3.2 Solving Inequalities
- Page ID
- 152961
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- solve basic linear inequalities
- solve quadratic and rational inequalities using positive/negative zone analysis
- solve simultaneous inequalities and ones involving absolute value
An inequality is also a statement that we want to be true, it's just looser than an equation. Instead of an equals sign, these will have inequality symbols in them, namely, \(<, \leq, >, \geq \), which mean "(strictly) less than, less than or equal to, (strictly) greater than, greater than or equal to," respectively. Here are some examples.
- \( x + 7 > 3 \) means "The quantity \( x + 7 \) is a bigger number than 3."
- \( x^2 + x - 2 \leq 0 \) means "The quantity \( x^2 + x - 2 \) is less than or equal to 0, aka is 0 or a negative number."
Solving an inequality is still just about isolating \(x\) on one side by performing legal inverse operations across the whole inequality.
- You can add or subtract the same thing on both sides of an inequality, and it won't change:
\[A < B \iff A + C < B + C \notag \] - You can multiply or divide something positive on both sides of an inequality, and it won't change: if \textcolor{blue}{\(C \geq 0\),}
\[ A < B \iff AC \textcolor{blue}{<} BC \notag \] - ALERT!!! If you multiply or divide both sides by a negative number, you must flip the direction of the inequality sign: if \(C \textcolor{red}{<} 0\),
\[ A<B \iff AC \textcolor{red}{>} BC \notag \]
Solving an inequality results in a whole slew of solutions. The statement "\(x \geq 3 \)" will be true if \(x\) takes the value 3, or the value 5, or the value 1000, or any of the infinitely many values bigger than 3. I want to show you the weird stuff that can happen when you solve inequalities via some examples, so read all of these carefully. First, the basic situation.
Solve the inequalities.
- \( 2x + 8 < 4 \)
- \( 9x - 3 \geq 0 \)
- \( 6 - x \leq 1 \)
- \( -\frac{x}{2} + 1 > 10\)
Solution
1. We undo all the stuff that happened to \(x\) using inverse operations just like with equations. He was multiplied by 2, and then added to 8, so we subtract 8 from both sides, and then divide by 2. Nothing needs to change on the inequality symbol.
\[ 2x + 8 < 4 \quad \longrightarrow \quad 2x < 4-8 \quad \longrightarrow \quad \frac{2x}{2} < \frac{-4}{2} \quad \longrightarrow \quad x < -2 \notag \]
Check your answers by taking some options from the solution statement and plugging them back in. For example, plugging in \(-3\) would give \(2(-3)+8 = 2 \), which is indeed less than 4. You can write your solution in interval notation (review in the domain section) as \( (-\infty, -2) \).
2. Again, undo the multiplication by 9 and subsequent subtraction by 3 by adding 3 and dividing by 9.
\[ 9x - 3 \geq 0 \quad \longrightarrow \quad 9x \geq 0 + 3 \quad \longrightarrow \quad \frac{9x}{9} \geq \frac{3}{9} \quad \longrightarrow \quad x \geq \frac{1}{3} \notag \]
Because \(x\) can be greater than or equal to \( \frac{1}{3}\), plugging in \(\frac{1}{3}\) itself should work. Test: \( 9\cdot \frac{1}{3} - 3 \) does indeed equal 0. In interval notation, the answer is \( \left[\frac{1}{3}, \infty\right) \).
3. Here as we go to undo the operations, we will have to divide by a \(-1\). At that point, we have to flip the inequality sign!
\[ 6 - x \leq 1 \quad \longrightarrow \quad -x \leq 1 - 6 \quad \longrightarrow \quad \frac{-x}{-1} \textcolor{red}{\geq} \frac{-5}{-1} \quad \longrightarrow \quad x \geq 5 \notag \]
Plug in something larger than \(5\) for \(x\) to test. In interval notation, we say \( [5, \infty) \).
4. As we isolate \(x\), we will need to multiply by a negative number, so again watch out for flipping the sign!
\[ -\frac{x}{2} + 1 > 10 \quad \longrightarrow \quad -\frac{x}{2} > 10 - 1 \quad \longrightarrow \quad -\frac{x}{2} \cdot (-2) < 9 \cdot (-2) \quad \longrightarrow \quad x < -18 \notag \]
Plug in something strictly greater than \(-18\) to check. In interval notation, we say \( (-\infty, -18) \).
Solve the inequalities.
- \( 12x - 144 \geq 0 \)
- \( 4 - \frac{3}{2}x < -5 \)
- Answer
-
- \( x \geq 12 \) or \( [12, \infty) \)
- \( x > 6\) or \( (6, \infty) \)
When more powers of \(x\) or factors are involved, we need to use logic and cases to cover all our bases. The easiest method is using a number line, so let me show you that technique.
Solve the inequality \( x^2 \geq 5x - 6 \).
Solution
If this were an equation, we would move everything to one side and factor, so let's start with that.
\[ x^2 - 5x + 6 \geq 0 \quad \longrightarrow \quad (x-3)(x-2) \geq 0 \notag \]
This is saying we want the product \( (x-3)(x-2) \) to turn out to be 0 or positive. For a product to be positive, we could have multiplied two positive quantities, or two negative quantities. If you follow this number line analysis method, you will cut to the chase on covering those cases. Start by marking 2 and 3 on a number line, the values that would have been solutions to this quadratic if it had been an equation instead.
These two points divide the number line into three zones: \( (-\infty, 2)\), \( (2,3)\), and \((3, \infty) \). In each zone, pick an easy test point to plug into the factors to see if the result is positive or negative. In the first zone, I pick 0 and plug in: \((0-3)(0-2) = (-3)(-2) = 6\) is positive. In the middle, I pick 2.5 and plug in: \( (2.5-3)(2.5-2) = (-0.5)(0.5) \) results in a negative number. In the last zone, I pick 4: \( (4-3)(4-2) = (1)(2) \) is positive. Mark these.
Since we want the product to be 0 or positive, we report the positive zones and include their endpoints, giving the answer \( (-\infty, 2] \cup [3, \infty) \). We would show this answer on a number line with closed dots, like so:
Solve the inequality \( x(x-1)^2 > 0 \).
Solution
Here we're already factored nicely, and we note that if this were an equation, the solutions would be \(x = 0, 1\). Mark them on a number line.
We are divided into three zones again: \((-\infty, 0), (0, 1), \) and \((1, \infty) \). Pick test points again and determine the positive and negative zones. I'm going to test...
- \(x = -1\) gives \( (-1)(-1-1)^2 = (-1)(-2)^2 = \) negative.
- \(x = 0.5\) gives \( (0.5)(0.5-1)^2 = (0.5)(-0.5)^2 = \) positive.
- \( x = 2 \) gives \( (2)(2-1)^2 = (2)(1)^2 = \) positive.
I want to report positive zones in my answer, but since the inequality is strict, I cannot include the endpoints! I report \( (0, 1) \cup (1, \infty) \), and I could draw this as:
Solve the inequality \( \dfrac{1+x}{1-x} \leq 1 \)
Solution
The first step is to get 0 on one side and a single rational expression on the other, so we can use positive/negative analysis as we have been doing. We subtract over the 1 and find a common denominator so we can create one fraction.
\[ \dfrac{1+x}{1-x} - 1 \leq 0 \quad \longrightarrow \quad \dfrac{1+x}{1-x} - \dfrac{1-x}{1-x} \leq 0 \quad \longrightarrow \quad \dfrac{1+x - (1-x)}{1-x} \leq 0 \notag \]
Simplifying gives us the inequality \( \dfrac{2x}{1-x} \leq 0\). Once again, look at the factors involved, \(2x\) and \( (1-x)\). These are zero if \(x\) is 0 or 1. Mark those points of interest.
Now we analyze zones by testing easy points to plug in (you try!) and get the following result.
We want to report the negative zones, but we still need to think about the endpoints. We plug in \(x = 0\) to the original inequality to test: yes, \(\frac{1+0}{1-0}\) is indeed less than or equal to 1. But at \(x = 1\), the left side of the inequality is undefined due to dividing by zero!! We leave him out. The answer is \( (-\infty, 0] \cup (1, \infty) \).
Solve the inequalities.
1. \( x^2 - 4x > 0 \)
2. \( \dfrac{x}{x+1} + \dfrac{1}{x} \leq 0 \) (Remember to check your answers against the original inequality.)
- Answer
-
- \( x < 0 \) or \(x > 4\), or in interval notation, \( (-\infty, 0) \cup (4, \infty) \).
- \( -1 < x < 0 \), or in interval notation, \( (-1,0)\).
Bookends and uhhhhh bookCases
Sometimes you might see inequalities that contain two signs, meaning that the quantity is supposed to satisfy two conditions. For example,
\[ 0 < x + 4 \leq 10 \notag \]
means that \( (x+4)\) is supposed to be a value greater than 0, but not more than 10. I think of these as bookends. You can still do the algebraic operations to this inequality, you just have to apply them across the entire thing.
Solve the simultaneous inequality \( 0 < x+4 \leq 10 \).
Solution
My goal is always to get \(x\) by himself. Currently there is an addition by 4 that I don't want, so to get rid of that, I subtract 4 from each part of the inequality and simplify.
\[ 0\textcolor{red}{ - 4} < x + 4\textcolor{red}{ - 4} \leq 10 \textcolor{red}{- 4} \quad \longrightarrow \quad -4 < x \leq 6 \notag \]
This tells me that the solution is for \(x\) to be bigger than \(-4\) but not more than \(6\). You can leave it like that, or you can report in interval notation, \( (-4, 6]\).
Remember the absolute value bars? We described the effect they have using cases. What matters is whether the inside is positive or negative. For example,
\[ |x + 4| = \begin{cases} x+4 & \text{ if } (x+4) \geq 0 \\ -(x+4) & \text { if } (x+4) < 0 \end{cases} \notag \]
A mistake I see ALL THE TIME is when people think the effect of absolute value bars is to individually change the sign of each term inside. "Absolute"ly not! (lol) You must take the opposite of the whole entire inside. Split into the two cases: "|inside|" will...
- spit out "inside" when the inside is nonnegative
- spit out "\(-\)(inside)" when the inside is negative
If you see absolute value bars appear in an inequality, all you gotta do is isolate that part, and then break it down into these two cases!
- For \( < \) and \(\leq \), split into cases using AND:
\[ |\text{inside}| < a \quad \implies \quad (\text{inside})<a \quad \text{AND}\quad -(\text{inside}) < a \notag \] - For \( >\) and \( \geq\), split into cases using OR:
\[ |\text{inside}| > a \quad \implies \quad (\text{inside})> a \quad \text{OR}\quad -(\text{inside}) > a \notag \]
Solve the inequalities.
1. \( |3x + 2| \leq 8 \)
2. \( |12x - 30 | -6 > 0 \)
Solution
1. The absolute value is by itself on one side, so we go ahead and look at the inequality sign. Since it's "\(\leq\)," we split into the two options with an AND:
\[ (3x + 2) \leq 8 \quad \text{ AND } \quad -(3x + 2) \leq 8 \notag \]
Now solve each inequality separately. We get as our answer, "\( x \leq 2 \) and \( x \geq -2 \)," which could also be written "\(-2 \leq x \leq 2 \)." In interval notation, this is \([-2, 2]\).
2. First isolate the absolute value part! Add 6 on both sides to get \( |12x - 30| > 6 \). Now split into cases with an OR:
\[ (12x - 30) > 6 \quad \text { OR } \quad -(12x - 30) > 6 \notag \]
Solving each inequality separately, we get "\( x > 3 \) or \( x< 2 \)," which cannot be written as one statement do not do it I will mail you a dead fish and you'll regret it. But you can write it in interval notation, as \( (-\infty, 2) \cup (3, \infty) \).
Now you try.
Solve the inequalities.
- \(-2 \leq 2x -1 \leq 5 \)
- \( |x - 10| - 5 < 0 \)
- \( |2x+4| \geq 6 \)
- Answer
-
- \( - \frac{1}{2} \leq x \leq 3 \), or \( \left[ -\frac{1}{2}, 3 \right] \)
- \( 5 < x < 15 \), or \( (5, 15) \)
- "\( x \geq 1 \) or \(x \leq -5\)," or \( (-\infty,5] \cup [1, \infty) \)