3 is a solution to the inequality
31 − 4  0 ?
−1  0
x  1 is a solution to the inequality.
2.
6−1  13 ≥ 13−1 − 2 ?
7 ≥ −15 
x  −1 is a solution to the inequality
6−5  13 ≥ 13−5 − 2 ?
− 27 ≥ −67 
x  −5 is a solution to the inequality
68  13 ≥ 138 − 2 ?
61 ≥ 102 X
x  8 is not a solution to the inequality
3.
−1 2 − 10−1  21  0 ?
1  10  21  0 ?
32  0 
x  −1 is a solution to the inequality
5 2 − 105  21  0 ?
25 − 50  21  0 ?
−4  0X
x  5 is not a solution to the inequality
12 2 − 1012  21  0 ?
144 − 120  21  0 ?
45  0 
4.
5.
6.
7.
©
x  12 is a solution to the inequality
−12, 7
9, 10
−4, 12
5, 8 (The inequality can be rewritten as 5 ≤ x  8 )
: Pre-Calculus
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8.
9.
: Pre-Calculus - Chapter 2B
−, 3 or 11,  Another way of writing this would be −, 3  11, 
This is not possible since we are looking for values of x which are both less than 5 and
greater than 7 (at the same time). The combined inequality implies that 5  7.
10.
4x − 12 ≥ 0
4x ≥ 12
x ≥ 3 or 3, 
11.
5x  7  2
5x  −5
x  −1 or −, −1
12.
− 5x − 9  6
− 5x  15
x  −3 or −, −3
Remember to switch the inequality when you divide by −5
13. We begin by multiplying by 20 (the LCD) to eliminate the fractions:
20 7 x − 20 3 ≤ 120
4
10
14x − 15 ≤ 20
14x ≤ 35
x ≤ 35
14
x ≤ 5 or −, 5 
2
2
14.
2x − 11 ≤ 12x  13
2x − 22 ≤ 12x  13
− 10x ≤ 35
x ≥ − 35
10
x ≥ − 7 or − 7 , 
2
2
15. We can solve both inequalities simultaneously:
− 1 ≤ 2x − 9  11
8 ≤ 2x  20
4 ≤ x  10 or 4, 10
16. We can solve the inequalities simultaneously:
9  −2x − 2  8
11  −2x  10
− 11  x  −5 or − 11 , −5
2
2
17. We must solve these inequalities separately:
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: Pre-Calculus - Chapter 2B
−3x  10 ≥ x and x ≥ −5x − 12
−4x ≥ −10
x ≤ 10
4
x≤ 5
2
and
6x ≥ −12
and
x ≥ −2
and
x ≥ −2
Therefore, the solution is −2 ≤ x ≤ 5 or −2, 5
2
2
18. Solving the inequalities separately:
−4x  11  6x − 10 and 6x − 10  9x − 7
−10x  −21
x  21
10
and
−3x  3
and
x  −1
The solution set is what is common to both intervals. If a number is larger than 21 (2. 1, it
10
will already be larger than −1. Therefore, our solution is x  21 or 21 , 
10
10
19. The inequality can be rewritten as
− 8  4x − 1  8
− 7  4x  9
− 7  x  9 or − 7 , 9
4
4
4 4
20. The inequality can be rewritten as
−6 ≤ 9−x ≤ 6
− 15 ≤ −x ≤ −3
15 ≥ x ≥ 3 or 3, 15
21. Isolate the absolute value term first before rewriting:
|5x| − 12  8
|5x|  20
− 20  5x  20
− 4  x  4 or −4, 4
22. The inequality must be split into two inequalities:
x − 2  8 or x − 2  −8
x  10
or
x  −6
Since these are ’or’ rather than ’and’, the solution is valid and can be written this way or
−, −6  10,  using interval notation.
23.
|2x − 5| − 9 ≥ 12
|2x − 5| ≥ 21
2x − 5 ≥ 21 or 2x − 5 ≤ −21
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: Pre-Calculus
2x ≥ 26
or
2x ≤ −16
x ≥ 13
or
x ≤ −8
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: Pre-Calculus - Chapter 2B
In interval notation, −, −8  13, .
24.
|3x − 7|  12 ≤ 9
|3x − 7| ≤ −3
Since the absolute value of an expression cannot be less than −3, there is no solution
25.
11  |2x  9| ≥ 10
|2x  9| ≥ −1
What happens here? Since the absolute value of an expression cannot be negative, it will
always be greater than −1. Therefore, the solution is all real numbers or −, 
26.
x2 − x − 6  0
x − 3x  2  0
The critical values are where the expression is equal to 0, namely, where x  3 or x  −2.
Since the inequality is strictly less than 0, we do not include the critical values. These
divide the number line into 3 intervals. The sign chart is given below:
Interval
x−3
x2
x − 3x  2
−, −2 negative negative
positive
−2, 3
negative positive
negative
3, 
positive positive
positive
Since the inequality is less than 0 (negative), the solution is −2, 3 (−2  x  3 )
27. Remember to first move all terms to one side:
x 2 − 4x − 45 ≤ 0
x − 9x  5 ≤ 0
The critical values are x  9 and x  −5. Since the inequality is less than or equal to, we
include both critical values in the solution. The critical values divide the number line into 3
intervals. The sign chart is given below:
Interval
x − 9
x  5
x − 9x  5
−, −5 negative negative
positive
−5, 9
negative positive
negative
9, 
positive positive
positive
Since the inequality is less than (negative) or equal to, the solution is −5, 9 (−5 ≤ x ≤ 9 )
28.
2x 2  x − 3 ≥ 0
2x  3x − 1 ≥ 0
The critical values are x  − 3 or x  1. Since the inequality is greater than or equal to,
2
these are included in the solution. The critical values divide the number line into 3 intervals.
The sign chart is given below:
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: Pre-Calculus
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: Pre-Calculus - Chapter 2B
2x  3
Interval
−, − 3 
2
3
− ,1
2
1, 
x − 1
2x  3x − 1
negative negative
positive
positive negative
negative
positive positive
positive
Since the inequality is greater than or equal to, the solution is −, − 3   1,  (x ≤ − 3
2
2
or x ≥ 1 )
29.
4x − 12  x 2
− x 2  4x − 12  0
− x 2 − 4x  12  0
− x − 6x  2  0
At this point, you can divide by the negative 1, but be sure to switch the inequality! If you
do not divide, you must take the negative 1 into account when you do the sign chart (critical
values 6 and 2 not included in the solution):
x − 6
Interval
x  2
−x − 6x  2
−, −2 negative negative negative b/c of -1
−2, 6
negative positive
positive
6, 
positive positive
negative
Since the inequality is less than 0, the solution is −, −2  6,  (x  −2 or x  6)
30.
x 3 − 2x 2 − 24x ≤ 0
xx 2 − 2x − 24 ≤ 0
xx − 6x  4 ≤ 0
The critical values are x  0, x  6, and x  −4, which are included in the solution.
The critical values divide the number line into 4 intervals. The sign chart is given below:
Interval
x
x − 6
x  4
−, −4 negative negative negative
xx − 6x  4
negative
−4, 0
negative negative positive
positive
0, 6
positive negative positive
negative
6, 
positive positive positive
positive
Since the inequality is less than or equal to 0, the solution is −, −4  0, 6 (x ≤ −4 or
0 ≤ x ≤ 6)
31.
x 4 − 2x 3 − 24x 2 ≤ 0
x 2 x 2 − 2x − 24 ≤ 0
x 2 x − 6x  4 ≤ 0
The critical values are x  0, x  6, and x  −4, which are included in the solution.
The critical values divide the number line into 4 intervals. The sign chart is given below:
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: Pre-Calculus
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: Pre-Calculus - Chapter 2B
Interval
x2
x − 6
x  4
−, −4 positive negative negative
xx − 6x  4
positive
−4, 0
positive negative positive
negative
0, 6
positive negative positive
negative
6, 
positive positive positive
positive
Since the inequality is less than or equal to 0, the solution is −4, 0  0, 6 (Note that since
0 is included, we can write this as −4, 6, or −4 ≤ x ≤ 6. If 0 were not included in the
solution, we would have to write it as 2 separate intervals).
32. Get a common denominator:
4 −1  0
x
4 − 1x  0
x
1 x
4−x  0
x
The critical values are x  4 (where the expression is 0) and x  0 (where the expression is
undefined). Neither critical value is included since the inequality is less than 0. The sign
chart is given below:
4−x
x
−, 0 positive negative negative
Interval
4−x
x
0, 4
positive positive positive
4, 
negative positive negative
Since the inequality is less than 0 (negative), the solution is −, 0  4,  (x  0 or
x  4)
33.
2x − 7 − 3 ≤ 0
x−5
2x − 7 − 3 x − 5 ≤ 0
x−5
1 x−5
2x − 7 − 3x − 15
≤0
x−5
2x − 7 − 3x  15 ≤ 0
x−5
−x  8 ≤ 0
x−5
The critical values are x  8 (where the expression equals 0-included) and x  5 (where the
expression is undefined-not included). The sign chart is given below:
−x  8
x−5
−, 5 positive negative negative
Interval
−x  8
x−5
5, 8
positive positive positive
8, 
negative positive negative
Since the inequality is less than (negative) or equal to 0, the solution is −, 5  8, 
(x  5 or x ≥ 8).
34.
©
: Pre-Calculus
©
: Pre-Calculus - Chapter 2B
3x  6 − 2 ≥ 0
x1
3x  6 − 2 x  1 ≥ 0
x1
1 x1
3x  6 − 2x − 2 ≥ 0
x1
x4 ≥ 0
x1
The critical values are x  −4 (where the expression is 0-included) and x  −1 (where the
expression is undefined-not included). The sign chart is given below:
x4
x1
−, −4 negative negative positive
Interval
x4
x1
−4, −1 positive negative negative
−1, 
positive positive positive
Since the inequality is greater than (positive) or equal to 0, the solution is −, −4  −1, 
(x ≤ −4 or x  −1).
35. You cannot cross-multiply to solve rational inequalities!
2 − 3 0
x−1
x1
LCD is x  1x − 1:
3x  1
2x − 1
−
0
x  1x − 1
x  1x − 1
2x − 2 − 3x  3
0
x  1x − 1
2x − 2 − 3x − 3  0
x  1x − 1
−x − 5
0
x  1x − 1
The critical values are x  −5 (where the expression is 0-not included since strictly greater
than), x  −1, and x  1 (where the expression is undefined-not included). The sign chart is
given below (note that the critical values divide the number line into 4 intervals):
Interval
−x − 5
x1
x−1
−x − 5
x  1x − 1
−, −5 positive negative negative
positive
−5, −1 negative negative negative
negative
−1, 1
negative positive negative
positive
1, 
negative positive positive
negative
Since the inequality is greater than (positive) 0, the solution is −, −5  −1, 1 (x  −5 or
−1  x  1 ).
36. Again move every term to one side and find the LCD (use on every term):
©
: Pre-Calculus
©
: Pre-Calculus - Chapter 2B
2 − 3 1
x
x−2
2x − 2
3x
1xx − 2
−

xx − 2
1xx − 2
x − 2x
2x − 4 − 3x  x 2 − 2x
xx − 2
x 2 − 3x − 4
xx − 2
x − 4x  1
xx − 2
 0 LCD xx − 2
 0 don’t forget the 1!
0
0
0
The critical values are x  4 , x  −1 (both where the expression is 0-not included in
solution since strictly greater than ), x  0, and x  2 (both where the expression is
undefined-not included in the solution): The sign chart (for 5 intervals!) is given below:
Interval x − 4 x  1
x
x−2
x − 4x  1
xx − 2
−, −1
neg
neg
neg
neg
positive
−1, 0
neg
pos
neg
neg
negative
0, 2
neg
pos
pos
neg
positive
2, 4
neg
pos
pos
pos
negative
4, 
pos
pos
pos
pos
positive
Since the inequality is greater than zero (positive), the solution is −, −1  0, 2  4, 
(x  −1 or 0  x  2 or x  4 ).
37. Solving the inequalities separately:
x−1  3
x5
x−1 −3  0
x5
x − 1 − 3x  5
0
x5
x − 1 − 3x − 15  0
x5
and
3 ≤ x2
and −x  3 ≤ 2
and
−x ≤ −1
and
x≥1
Continuing to solve the first inequality yields −2x − 16  0. The critical values are x  −8
x5
and x  −5 (both not included since we have a strict inequality). The sign chart is given
below:
−, −8 positive negative
−2x − 16
x5
negative
−8, −5 negative negative
positive
−5, 
negative
Interval −2x − 16
x5
negative positive
Since the inequality is less than 0 (negative), the solution for this inequalitiy is
−, −8  −5,  . But the second inequality was x ≥ 1. Since x must satisfy both
inequalities, the solution is x ≥ 1 OR 1, .
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: Pre-Calculus