Lesson SOLVING ABSOLUTE VALUE EQUATIONS

This lesson provides an overview of SOLVING ABSOLUTE VALUE EQUATIONS

REFERENCES

Information for this tutorial was obtained from the following websites:

http://www.wtamu.edu/academic/anns/mps/math/mathlab/col_algebra/col_alg_tut21_abseq.htm
http://tutorial.math.lamar.edu/Classes/Alg/SolveAbsValueEqns.aspx
http://www.purplemath.com/modules/solveabs.htm

These references provide many examples and exercises. You are encouraged to look at them if you have time. If you still need more then just go to www.yahoo.com or www.google.com and do a search on "absolute value equations".

TERMINOLOGY

x and y or any other variable name used in this lesson can be positive or negative.
If the variable represents a negative number, then minus the variable represents a positive number.
Example:
-(-7) = +7 = 7

ABSOLUTE VALUE SYMBOLS

Absolute value of an expression is represented by the expression being enclosed by vertical bars that look like this: ||
Example:
Absolute value of (x^2 + 2x + 3) is represented by enclosing it in vertical bars as shown:
|x^2 + 2x + 3|.

On your keyboard, the | symbol is usually found on the right hand side second row down from the keyboard numbers and is the SHIFT option of the \ key. If you look at the key, you should see \ underneath and | on top although the | may look like it's separated in the middle which can make it look like 2 vertical lines on top of each other rather than 1.

BASIC DEFINITION OF ABSOLUTE VALUE

THE ABSOLUTE VALUE OF AN EXPRESSION IS EQUAL TO THE VALUE OF THE EXPRESSION IF THE EXPRESSION IS GREATER THAN OR EQUAL TO 0.

Examples:

|5| = 5
|x| = x when x is >= 0
|y| = y when y is >= 0
|x+7| = (x+7) when (x+7) >= 0
|3x-3| = (3x-3) when (3x-3) >= 0

NOTE:
It is the value of the EXPRESSION that is positive, not just the variable within the expression.
Example:
The expression (x+7) is positive when x is >= -7.

THE ABSOLUTE VALUE OF AN EXPRESSION IS EQUAL TO MINUS THE VALUE OF THE EXPRESSION IF THE EXPRESSION IS LESS THAN 0.

Examples:

|-5| = -(-5) = +5 = 5
|x| = -x when x < 0
|y| = -y when x < 0
|x+7| = -(x+7) when (x+7) < 0
|3x-3| = -(3x-2) when (3x-2) < 0

NOTE:
It is the value of the EXPRESSION that is negative, not just the variable within the expression.
Example:
The expression (x+5) is negative when x is less than -5.

SOLVING ABSOLUTE VALUE EQUATIONS

Understanding the basic definition of absolute value is the key to solving absolute value equations.

Each Absolute Value Expression will have 2 possible solution sets.
Within each possible solution set, there will be one or more possible solutions.
Some of the solutions can be the same as others and are discarded to avoid redundancy.
Some of the solutions can be invalid and are discarded.
You will see how this works when we go through the examples.

One possible solution set will be when the expression within the absolute value sign is positive.
The other possible solution set will be when the expression within the absolute value sign is negative.

When the expression within the absolute value signs is positive, then the expression replaces the absolute value of the expression in the equation to be solved.

When the expression within the absolute value signs is negative, then MINUS the expression replaces the absolute value of the expression in the equation to be solved.

It is very important to analyze the absolute value equation prior to attempting to solve it. This is done to see if there are any possible solutions at all and to get an idea of where the range of the possible solutions will lie.

When you analyze the absolute value equation, simply remember that the value within the absolute value symbols must be greater than or equal to 0.

ARITHMETIC SYMBOLS

In this lesson, the following logical operators will be used:

equal: =
greater than: >
less than: <
greater than or equal: >=
less than or equal: <=

The following arithmetic / algebraic operators may or may not be used where applicable.

x^n: x raised to the nth power
root (n,x): nth root of x
sqrt(x): square root of x
x*y: x multiplied by y
x/y: x divided by y
x+y: y added to x
x-y: y subtracted from x

DERIVATION OF RULE SETS FOR ABSOLUTE VALUE EQUATIONS

Some lessons other than this one will provide rule sets for absolute value equations.

Example:
One such rule set would be:

|x| >= y means that:
x >= y when x >= 0
or:
x <= -y when x < 0

In this lesson, rather than giving you the rule set, I work directly from the basic definition of absolute view to derive the rule set.

The same equation in this lesson will appear as:

|x| >= y means that:
x >= y when x >= 0
or:
-x >= y when x < 0
Multiply both sides of -x >= y by -1 to get:
x <= -y

Your equation becomes:
x >= y when x >= 0
or:
x <= -y when x < 0

I go through this extra set of deriving the rule set each time to reinforce the concept that the basic definition of absolute values provides you. Furthermore, there are equations where the rule set cannot be applied directly. In those situations, reverting to the basic definition of absolute values makes analyzing and solving those equations easier.

AND / OR LOGIC.

AND means both conditions must be applicable at the same time.
OR means one of the conditions must be applicable, but not at the same time as the other.

ADDITIONAL WORD ABOUT OR LOGIC.

If you study logic not connected with this lesson, AND will still mean both conditions need to be applicable at the same time. OR, however, can have different meanings. OR can mean only one of these conditions are applicable at the same time as expressed above. OR can also mean one or both of these conditions can be applicable at the same time. My definition of OR above is more in line with exclusive OR. One or the other of the conditions is applicable, but not at the same time. Just for your own information, this exclusive type of OR is called Exclusive OR, abbreviated as XOR.

OR LOGIC APPLICATION

If the absolute value equation is > or >= or =, then OR logic applies.

Example 1:
|x| = 5 means that:
x = 5 OR x = -5
Both conditions cannot be satisfied at the same time.

Example 2:
|x| > 5 means that:
x > 5 OR x < -5
Both conditions cannot be satisfied at the same time.

Example 3:
|x| >= 5 means that:
x >= 5 OR x <= -5
Both conditions cannot be satisfied at the same time.

AND LOGIC APPLICATION

If the absolute value equation is < or <=, then AND logic applies.

Example 1:
|x| < 5 means that:
x < 5 AND x > -5
Both conditions must be satisfied at the same time.
This can also be written as:
-5 < x < 5

Example 2:
|x| <= 5 means that:
x <= 5 AND x >= -5
Both conditions must be satisfied at the same time.
This can also be written as:
-5 <= x <= 5

EXAMPLE NUMBER 1

|x| >= 0

We determine that the absolute value expression is (x).

We do an up front check to determine if our absolute value equation is valid or not.

We determine that it is a valid equation because the absolute value of any expression will always be greater than or equal to 0.

We further determine that this particular equation will be true for all real values of x because no matter what the value of x is, the absolute value of that will always be greater than or equal to 0.

Our answer is that x can be any real number.

EXAMPLE NUMBER 2

|x+5| >= 0

We determine that the absolute value expression is (x+5).

We do an up front check to determine if our absolute value equation is valid or not.

We determine that it is a valid equation because the absolute value of any expression will always be greater than or equal to 0.

We further determine that this equation will be true for all real values of x because no matter what the value of x is, the absolute value of (x+5) will always be greater than or equal to 0.

Our answer is that x can be any real number.

EXAMPLE NUMBER 3

|x-5| >= 0

We determine that the absolute value expression is (x-5).

We do an up front check to determine if our absolute value equation is valid or not.

We determine that it is a valid equation because the absolute value of any expression will always be greater than or equal to 0.

We further determine that this equation will be true for all real values of x because no matter what the value of x is, the absolute value of (x-5) will always be greater than or equal to 0.

Our answer is that x can be any real number.

EXAMPLE NUMBER 4

|x| < 0

We determine that the expression within the absolute value signs is x.

We do an up front check to determine if our absolute value equation is valid or not.

We determine that it is NOT a valid equation because the absolute value of any expression can never be negative.

Our answer is that there is no possible solution to this equation.

EXAMPLE NUMBER 5

|x+5| <= 0

We determine that the absolute value expression is (x+5).

We do an up front check to determine if our absolute value equation is valid or not.

We determine that it is a valid equation because the absolute value of any expression can be equal to 0 even if it can never be negative.

We will solve for |x+5| = 0 only.

We will not solve for |x+5| < 0 because that part of the equation is invalid.

If (x+5) is positive, then |x+5| = 0 means that (x+5) = 0.
Solving for x, we get:
x = -5

If (x+5) is negative, then |x+5| = 0 means that -(x+5) = 0.
Solving for x, we multiply both sides of this equation by -1 to get:
(x+5) = -0 which is the same as (x+5) = 0
We then solve for x to get:
x = -5

We have 2 possible solutions.
x = -5
or:
x = -5

Since these are both the same solution, they reduce to the one possible solution of:
x = -5

That value of x is our only solution to this problem and it will only satisfy the part of the equation that states that |x+5| = 0.

We do, however, include the < part of the original equation when we state our answer. This is because OR logic applies and only one of the conditions needs to be applicable, not necessarily both.

Our answer is that |x+5| <= 0 is true when x = -5 only.

EXAMPLE NUMBER 6

|5-3x| <= 7

We determine that that the absolute value expression is (5-3x)

We do an up front check to determine if our absolute value equation is valid or not.

We determine that this absolute value equation is valid since the solution we are looking can be greater than or equal to zero even if it has to be less than 7.

|5-3x| <= 7 means that:
(5-3x) <= 7 if (5-3x) >= 0
and:
-(5-3x) <= 7 if (5-3x) < 0

Multiply both sides of -(5-3x) <= 7 by -1 to get:
(5-3x) >= -7 and our two possible solution sets become:

(5-3x) <= 7 if (5-3x) >= 0
and:
(5-3x) >= -7 if (5-3x) < 0

Note the use of AND logic.

The first possible solution set of (5-3x) <= 7 is solved as follows:

Subtracting 5 from both sides of this equation gets:
-3x <= 2
Multiplying both sides of this equation by (-1) gets:
3x >= -2
Dividing both sides by 3 gets:
x >= -2/3

The second possible solution set of (5-3x) >= -7 is solved as follows:

Subtracting 5 from both sides of this equation gets:
-3x >= -12
Multiplying both sides of this equation by (-1) gets:
3x <= 12
Dividing both sides by 3 gets:
x <= 4

Our answer is:

x >= -2/3
and:
x <= 4

This can also be written as:
-2/3 <= x <= 4

We plug this solution into the original equation to confirm its validity.
Original equation is:
|5-3x| <= 7

When x = -2/3, our original equation of |5-3x| <= 7 becomes:
|5-3*(-2/3)| <= 7
This becomes:
|5+(6/3)| <= 7
This becomes:
|5+2| <= 7 which becomes |7| <= 7 which becomes 7 <= 7 confirming that x >= (-2/3) is valid.

When x = 4, our original equation of |5-3x| <= 7 becomes:
|5-(3*4)| <= 7
This becomes:
|5-12| <= 7 which becomes |-7| <= 7 which becomes 7 <= 7 confirming that x <= (4) is valid.

To fully confirm, we would need to pick a value of x between the range of -2/3 to 4 to confirm the equation is still true. We would also need to pick a value of x above this range and below this range to confirm that the equation is false if we are not within the specified guidelines of the answer.

let x = 0 (within the range)
|5-3x| <= 7 becomes |5| <= 7 which becomes 5 <= 7 which is true.
let x = 5 (above the range)
|5-3x| <= 7 becomes |-10| <= 7 which becomes 10 <= 7 which is false (not true).
let x = -1 (below the range)
|5-3x| <= 7 becomes |8| <= 7 which becomes 8 <= 7 which is false (not true).

The answer is fully confirmed as good since following the rules makes the equation true and not following the rules makes the equation false.

EXAMPLE NUMBER 7

|x^2 + x - 9| = 3

We determine that the absolute value expression is (x^2 + x - 9).

We do an up front check to determine if our absolute value equation is valid or not.

We determine that this absolute value equation is valid since the solution we are looking will be 3 which is greater than or equal to 0.

|x^2 + x - 9| = 3 means that:

(x^2 + x - 9) = 3 if (x^2 + x - 9) >= 0
or:
-(x^2 + x - 9) = 3 if (x^2 + x - 9) < 0

We multiply both sides of -(x^2 + x - 9) = 3 by -1 to get:
(x^2 + x - 9) = -3

Our possible solution sets now become:
(x^2 + x - 9) = 3 if (x^2 + x - 9) >= 0
or:
(x^2 + x - 9) = -3 if (x^2 + x - 9) < 0

Note the use of OR logic.

We solve for the first possible solution set as follows:

Equation to solve is:
(x^2 + x - 9) = 3

We subtract 3 from both sides of this equation to get:
x^2 + x - 12 = 0

We factor this quadratic equation to get:
(x+4) * (x-3) = 0

The possible solutions for the first possible solution set are:

x = -4
or:
x = 3

We solve for the second possible solution set as follows:

Equation to solve is:
(x^2 + x - 9) = -3

Adding 3 to both sides of this equation gets:
x^2 + x - 6 = 0
We factor this quadratic equation to get:
(x+3) * (x-2) = 0

The possible solutions for the second possible solution set are:

x = -3
or:
x = 2

We have 4 possible solutions to this absolute value equation.
They are:

x = -4
or:
x = 3
or:
x = -3
or:
x = 2

Plugging these values into the original absolute value equation to confirm whether they are valid or not gets:

x = -4:

|x^2 + x - 9| = 3 becomes |(-4)^2 + (-4) - 9| = 3 which becomes |16-4-9| = 3 which becomes |3| = 3 which becomes 3 = 3 confirming that x = -4 is a valid solution.

x = 3:

|x^2 + x - 9| = 3 becomes |3^2 + 3 - 9| = 3 which becomes |9+3-9| = 3 which becomes |3| = 3 which becomes 3 = 3 confirming that x = 3 is a valid solution.

x = -3:

|x^2 + x - 9| = 3 becomes |(-3)^2 + (-3) - 9| = 3 which becomes |9-3-9| = 3 which becomes |-3| = 3 which becomes 3 = 3 confirming that x = -3 is a valid solution.

x = 3:

|x^2 + x - 9| = 3 becomes |2^2 + 2 - 9| = 3 which becomes |4+2-9| = 3 which becomes |-3| = 3 which becomes 3 = 3 confirming that x = 2 is a valid solution.

All the possible solutions check out ok.

The possible solutions will not always be valid. Confirming that the solution you derived is valid is a very important step. Confirming weeds out invalid solutions and also gives you an indication whether you solved the problem correctly or even whether you analyzed it correctly up front.

EXAMPLE NUMBER 8

|x–3| = |3x+2|–1

This equation has 2 absolute value expressions rather than 1.

This is a complication but it is not unmanageable.

We will use the same procedure as before to solve this problem.

We determine that the absolute value expressions are (x-3) and (3x+2).

We do an up front check to determine if our absolute value equation is valid or not.

The equation is:
|x–3| = |3x+2|–1

|x-3| will always be greater than or equal to 0 for any value of x. This means that the right side of this equation must always be greater than or equal to 0.

|3x+2|–1 will always be greater than or equal to 0 if |3x+2| >= 1

Since this is possible for at least some values of x, this equation is considered to be valid and can possibly be solved.

By the basic definition of absolute values, there are a minimum of 2 possible solutions for each absolute value expression which means that there are a minimum of 4 possible solution sets to this equation.

The original equation is:
|x–3| = |3x+2|–1

The 4 possible solution sets are:
(x-3) is positive and (3x+2) is positive
(x-3) is positive and (3x+2) is negative
(x-3) is negative and (3x+2) is positive
(x-3) is negative and (3x+2) is negative

We also need to determine when the expression within the absolute value signs is positive and when the expression within the absolute value signs is negative.

The first expression within the absolute value signs is (x-3).
We set it to (x-3) >= 0 to see that when x >= 3 this expression is positive.

The second expression within the absolute value signs is (3x+2).
We set it to (3x+2) >= 0 to see that when x >= -2/3 this expression is positive.

The first possible solution set is (x-3) is positive and (3x+2) is positive:
Original equation of |x–3| = |3x+2|–1 becomes:
x–3 = 3x+2–1
We solve this to get:
x = -2
To confirm this answer is correct, we plug it into the original equation to get:
5 = 3 which is NOT correct so this solution is rejected as invalid.

This answer is NOT correct because we are looking at the equation when the expression (x-3) is positive and the value of x = -2 is NOT >= 3 which is when the expression (x-3) is positive.

The second possible solution set is (x-3) is positive and (3x+2) is negative
Original equation of |x–3| = |3x+2|–1 becomes:
x-3 = -(3x+2)–1
We solve this to get:
x = 0
To confirm this answer is correct, we plug it into the original equation to get:
3 = 1 which is not correct so this solution is rejected as invalid.

Once again, this answer is NOT correct because we are looking at the equation when the expression (x-3) is positive and the value of x = 0 is NOT >= 3 which is when the expression (x-3) is positive.

The third possible solution set is (x-3) is negative and (3x+2) is positive
Original equation of |x–3| = |3x+2|–1 becomes:
-(x–3) = 3x+2–1
We solve this to get:
x = 1/2
To confirm this answer is correct, we plug it into the original equation to get:
2.5 = 2.5 which is correct so this solution is valid.

The reason this answer is correct is because we are in the acceptable range for when |x-3| is negative (x < 3), and we are in the acceptable range for when (3x+2) is positive (x >= -2/3).

The fourth possible solution set is (x-3) is negative and (3x+2) is negative
Original equation of |x–3| = |3x+2|–1 becomes:
-(x–3) = -(3x+2)–1
We solve this to get:
x = -3
To confirm this answer is correct, we plug it into the original equation to get:
6 = 6 which is correct so this solution is valid.

The reason this answer is correct is because we are in the acceptable range for when (x-3) is negative (x < 3), and we are in the acceptable range for when (3x+2) is negative (x < -2/3).

Out of the 4 possible solutions, only 2 were valid.
They are:
x = 1/2
or:
x = -3

That we had any potential valid solutions at all was determined up front by the preliminary analysis.

It is very important to check your possible solutions by substituting in the original equation. Just because you have a possible solution doesn't mean that the solution is valid. In this particular case, we had 2 solutions that were valid and 2 solutions that were not.

It is also every important to determine the range in which the expression within the absolute value signs is positive and when it is negative. This tells you if the answer you derived will be an acceptable answer or not. We had two invalid solutions and both of those solutions were invalid because the answer was not in the acceptable range for the conditions that were being tested.

EXAMPLE NUMBER 9

|x| + |y| <= 1

We determine that the absolute value expressions are (x) and (y).

We do an up front check to determine if our absolute value equation is valid or not.

Since the answer can be positive then we have possible solutions to this problem.

Looking at this equation, it becomes clear that |x| and |y| must both be <= 1.

this is because if either |x| or |y| were > 1, that would force the other absolute value to be < 0 which is impossible with absolute values since they are always >= 0.

If we solve for |x|, we will see that:
|x| <= 1 - |y|

If we solve for |y|, we will see that:
|y| <= 1 - |x|

Because we have one equation in two unknowns, unless one of the values, either x or y, is given, we will not be able to solve this equation. We can, however, find a range of values for x and y in which the solution of this equation can be found.

We already have one possible solution set.
That is when:
|x| <= 1

|x| <= 1 means that:
x <= 1
and:
-x <= 1
Multiply this second part by -1 to get:
x >= -1

Our possible solution set becomes:

x <= 1
and:
x >= -1

This can also be written as:
-1 <= x <= 1

Our second possible solution set is when |y| <= 1

Solving for |y| <= 1, we get:

y <= 1
and:
y >= -1

This can also be written as:

-1 <= y <= 1

Our possible solution sets are:
-1 <= x <= 1
and:
-1 <= y <= 1

To show that this solution set will work, we can pick any value of x that satisfies the criteria and then solve for y.

Example 1:
let x = 0
The original equation of |x| + |y| <= 1 becomes |y| <= 1
If y >= 0, this equation becomes y <= 1
If y < 0, this equation becomes -y <= 1 which becomes y >= -1
Solutions for y are good when:
-1 <= y <= 1 which is within the range of our possible solution set for y confirming that if x and y are within the range of their possible solution sets, we can find an answer to this problem that is valid.

Example 2:
let x = -.7
The original equation of |x| + |y| <= 1 becomes:
|-.7| + |y| <= 1
This becomes:
.7 + |y| <= 1
This becomes:
|y| <= .3
If y >= 0, this becomes y <= .3
If y < 0, this becomes -y <= .3 which becomes y >= -.3
Solutions for y are good when:
-.3 <= y <= .3 which is within the range of our possible solution set for y confirming that if x and y are within the range of their possible solution sets, we can find an answer to this problem that is valid.

To complete the analysis, we will pick a value for x that is NOT within the requirements.
Take x = 2.
The original equation of |x| + |y| <= 1 becomes:
|2| + |y| <= 1
This equation becomes:
2 + |y| <= 1
Subtract 2 from both sides of this equation to get:
|y| <= -1
Since absolute value of anything can never be < 0, this equation is invalid and there are no possible solutions for it.

EXAMPLE NUMBER 10



We determine that the absolute value expressions is (x+1)/(2x-3).

We do an up front check to determine if our absolute value equation is valid or not.

Since the answer can be positive then we have possible solutions to this problem.

We look a little further to see that the value of x cannot be equal to 3/2 because that would make the denominator of this equation equal to 0 which would result in an undefined value for the equation.

We further determine that a division within the absolute value sign will cause some complications that will need to be dealt with.

Those complications are:

It has to be determine when the expression within the absolute value signs is positive or negative.
In order to determine that, it has to be determined when the expression in the numerator will be positive and when the expression in the denominator will be positive. Once that is determined, it can be determined when the whole expression within the absolute value sign will be positive and when it will be negative. You must remember that the expression will be positive when both the numerator and denominator are positive and when both the numerator and denominator are negative (minus divided by a minus is a plus).

You will encounter one more complication when you are multiplying both sides of the equation by the denominator in order to solve the equation. That complication is that you have to take into account when the denominator is positive and when the denominator is negative. When the denominator is positive, the multiplication is normal. When the denominator is negative, the multiplication will reverse the sign of the inequality. You will see how that works when the time comes.

The complication when multiplying by a negative denominator comes from the fact you can't see that it is negative. You have to know that it will be and when it will be and you have to take that into account. The other things you have to know is that when you do multiply by the negative expression, you change the inequality but nothing else changes, i.e. no other signs are reversed during that operation.

Example:
Equation is:
(x+1)/(2x-3)) > y

While (2x-3) is positive, multiply both sides of this equation by (2x-3).

Your answer will be:
(x+1) > y * (2x-3)
Nothing happened other than you wound up moving the denominator over to the right hand side of the equation as you would normally do when you multiply both sides of an equation by a denominator on one side of the equation. The inequality is still greater than.

While (2x-3) is NEGATIVE, multiply both sides of this equation by (2x-3).

Your answer will be:
(x+1) < y * (2x-3)
The same thing happened EXCEPT now the inequality is reversed. The inequality is now smaller than, when before it was greater than.

You will see all this happening as we go through the example.

APPLY BASIC DEFINITION OF ABSOLUTE VALUES TO GET THE POSSIBLE SOLUTION SETS

by the basic definition of absolute values, |(x+1)/(2x-3)| < 2 becomes:

when the expression within the absolute value signs is positive.
and:
when the expression within the absolute value signs is negative.

When you multiply by -1 on both sides of the equation, it becomes:
when the expression within the absolute value signs is negative.

POSSIBLE SOLUTION SETS

You have 2 possible solution sets.

set 1 is when the expression within the absolute value signs is positive and is:


set 2 is when the expression within the absolute value signs is negative and is:


DETERMINE WHEN THE EXPRESSION WITHIN THE ABSOLUTE VALUE SIGN IS POSITIVE AND WHEN THE EXPRESSION WITHIN THE ABSOLUTE VALUE SIGN IS NEGATIVE.

The numerator is positive when which becomes when
The denominator is positive when which becomes when .

Note that x cannot be equal to 3/2 because then the denominator in the equation is equal to 0 which is not a real number and therefore can't be part of any solution to this problem.

Numerator is >= 0 when x >= -1
Denominator is >= 0 when x > 3/2 (= 3/2 not allowed).

The expression is positive when both the numerator and the denominator are both positive. This happens when .

The expression is also positive when both the numerator and the denominator are both negative.

this happens when .

The expression is positive when:
or

This means the expression is negative when:
and
This can be written as:


remember that x cannot be equal to 3/2.

SOLVE FOR WHEN THE EXPRESSION WITHIN THE ABSOLUTE VALUE SIGNS IS POSITIVE.

When the expression within the absolute sign is positive, the possible solution set is:


You have two possible scenarios within this solution set.
The first scenario is when the denominator is positive.
The second scenario is when the denominator is negative.

If the denominator is positive, then you would solve this is as follows:

Equation to solve is:

Multiply both sides of this equation by (2x-3) to get:

Remove parentheses to get:

subtract 1 from both sides and subtract 3x from both sides to get:

divide both sides by 3 and multiply both sides by -1 to get:


If the denominator is negative, then you would solve this as a follows:

Equation to solve is:

Multiply both sides of this equation by (2x-3) to get:


Note that the inequality reversed because we were multiplying by a negative number which is represented by the expression (2x-3) in the denominator.

Remove parentheses to get:

subtract 4x from both sides of the equation and subtract 1 from both sides of the equation to get:

multiply both sides of this equation by -1 and divide both sides of this equation by 3 to get:


You have 2 possible solutions when the expression within the absolute value sign is positive.

Those 2 possible solutions are:
when the denominator is positive
when the denominator is negative

When the expression is positive and the denominator is positive, x has to be greater than 3/2, so x > 7/3 does not have to be modified and can stand as is.

When the expression is positive and the denominator is negative, x has to be less than -1 so x < 7/3 has to be modified to say x < -1

Your possible solutions for when the absolute value expression is positive are:
or

SOLVE FOR WHEN THE EXPRESSION WITHIN THE ABSOLUTE VALUE SIGNS IS NEGATIVE

When the expression within the absolute value signs is negative, the possible solution set is:


The boundaries for this occur when:
and

The denominator of 2x-3 is negative throughout this boundary so we only have to solve for when the denominator is negative and do not have to solve for when the denominator is positive.

When the denominator is negative you would solve as follows:

Equation to solve is:

Multiply both sides of this equation by (2x-3) to get:


Note that (2x-3) is a negative number so multiplying both sides of this equation by a negative number reverses the inequality.

Remove parentheses to get:

Subtract 1 from both sides of this equation and add 4x to both sides of this equation to get:

divide both sides of this equation by 5 to get:
when the denominator is negative.

Since the solution set is negative when and , then the possible solution when the expression within the absolute value sign is negative is:
and

this can be written as:


PUT ALL THE INFORMATION TOGETHER SO YOU CAN ANALYZE IT TO CONFIRM THAT THE ANSWER IS GOOD.

When the expression within the absolute value signs is positive, the possible solutions are:
or
When the expression within the absolute value signs is negative, the possible solutions are:


These 2 solutions can be combined to show as:
or
We test the intervals to see if they are accurate by taking values within and without those intervals to see if the equation is true or not.

Test values for x will be:
-2 (should be good)
-1 (should be good)
0 (should be good)
1 (should not be good)
7/3 (should not be good)
8/3 (should be good)

When x = -2:
becomes .14 < 2 so this is good as it should be.
When x = -1:
becomes 0 < 2 so this is good as it should be.
When x = 0:
becomes .33 so this is good as it should be.
When x = 1:
becomes 2 NOT < 2 so this is NOT good as it should be.
When x = 7/3:
becomes 2 so this is NOT good as it should be.
When x = 8/3:
becomes 1.57 so this is good as it should be.

Questions or comments regarding this lesson can be directed to dtheophilis@yahoo.com