Solving Linear Inequalities Alternative Format

Lesson Part 1

In This Module

• We will explore how to solve linear inequalities

In past experiences, we have solved a variety of different equations using a variety of techniques. In particular, we have solved linear and quadratic equations.

We have discovered that we can perform the same operation on both sides of the equation without affecting the equality.

Is the same true for inequalities?

Operations on Linear Inequalities

Let's look at some specific inequalities and perform operations to both sides to determine if the inequality still holds.

Inequality	Operation	Left Side	Right Side	Inequality still true?
\(-2\lt5\)	add \(3\) to both sides	\(1\)	\(8\)	\(1\lt8\)	True
\(-2\lt5\)	subtract \(3\) from both sides	\(-5\)	\(2\)	\(-5\lt2\)	True
\(-8\gt-12\)	multiply both sides by \(2\)	\(-16\)	\(-24\)	\(-16\gt-24\)	True
\(-8\gt-12\)	divide both sides by \(2\)	\(-4\)	\(-6\)	\(-4\gt-6\)	True
\(-8\lt-6\)	multiply both sides by \(-2\)	\(16\)	\(12\)	\(16\lt12\)	False
\(8\gt-12\)	divide both sides by \(-2\)	\(-4\)	\(6\)	\(-4\gt6\)	False

When solving inequalities, you can add or subtract the same number from each side of the inequality and the inequality will remain true.

You can multiply or divide both sides of the inequality by the same positive number and the inequality will remain true.

However, when you multiply or divide both sides of an inequality by the same negative number, the resulting inequality becomes false.

To make the inequality true again, reverse the direction of the inequality symbol.

Lesson Part 2

Examples

Example 1

Solve \(-3x+1\gt-8\), where \(x\in \mathbb{R}\). Write the solution using interval notation and illustrate the solution on a number line.

Solution

In this solution we will write a descriptor to the left to remind you what has been done.

\[-3x+1 \gt-8 \]

Subtract \(1\) from both sides:

Divide both sides by \(\color{BrickRed}-3\):

This leaves us with a solution of \(x \lt 3\).

So we'll draw the number line for this solution. \(3\) is not included in the interval. But all points to the left are. So we put an open dot at \(3\) and an arrow going to the left.

Using interval notation, \(x\in (-\infty,3),\ x\in \mathbb{R}\).

If we choose any value of \(x\) less than \(3\), the inequality will be true.

We should be able to check values of \(x\) into the original inequality, \(-3x + 1 \gt -8\), to see whether or not they are actually true.

If we choose any value of \(x\) greater than or equal to \(3\), the inequality will be false.

\[ \begin{align*} \text{When} \quad x=1,\quad -3(1)+1 &\gt -8 \\ -2 &\gt -8 \quad \color{BrickRed}\text{TRUE} \end{align*} \]\[ \begin{align*} \text{When} \quad x = 5, \quad -3(5)+1 &\not\gt-8 \\ -14 &\not\gt -8 \quad \color{BrickRed} \text{FALSE} \end{align*} \]

You might wish to take some more time to test other values for yourself before proceeding.

Check Your Understanding A

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Example 2

Solve \(-4x-1\le 2x+5 \), where \(x\in \mathbb{R}\). Write the solution using interval notation and illustrate the solution on a number line.

Solution

Looking at the solution, you will see some editorial comments to the left of each line. And you'll see the solution presented before you.

\[-4x-1 \le 2x+5\]

Subtract \(2x\) from both sides:

\[-6x-1 \le 5\]

Add \(1\) to both sides:

\[-6x \le 6\]

Divide both sides by \(\color{BrickRed}-6\):

\[x \color{BrickRed}\ge\color{black}-1\]

Using interval notation, \( x\in [-1,+\infty),\ x\in \mathbb{R}\).

Check Your Understanding B

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Lesson Part 3

Examples

In this next example we will look at two different ways to solve an inequality.

Example 3

Solve \(-1\lt\dfrac{4-x}{2}\lt 3,\ x\in \mathbb{R}\).

Write the solution using interval notation and illustrate the solution on a number line.

Solution 1

In solution 1, all we're going to do is break the inequality into two separate inequalities.

We want \(-1\lt\dfrac{4-x}{2}\) and \(\dfrac{4-x}{2}\lt 3\).

Whatever solution we find must satisfy both inequalities.

For \(x\lt 6\), we have an open dot at \(6\) and an arrow to the left.

For \(x\gt -2\), we have an open dot at \(-2\) and an arrow going to the right.

We want the intersection of these two solutions. That is, we want all numbers that satisfy both inequalities at the same time.

So in this case, if you look at the two number lines, you will see that numbers from \(-2\) to the left are not on both number lines. You will see that numbers on the first number line from \(6\) to the right are not on both number lines.

But between \(-2\) and \(6\), the numbers are on both. Therefore, any number between \(-2\) and \(6\) is in the solution.

Note that \(-2\) is not in both solutions. And note that \(6\) is not in both solutions.

So when we compress the two number lines into one, we see an open dot at \(-2\), an open dot at \(6\), and a line connecting the two.

Using interval notation, \( x\in (-2,6),\ x\in \mathbb{R}\).

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Example 3

Solve \(-1\lt\dfrac{4-x}{2}\lt 3,\ x\in \mathbb{R}\).

Let's revisit this example.

Solution 2

In this solution, we will work directly with the three parts of the inequality. When we perform an operation, we will do it to each part of the inequality.

\[-1 \lt \dfrac{4-x}{2} \lt 3 \]

Multiply each part by \(2\).

\[-2 \lt 4-x \lt 6\]

Subtract \(4\) from each part.

\[-6\lt -x \lt 2\]

Multiply each part by \(-1\).

\[6\textcolor{BrickRed}{\gt} x \textcolor{BrickRed}{\gt} -2\]

Using interval notation, \( x\in (-2,6),\ x\in \mathbb{R}\).

This results in the same solution. Sometimes we're able to solve inequalities this way.

Example 4

Solve \(9+4x\lt3x+2\le -1,\ x\in \mathbb{R}\). Write the solution using interval notation and illustrate the solution on a number line.

Solution

This example, in some ways, look similar to the previous example. However, there's a variable term in more than one part of the inequality. So we have to use the procedure seen in solution \(1\).

We will separate the inequality into two separate parts.

We want \(9+4x \lt 3x+2\) and \(3x+2\le -1\). Whatever solution we find must satisfy both inequalities.

For \(x\lt -7\), we have an open dot at \(-7\), a line drawn to the left.

For \(x\le -1\), we have a solid dot at \(-1\) and a line drawn to the left.

We want the intersection of these two solutions. That is, we want all numbers that satisfy both inequalities at the same time.

Any number less than \(-7\) satisfies both inequalities.

Using interval notation, \( x\in (-\infty,-7),\ x\in \mathbb{R}\).

Check Your Understanding C

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Lesson Part 4

Extension

Let's look at an inequality that's not linear.

Example 5

Solve \((2x+5)(x-1)\gt 0,\ x\in \mathbb{R}\).

Write the solution using interval notation and illustrate the solution on a number line.

Solution

This problem doesn't look very familiar to anything we've been doing. But if we examine the inequality, we see that we have a binomial times a binomial such that the product is greater than \(0\).

To obtain a product greater than zero, we multiply two positive quantities or two negative quantities.

We want

Note
Notice that if \(2x+5\gt 0\) and \(x-1\lt 0\), then \((2x+5)(x-1)\lt 0\).
Similarly, if \(2x+5\lt 0\) and \(x-1\gt 0\), then \((2x+5)(x-1)\lt 0\).

From the first pair of inequalities, we want \(x\gt -\dfrac{5}{2}\) and \(x\gt 1\) at the same time. Therefore, \(x\gt 1\).

From the second pair of inequalities, we want \(x\lt -\dfrac{5}{2}\) and \(x\lt 1\) at the same time. Therefore, \(x \lt -\frac{5}{2}\).

If you're stuck, draw \(3\) number lines and you'll see what the intersection is.

Since \(x\gt 1\) or \(x\lt -\dfrac{5}{2}\), then in interval notation, \(x\in (-\infty,-\dfrac{5}{2}) \cup (1,\infty),\ x\in \mathbb{R}\).

Lesson Part 5

Further Extension

One final little extension to our inequalities.

Example 6

Solve \(\dfrac{3}{x} \lt 5,\ x\in \mathbb{R}\).

Write the solution using interval notation and illustrate the solution on a number line.

Solution

Ideally, we would like to multiply both sides of the inequality by \(x\) to clear the fraction.

When \(x\gt 0\), there is no problem.

However, when \(x\lt 0\), we are multiplying both sides of the inequality by a negative number and the direction of the inequality symbol must switch.

We also know that \(x\not =0\) since division by zero is undefined.

Let's break the solution into cases.

From Case 1, we have \(x\gt\dfrac{3}{5}\).
From Case 2, we have \(x\lt 0\).

It is either the solution from Case 1 or the solution from Case 2, so we want the union of the two solutions.

On a number line the solution is:

Using interval notation, \(x\in (-\infty,0)\cup \left(\dfrac{3}{5},\infty\right),\ x\in \mathbb{R}\).

To verify the solution, substitute values of \(x\) less than \(0\) or greater than \(\dfrac{3}{5}\) into the inequality. The resulting inequality will be true.

If you substitute values for \(x\) greater than \(0\) and less than or equal to \(\dfrac{3}{5}\), then the resulting inequality will be false.