I’ve been getting a number of questions lately about inequalities. So I think it’s time to add a post that can serve as a reference that the student can use to get them back on track, or just to get them started. This will be more about the mechanics of solving inequalities than going into the why of things. My goal here is to help the students who are short on time, working through problem sets, and appreciate some quick guidance.
Inequalities can be a stumbling block for math students because the rules are similar to how we solve equations, but with a couple of twists. We will start by talking about solution sets, then move into the different forms of inqualities students most often encounter.
Solution Sets
Solution sets contain all of the values that make the inequality true when you replace the variable, or variables, with those values. For instance, consider, \( x > 5 \):
- Replacing x with 6 give us, \( 6 > 5 \), which is a true statement because 6 is definitely a greater number than 5. That means 6 is in the solution set of this inequality.
- Replacing x with 1 we get, \( 1 > 5 \), which is not a true statement because 1 less than 5, not greater than. That means 1 is NOT in the solution set of this inequality.
Interval Notation
We can express solution sets several in different ways. In high school or college algebra we most often express it in interval form such as, \( (5, \infty) \), \((-\infty, 5]\), or \([3, 6]\). The number on the left represents the lower bound of the solution set. The number on the right represents the upper bound. All real values between are part of the solution set. If the solution set includes the value on the left we start with the opening bracket, [. If it includes the value on the right we end with the closing brack, ]. Otherwise we use the open or close parenthesis, ( ). If the solution set extends all the way to plus or minus infinity we place parenthesis next to the appropriate symbol, \(+\infty\) or \(-\infty\).
Set Builder Notation
In set builder notation, a variable is separated from a list of constraints with a vertical bar. The vertical bar is symbol that represents the phrase “such that”. The list of constraints define properties that the variable must take on in order to be included as an element of the set, or as part of the solution.
Here are a couple of examples:
“A contains the values of x such that x is a real number and x is less than 5“
\[A = \{x \mid x \in \mathbb{R}, x < 5\}\]
“A contains the values of x such that x is a natural number”
\[A = \{x \mid x \in \mathbb{N}\}\]
Graphing on Number Line
Stating the solution of an inequality on a number line can gives us an intuitive understanding what the inequality represents. Here is a list of rules we use to display a solution on the number line:
- If a line or line segment has an open circle on it, the value at that point is not part of the solution. Sometimes you will see an open or close parenthesis used instead.
- If a line or line segment has a closed circle on it, the value at that point is part of the solution. Sometimes you will see an open or close square bracket used instead.
- If a line has an arrow on the end pointing to the right, the set extends from the lower bound to \(+\infty\).
- If a line has an arrow on the end pointing to the left, the set extends from the upper bound to \(-\infty\).
Here is a table of solution sets expressed in interval and set builder notation, along with a number line representation:
Interval Notation | Set Builder Notation | Number Line |
---|---|---|
\((a, b)\) | \(\{x \mid a < x < b \}\) | |
\([a, b) ]\) | \(\{x \mid a \le x \le b \}\) | |
\([a, b)\) | \(\{x \mid a \le x < b \}\) | |
\((a, b]\) | \(\{x \mid a < x \le b \}\) | |
\((a, \infty)\) | \(\{x \mid a < x < \infty \}\) | |
\([a, \infty)\) | \(\{x \mid a \le x < b \}\) | |
\((-\infty, b)\) | \(\{x \mid -\infty < x < b \}\) | |
\((-\infty, b)]\) | \(\{x \mid -\infty < x \le b \}\) | |
\((-\infty, \infty)\) | \(\{x \mid -\infty < x < \infty \}\) |
Solving Inequalities
The example, \( x > 5 \), is the simplest form of a linear inequality. The solution set is given to us. We don’t have to rearrange terms to figure out what the upper and lower bounds are. Inequalities can get progressively more intricate though.
The following examples demonstrate the rules we use to getting x by itself on one side of an inequality.
- When adding and subtracting we apply the same rules we do when solving an equality.
- When multiplying or dividing by a positive number we apply the same rules we do when solving an equality.
- When multiplying or dividing by a negative number we apply the same rules we do when solving an equality, but we have to flip the direction of the inequality as well.
Examples
- \( 3+2x < 11 \)
- \( 2x < 8 \)
- \( x < 4 \)
- \( (-\infty, 4) \)
- \( 2+3x \leq 6 \)
- \( 3x \leq 6 \)
- \( x \leq 2 \)
- \( (-\infty, 2] \)
- \( x+6 > 2x-3 \)
- \( -x+6 > -3 \)
- \( -x > -9 \)
- \( x < 9 \)
- \( (-\infty, 9) \)
- \( 20 \geq 5x+5 \)
- \( 15 \geq 5x \)
- \( 3 \geq x \)
- \( x \leq 3 \)
- \( (-\infty, 3] \)
- \( -2x-4x > -x + 5 \)
- \( -6x > -x + 5 \)
- \( -5x > 5 \)
- \( x < -11 \)
- \( (-\infty, -11) \)
- Subtract 3 from both sides
- Divide both sides by positive 2
- Solution in interval form
- Subtract 2 from both sides
- Divide both sides by positive 3
- Solution set includes 2
- Subtract 2x from both sides
- Subtract 6 from both sides
- Multiply both sides by -1, flip inequality
- Solution set in interval notation.
- Subtract 5 from both sides..
- Divide both sides by positive 5.
- Switching sides changes direction inequality
- Combine like terms.
- Add +x to both sides.
- Dividing both sides by -5 flips inequality.
Compound Inequalities
The following are examples of compound equalities. Note that our operations are performed on all three parts of these inequalities. The goal is to end up with the variable to be by itself between in the middle.
Examples
- \( -5 < 2x + 1 \leq 9 \)
- \( -6 < 2x \leq 8 \)
- \( -3 < x \leq 4 \)
- \( (-3, 4] \)
- \( -3 \leq \frac{2}{3}x – 5 < -1 \)
- \( -9 \leq 2x – 15 < -1 \)
- \( 6 \leq 2x < 14 \)
- \( 3 \leq x < 7 \)
- \( [-2, 2) \)
- \( -6 \leq -\frac{1}{2}x – 4 \leq -3 \)
- \( -12 \leq -x – 8 \leq -6 \)
- \( -4 \leq -x \leq 2 \)
- \( 4 \geq x \geq -2 \)
- \( -2 \leq x \leq 4 \)
- \( [-2, 4] \) .
- Subtract 1 from each part
- Divide all three parts by 2
- Solution set in interval notation
- Multiply all three parts by 3
- Add 15 to all three parts
- Divide all three parts by 2
- Solution set includes -2 but not 2
- Multiply all three parts by 2
- Add 8 to all three parts
- Multiply each part by -1 and flip
- Exchange outer parts and flip
- Solution includes both bounds
Absolute Value Inequalities
Absolute Value Inequalities represent the intersection or union of two solutions. This is due to the fact that the absolute value of a number, as well as it’s negative, is the magnitude of that number. That means there are two values which satisfy the equation. So, to solve an absolute value inequality we have to create two separate equations. Then we take the intersection or union of the solutions to those two equations depending on which inequality is being used.
Intersection of Two Solutions
Absolute value inequalities with less than symbols, \( < \) and \( \leq \), get rewritten as compound inequality. For example \( (\mid a \mid \leq b) \) is rewritten as \( (-b \leq a \leq b) \):
- Remove absolute value symbol
- Value on the right stays the same
- Value on the left changes sign.
Examples
- \( \mid 2x- 6 \mid < 8 \)
- \( -8 < 2x – 6 < 8 \)
- \( -2 < 2x < 14 \)
- \( -1 < x < 7 \)
- \( (-1, 7) \)
- \( -8 < 2x – 6 < 8 \)
- \( \mid \frac{2x+6}{3} \mid < 2 \)
- \( -2 < \frac{2x+6}{3} < 2 \)
- \( -6 < 2x + 6 < 6 \)
- \( -12 < 2x < 0 \)
- \( -6 < x < 0 \)
- \( (-6, 0) \)
- \( -2 < \frac{2x+6}{3} < 2 \)
- \( -3 \mid x + 7 \mid – 3 \geq -30 \)
- \( -3 \mid x + 7 \mid \geq -27 \)
- \( \mid x + 7 \mid \leq 9 \)
- \( -9 \leq x + 7 \leq 9 \)
- \( -16 \leq x \leq 2 \)
- \( [-16, 2] \)
- Rewrite as compound inequality
- Add 6 to all three parts
- Divide all three parts by 2
- Rewrite as compound inequality
- Multiply all three parts by 3
- Subtract 6 from all three parts
- Divide all three parts by 2
- Get absolute value expression by itself
- Add 3 to both sides of inequality
- Divide both sides by -3 and flip arrow
- Rewrite as a compound expression
- Subtract 7 from all three parts
Union of Two Solutions
For \( > \) and \( \geq \) inequalities we rewrite the absolute inequalities as two separate inqualities without the absolute value symbol. For example \( (\mid a \mid \geq b) \) is rewritten as \( (a \geq b) \) and \( (a \leq -b) \). The final solution set will be the union, \( \cup \), of the solutions we arrive at for each separate inequality.
Examples
- \( \mid x + 3 \mid \geq 4 \)
- \( x + 3 \geq 4 \)
- \( x \geq 1 \)
- \( [1, \infty] \)
- \( x + 3 \leq -4 \)
- \( x \leq -7 \)
- \( (-\infty, -7] \)
- \( (-\infty, -7] \cup [1, \infty] \)
- \( x + 3 \geq 4 \)
- \( 5 \mid 2x + 1 \mid – 3 \geq 9 \)
- \( 5 \mid 2x + 1 \mid \geq 12 \)
- \( \mid 2x + 1 \mid \geq \frac{12}{5} \)
- \( 2x + 1 \geq \frac{12}{5} \)
- \( 2x \geq \frac{7}{5} \)
- \( x \geq \frac{7}{10} \)
- \( [\frac{7}{10}, \infty) \)
- \( 2x + 1 \leq -\frac{12}{5} \)
- \( 2x \leq -\frac{17}{5} \)
- \( x \leq -\frac{17}{10} \)
- \( (-\infty, -\frac{17}{10}]\)
- \( (-\infty, -\frac{17}{10}] \cup [\frac{7}{10}, \infty) \)
- Create first of two inequalities
- Subtract 3 from both sides
- First solution set
- Create second inequality
- Subtract 3 from both sides
- Second solution set
- Union of first and second solutions
- Isolate the absolute expression
- Add three to both sides
- Divide both sides by 5
- Create first of two inequalities
- Add 1 to both sides
- Divide both sides by 2
- First solution set
- Create second inequalities
- Subtract 1 from both sides
- Divide both sides by 2
- Second solution set
- Union of both solution sets
Solutions Containing All Real Numbers
- \( x -1 < x\)
- \( -1 < 0 \)
- \( (-\infty, \infty) \)
- \( x+1 > x \)
- \(1 > 0 \)
- \( (-\infty, \infty) \)
- \( 3(x+1) > 3x+2 \)
- \( 3x+3 > 3x+2 \)
- \( 3 > 2 \)
- \( (-\infty, \infty) \)
- Subtract the variable, x, from both sides.
- True for all values of x, \(-\infty\) to \(+\infty\)
- Subtract x from both sides.
- True for all values of x, \(-\infty\) to \(+\infty\)
- Distribute the 3 across expression on left
- Subtract 3x from both sides
- True for all values of x, \(-\infty\) to \(+\infty\)
Note: the union of inequalities, \( x > a\), and \(x < b\), where \( a < b \), is the set of all real numbers.
Solutions Containing the Empty Set
- \( x+1 < x \)
- \( 1 < 0 \)
- \( (\varnothing) \)
- \( x+5 \leq x-6 \)
- \( 5 \leq -6 \)
- \( (\varnothing) \)
- Subtract x from both sides.
- 1 is never < 0, so solution is the empty set.
- Subtract x from both sides.
- 5 is never < -6, so the solution is the empty set.
Note: \( (\mid a \mid < b) \), where \(b \leq 0 \) will always be an empty set, \(\varnothing\).