Linear Equations and Inequalities in One Variable

Equality and Inverse Operations are the foundation of solving linear equations. When you solve an equation, your goal is to find the value of the variable that makes the equation true.

An equation is a math sentence that says two expressions are equal. For example, in the equation \(x + 5 = 12\), the left side and the right side have the same value.

To solve an equation, you must keep it balanced. You can think of an equation like a balanced scale: if you do something to one side, you must do the same thing to the other side so the balance stays equal.

This idea is called the properties of equality. The most important ones for solving equations are:

• Addition Property of Equality: If \(a = b\), then \(a + c = b + c\).
• Subtraction Property of Equality: If \(a = b\), then \(a - c = b - c\).
• Multiplication Property of Equality: If \(a = b\), then \(ac = bc\).
• Division Property of Equality: If \(a = b\) and \(c \ne 0\), then \(\frac{a}{c} = \frac{b}{c}\).

These properties let you change an equation without changing its truth. That is why they are so useful when solving for a variable.

An inverse operation is an operation that undoes another operation.

• Addition and subtraction are inverse operations.
• Multiplication and division are inverse operations.

For example:

• Adding \(7\) is undone by subtracting \(7\).
• Multiplying by \(4\) is undone by dividing by \(4\).

When solving equations, you use inverse operations to isolate the variable. Isolate means to get the variable by itself on one side of the equation.

Here is the basic strategy:

1. Look at what is happening to the variable.
2. Use the inverse operation to undo it.
3. Do the same operation to both sides of the equation.
4. Keep going until the variable is alone.
5. Check your answer by substituting it back into the original equation.

It is important to undo operations in a sensible order. If more than one operation is affecting the variable, usually undo addition or subtraction first, then undo multiplication or division.

Worked Example 1: One-step equation

Solve \(x + 6 = 14\).

The variable has \(+6\), so use the inverse operation, which is subtracting \(6\).

$$ x + 6 = 14 $$ $$ x + 6 - 6 = 14 - 6 $$ $$ x = 8 $$

Check:

$$ 8 + 6 = 14 $$

This is true, so the solution is \(x = 8\).

Worked Example 2: One-step equation with multiplication

Solve \(5x = 35\).

The variable is being multiplied by \(5\), so use the inverse operation, which is dividing by \(5\).

$$ 5x = 35 $$ $$ \frac{5x}{5} = \frac{35}{5} $$ $$ x = 7 $$

Check:

$$ 5(7) = 35 $$ $$ 35 = 35 $$

The solution is \(x = 7\).

Worked Example 3: Two-step equation

Solve \(3x - 4 = 20\).

First, look at the operations on \(x\). The variable is multiplied by \(3\), and then \(4\) is subtracted. Undo the subtraction first by adding \(4\) to both sides.

$$ 3x - 4 = 20 $$ $$ 3x - 4 + 4 = 20 + 4 $$ $$ 3x = 24 $$

Now divide both sides by \(3\).

$$ \frac{3x}{3} = \frac{24}{3} $$ $$ x = 8 $$

Check:

$$ 3(8) - 4 = 24 - 4 = 20 $$

The solution is \(x = 8\).

Worked Example 4: Variable on both sides

Solve \(2x + 5 = x + 11\).

When a variable appears on both sides, move the variable terms to one side and the number terms to the other side. Subtract \(x\) from both sides first.

$$ 2x + 5 = x + 11 $$ $$ 2x - x + 5 = x - x + 11 $$ $$ x + 5 = 11 $$

Now subtract \(5\) from both sides.

$$ x + 5 - 5 = 11 - 5 $$ $$ x = 6 $$

Check:

$$ 2(6) + 5 = 12 + 5 = 17 $$ $$ 6 + 11 = 17 $$

Both sides are equal, so the solution is \(x = 6\).

Important reminders when using equality and inverse operations:

• Whatever you do to one side, do to the other side.
• Use inverse operations to undo what is happening to the variable.
• Work step by step and keep the equation balanced.
• Simplify each side carefully.
• Always check your solution in the original equation.

Common mistakes to avoid:

• Only changing one side: For example, subtracting \(3\) from the left side but not the right side breaks the equality.
• Using the wrong inverse operation: If the variable is multiplied by \(4\), you should divide by \(4\), not subtract \(4\).
• Doing steps out of order: In \(2x + 7 = 15\), subtract \(7\) first, then divide by \(2\).
• Forgetting to check: A quick check can catch mistakes.

Why this matters:

Equality and inverse operations are used in almost every equation-solving problem in algebra. Once you understand how to keep an equation balanced and undo operations, you can solve many different kinds of linear equations with confidence.

Summary

An equation shows that two expressions are equal. To solve it, you use the properties of equality to do the same thing to both sides and keep the equation balanced.

Inverse operations help you undo addition, subtraction, multiplication, and division so you can isolate the variable. By working carefully and checking your answer, you can solve linear equations correctly and understand why each step works.

Multi-Step Linear Equations are equations that take more than one step to solve. Instead of doing just one operation, you may need to distribute, combine like terms, and move variables and constants to different sides before you can find the value of the variable.

Learning how to solve multi-step linear equations is important because many real problems do not come in a simple form like \(x + 3 = 7\). They often look more complicated, but the goal is always the same: get the variable by itself.

In this lesson, you will learn how to solve equations such as \(3(x+2)-5=10\) and \(4x-7=2x+9\). You will also learn how to check your answer to make sure it is correct.

What is a linear equation?

A linear equation is an equation where the variable has a power of 1. Examples include \(2x+5=11\) and \(3(x-4)+2=2x+7\). When solved, these equations usually have one solution.

Main idea: Use inverse operations to undo what is happening to the variable.

• Addition is undone by subtraction.
• Subtraction is undone by addition.
• Multiplication is undone by division.
• Division is undone by multiplication.

For multi-step equations, it helps to follow a clear process.

Steps for solving multi-step linear equations

1. Simplify both sides of the equation if needed.
   • Use the distributive property.
   • Combine like terms.
2. Move variable terms so that all variables are on one side.
3. Move constant terms so that all numbers are on the other side.
4. Solve by dividing or multiplying to isolate the variable.
5. Check your answer by substituting it back into the original equation.

1. Distributive Property

If you see parentheses, you may need to distribute first. The distributive property means multiplying the number outside the parentheses by each term inside.

For example,

$$3(x+2)=3x+6$$

and

$$-2(x-5)=-2x+10$$

Be careful with negative signs. A negative outside the parentheses changes the sign of each term inside.

2. Combining Like Terms

Like terms have the same variable part. For example, \(4x\) and \(-2x\) are like terms, but \(4x\) and \(4\) are not.

Examples:

• \(3x+5x=8x\)
• \(7-2=5\)
• \(6x+3-2x+4=4x+7\)

Combining like terms makes the equation simpler and easier to solve.

3. Variables on Both Sides

Sometimes the variable appears on both sides of the equation. When that happens, move the variable terms to one side by adding or subtracting.

For example, in \(5x+2=3x+10\), subtract \(3x\) from both sides:

$$5x+2=3x+10$$

$$2x+2=10$$

Then continue solving as usual.

Worked Example 1: Two-step review

Solve:

$$2x+7=19$$

Step 1: Subtract 7 from both sides.

$$2x+7-7=19-7$$

$$2x=12$$

Step 2: Divide both sides by 2.

$$x=6$$

Check:

Substitute \(x=6\) into the original equation.

$$2(6)+7=12+7=19$$

The answer is correct.

Worked Example 2: Equation with distribution

Solve:

$$3(x+4)-2=16$$

Step 1: Distribute the 3.

$$3x+12-2=16$$

Step 2: Combine like terms.

$$3x+10=16$$

Step 3: Subtract 10 from both sides.

$$3x=6$$

Step 4: Divide by 3.

$$x=2$$

Check:

$$3(2+4)-2=3(6)-2=18-2=16$$

The answer is correct.

Worked Example 3: Variables on both sides

Solve:

$$4x-7=2x+9$$

Step 1: Move variable terms to one side. Subtract \(2x\) from both sides.

$$4x-2x-7=2x-2x+9$$

$$2x-7=9$$

Step 2: Add 7 to both sides.

$$2x=16$$

Step 3: Divide by 2.

$$x=8$$

Check:

Left side: \(4(8)-7=32-7=25\)

Right side: \(2(8)+9=16+9=25\)

Both sides are equal, so \(x=8\).

Worked Example 4: Distribution, combining like terms, and variables on both sides

Solve:

$$2(x+5)+3x=4(x-1)+6$$

Step 1: Distribute on both sides.

$$2x+10+3x=4x-4+6$$

Step 2: Combine like terms.

$$5x+10=4x+2$$

Step 3: Move variable terms to one side. Subtract \(4x\) from both sides.

$$x+10=2$$

Step 4: Subtract 10 from both sides.

$$x=-8$$

Check:

Original equation:

$$2(x+5)+3x=4(x-1)+6$$

Substitute \(x=-8\):

Left side: \(2(-8+5)+3(-8)=2(-3)-24=-6-24=-30\)

Right side: \(4(-8-1)+6=4(-9)+6=-36+6=-30\)

Both sides are equal, so the solution is correct.

Tips for success

• Write each step clearly. This helps prevent mistakes.
• Do the same operation to both sides. This keeps the equation balanced.
• Distribute carefully, especially with negatives.
• Combine like terms before solving.
• Always check your answer.

Common mistakes to avoid

• Forgetting to distribute to every term.
  Example: \(2(x+3)\) should become \(2x+6\), not \(2x+3\).
• Combining terms that are not like terms.
  For example, \(3x+4\) cannot be simplified to \(7x\).
• Moving terms without using inverse operations.
  Instead of saying a term “changes sides and changes sign,” think about the actual operation you are doing to both sides.
• Not checking the final answer.
  A quick check can catch errors.

Practice on your own

Try solving these equations:

1. $$5x+9=24$$
2. $$4(x-2)+3=15$$
3. $$7x-5=3x+11$$
4. $$3(x+1)+2x=2(x+4)+7$$

Answers

1. \(x=3\)
2. \(x=3\)
3. \(x=4\)
4. \(x=12\)

Summary

To solve a multi-step linear equation, first simplify both sides by distributing and combining like terms. Then move variables to one side and constants to the other. Finally, isolate the variable and check your answer in the original equation.

These equations may look complicated at first, but if you work step by step and keep the equation balanced, they become much easier to solve.

Equations with Rational Coefficients

In algebra, a rational number is any number that can be written as a fraction. This includes fractions like \(\frac{3}{4}\), whole numbers like \(5\) (because \(5 = \frac{5}{1}\)), and decimals that end or repeat, like \(0.6\) or \(1.25\).

An equation with rational coefficients is an equation where the numbers multiplying the variable, or the constant terms, are fractions or decimals. These equations may look harder at first, but they are solved using the same idea as other linear equations: isolate the variable.

The main extra skill is learning how to clear fractions or decimals so the equation becomes easier to work with.

Why clear fractions and decimals?

Fractions and decimals can make an equation look messy. If we rewrite the equation using whole numbers, it becomes much easier to solve. We do this by multiplying every term on both sides by the same number.

This works because multiplying both sides of an equation by the same nonzero number keeps the equation balanced.

Main idea 1: Clearing fractions

If an equation has fractions, find the least common multiple (LCM) of the denominators. Then multiply every term in the equation by that LCM.

This removes the denominators and turns the equation into one with whole-number coefficients.

For example, in the equation

$$\frac{1}{2}x + \frac{1}{3} = \frac{5}{6},$$

the denominators are \(2\), \(3\), and \(6\). Their LCM is \(6\). So we multiply every term by \(6\):

$$6\left(\frac{1}{2}x\right) + 6\left(\frac{1}{3}\right) = 6\left(\frac{5}{6}\right).$$

Then simplify:

$$3x + 2 = 5.$$

Now solve as usual:

$$3x = 3$$

$$x = 1.$$

Main idea 2: Clearing decimals

If an equation has decimals, multiply every term by a power of 10, such as \(10\), \(100\), or \(1000\), depending on how many decimal places appear.

For example, in

$$0.4x - 1.2 = 2.8,$$

each decimal has 1 digit after the decimal point. Multiply every term by \(10\):

$$10(0.4x) - 10(1.2) = 10(2.8).$$

Simplify:

$$4x - 12 = 28.$$

Now solve:

$$4x = 40$$

$$x = 10.$$

Main idea 3: Multiply every term

A very common mistake is forgetting to multiply one of the terms. When clearing fractions or decimals, every term on both sides must be multiplied.

For example, if you have

$$\frac{2}{3}x - \frac{1}{6} = \frac{1}{2},$$

the denominators are \(3\), \(6\), and \(2\), so the LCM is \(6\). Multiply every term by \(6\):

$$6\left(\frac{2}{3}x\right) - 6\left(\frac{1}{6}\right) = 6\left(\frac{1}{2}\right).$$

That becomes

$$4x - 1 = 3.$$

Then solve:

$$4x = 4$$

$$x = 1.$$

Steps for solving equations with rational coefficients

1. Look at the equation and decide whether it has fractions, decimals, or both.

2. If there are fractions, find the LCM of all denominators.

3. If there are decimals, choose a power of 10 that clears all decimals.

4. Multiply every term on both sides by that number.

5. Simplify the equation.

6. Solve the resulting linear equation.

7. Check your answer by substituting it back into the original equation.

Worked Example 1: Simple equation with fractions

Solve:

$$\frac{3}{4}x = 6$$

The denominator is \(4\), so multiply both sides by \(4\):

$$4\left(\frac{3}{4}x\right) = 4(6)$$

Simplify:

$$3x = 24$$

Now divide by \(3\):

$$x = 8$$

Check:

$$\frac{3}{4}(8) = 6$$

Since \(\frac{3}{4} \cdot 8 = 6\), the solution is correct.

Worked Example 2: Fractions on both sides

Solve:

$$\frac{1}{2}x + \frac{1}{4} = \frac{3}{4}$$

The denominators are \(2\) and \(4\), so the LCM is \(4\).

Multiply every term by \(4\):

$$4\left(\frac{1}{2}x\right) + 4\left(\frac{1}{4}\right) = 4\left(\frac{3}{4}\right)$$

Simplify:

$$2x + 1 = 3$$

Subtract \(1\) from both sides:

$$2x = 2$$

Divide by \(2\):

$$x = 1$$

Check:

$$\frac{1}{2}(1) + \frac{1}{4} = \frac{1}{2} + \frac{1}{4} = \frac{3}{4}$$

The solution is correct.

Worked Example 3: Equation with decimals

Solve:

$$1.5x + 0.7 = 4.3$$

Each decimal has 1 digit after the decimal point, so multiply every term by \(10\):

$$10(1.5x) + 10(0.7) = 10(4.3)$$

Simplify:

$$15x + 7 = 43$$

Subtract \(7\):

$$15x = 36$$

Divide by \(15\):

$$x = \frac{36}{15}$$

Simplify the fraction:

$$x = \frac{12}{5}$$

Or as a decimal,

$$x = 2.4$$

Check:

$$1.5(2.4) + 0.7 = 3.6 + 0.7 = 4.3$$

The solution is correct.

Worked Example 4: Fractions with variable terms on both sides

Solve:

$$\frac{2}{3}x - \frac{1}{2} = \frac{1}{6}x + 1$$

The denominators are \(3\), \(2\), and \(6\). The LCM is \(6\).

Multiply every term by \(6\):

$$6\left(\frac{2}{3}x\right) - 6\left(\frac{1}{2}\right) = 6\left(\frac{1}{6}x\right) + 6(1)$$

Simplify:

$$4x - 3 = x + 6$$

Subtract \(x\) from both sides:

$$3x - 3 = 6$$

Add \(3\) to both sides:

$$3x = 9$$

Divide by \(3\):

$$x = 3$$

Check:

Left side:

$$\frac{2}{3}(3) - \frac{1}{2} = 2 - \frac{1}{2} = \frac{3}{2}$$

Right side:

$$\frac{1}{6}(3) + 1 = \frac{1}{2} + 1 = \frac{3}{2}$$

Both sides are equal, so the solution is correct.

Helpful tips

• Before solving, identify all denominators or decimal places.

• Use the smallest number that clears the fractions or decimals. This keeps the numbers manageable.

• Be careful with negative signs when multiplying and simplifying.

• After clearing fractions or decimals, solve like any other linear equation.

• Always check your answer in the original equation, not just the simplified one.

Common mistakes to avoid

• Not multiplying every term: each term on both sides must be multiplied.

• Using the wrong LCM: list the denominators carefully before choosing the LCM.

• Dropping a sign: for example, changing \(-\frac{1}{2}\) into \(+\frac{1}{2}\).

• Stopping too early: after clearing fractions, you still need to solve the equation fully.

Summary

Equations with rational coefficients contain fractions or decimals. To solve them, first clear the fractions by multiplying by the LCM of the denominators, or clear the decimals by multiplying by a power of 10. Then simplify and solve the new linear equation just like any other one-variable equation.

The key rule is simple: whatever number you use to clear fractions or decimals, multiply every term on both sides by that number. With practice, these equations become much easier to solve.

Literal Equations and Formula Rearrangement

In earlier algebra, you often solved equations with one variable, such as \(3x+5=17\). In this lesson, you will learn how to work with literal equations, which are equations that contain more than one variable.

A literal equation is often a formula, like the area of a rectangle \(A=lw\) or the distance formula \(d=rt\). Sometimes you know some values in the formula, but other times you need to rearrange the formula to make a different variable the subject.

For example, if you know \(A=lw\) and want to solve for \(w\), you rearrange the equation so that \(w\) is alone. This process uses the same algebra rules you already know: add, subtract, multiply, divide, and sometimes factor.

Goal of formula rearrangement: isolate the variable you are asked to solve for.

Main idea: Whatever you do to one side of an equation, you must do to the other side to keep the equation balanced.

Why this matters: Literal equations are used in science, geometry, finance, and many real-life situations. Being able to rearrange formulas helps you find missing measurements, rates, times, and other important quantities.

Key vocabulary

• Literal equation: an equation with two or more variables.
• Formula: a rule written as an equation.
• Isolate a variable: get the variable by itself on one side of the equation.
• Rearrange: rewrite the equation so a different variable is isolated.

Steps for rearranging a formula

1. Decide which variable you need to solve for.
2. Look at what is being done to that variable.
3. Undo those operations in reverse order.
4. Do the same operation to both sides of the equation.
5. Check that the variable is completely isolated.

It helps to think of solving as undoing. If the variable is being added, subtract. If it is being multiplied, divide. If it is inside parentheses, sometimes distribute or factor first, depending on the equation.

Important note: When dividing by a variable expression, make sure it is not zero. In 9th grade, it is enough to remember that division by zero is not allowed.

Common patterns

• If the variable is added: subtract the other term.
• If the variable is subtracted: add the other term.
• If the variable is multiplied by something: divide by that thing.
• If the variable is in a fraction: multiply both sides to clear the fraction.

Let’s look at some examples, starting simple and building up.

Worked Example 1: Solve for a variable in a multiplication formula

Rearrange \(A=lw\) to solve for \(w\).

We want \(w\) by itself. In the equation, \(w\) is being multiplied by \(l\).

Undo multiplication by dividing both sides by \(l\):

$$ A=lw $$ $$ \frac{A}{l}=\frac{lw}{l} $$ $$ \frac{A}{l}=w $$

So the rearranged formula is

$$ w=\frac{A}{l} $$

Check: If \(A=20\) and \(l=4\), then

$$ w=\frac{20}{4}=5 $$

This makes sense because \(A=lw=4\cdot5=20\).

Worked Example 2: Solve for a variable in an addition formula

Rearrange \(P=2l+2w\) to solve for \(w\).

We want to isolate \(w\). Right now, \(2w\) is added to \(2l\).

First subtract \(2l\) from both sides:

$$ P=2l+2w $$ $$ P-2l=2w $$

Now divide both sides by \(2\):

$$ \frac{P-2l}{2}=w $$

So the rearranged formula is

$$ w=\frac{P-2l}{2} $$

Tip: You could also write this as

$$ w=\frac{1}{2}(P-2l) $$

Both forms mean the same thing.

Worked Example 3: Solve for a variable that appears in a fraction

Rearrange \(d=rt\) to solve for \(t\).

We want \(t\) by itself. In the equation, \(t\) is multiplied by \(r\).

Divide both sides by \(r\):

$$ d=rt $$ $$ \frac{d}{r}=t $$

So the formula solved for \(t\) is

$$ t=\frac{d}{r} $$

This is a common science and motion formula: time equals distance divided by rate.

Worked Example 4: Solve for a variable when it appears in more than one step

Rearrange \(C=2\pi r+2\pi h\) to solve for \(h\).

The variable \(h\) is part of the term \(2\pi h\). To isolate it, first move the other term to the other side.

Subtract \(2\pi r\) from both sides:

$$ C=2\pi r+2\pi h $$ $$ C-2\pi r=2\pi h $$

Now divide both sides by \(2\pi\):

$$ \frac{C-2\pi r}{2\pi}=h $$

So the rearranged formula is

$$ h=\frac{C-2\pi r}{2\pi} $$

This example shows that when the variable is not alone, you often need more than one step.

How to decide what to do first

When rearranging formulas, follow the order of operations in reverse.

• If something is added or subtracted outside the variable term, undo that first.
• Then undo multiplication or division.
• Keep going until the variable is alone.

For example, in \(P=2l+2w\), the \(2w\) term is not alone because \(2l\) is added. So subtract first, then divide.

A useful strategy: keep the target variable on one side

Try to avoid moving the target variable back and forth. Instead, move the other terms away from it. This usually makes the work cleaner and easier to follow.

Common mistakes to avoid

• Forgetting to do the same thing to both sides. Equations must stay balanced.
• Only dividing part of a side. If you divide a whole side by a number, every term on that side must be divided, unless the side is written as one grouped expression.
• Dropping signs. Be careful with subtraction and negative signs.
• Stopping too early. Make sure the requested variable is fully isolated.

Example of a common error

Suppose you have

$$ P=2l+2w $$

If you subtract \(2l\), you get

$$ P-2l=2w $$

To solve for \(w\), you must divide the entire left side by \(2\):

$$ w=\frac{P-2l}{2} $$

It would be incorrect to write \(w=P-2l/2\), because that only divides the \(2l\) term by 2, not the whole expression.

Practice thinking

Ask yourself these questions as you work:

• Which variable am I solving for?
• What operations are attached to that variable?
• What is the opposite of those operations?
• Did I apply each step to both sides?
• Is the variable completely alone at the end?

More quick examples

1. Solve \(V=lwh\) for \(h\):

$$ h=\frac{V}{lw} $$

2. Solve \(y=mx+b\) for \(x\):

$$ y-b=mx $$ $$ x=\frac{y-b}{m} $$

3. Solve \(I=Prt\) for \(r\):

$$ r=\frac{I}{Pt} $$

Each of these uses the same idea: undo what is happening to the variable.

Summary

Literal equations are equations with more than one variable. Rearranging a formula means solving for one chosen variable by using inverse operations and keeping the equation balanced.

To do this well, identify the target variable, move other terms away from it, and undo operations one step at a time. With practice, you will see that rearranging formulas is just like solving regular equations, but with letters standing for different quantities.

Identity, Contradiction, and Conditional Equations are three ways to describe equations based on how many solutions they have.

When you solve an equation, you are trying to find which value or values of the variable make the equation true. Sometimes there is one solution. Sometimes every value works. Sometimes no value works.

Learning to tell the difference is important because two equations can look similar at first, but their solution sets can be very different.

Let’s begin with the three types.

• Conditional equation: true for only certain value(s) of the variable. In 9th grade linear equations in one variable, this usually means one solution.
• Identity: true for all values of the variable.
• Contradiction: true for no values of the variable.

Another way to think about this is by looking at the solution set.

• A conditional equation has one value in its solution set, such as \(\{4\}\).
• An identity has infinitely many solutions, meaning every real number works.
• A contradiction has an empty solution set, written as no solution.

1. Conditional Equations

A conditional equation becomes true only when the variable has a particular value.

Example: \(2x+3=11\)

To solve, isolate the variable.

$$2x+3=11$$$$2x=8$$$$x=4$$

If we check the answer:

$$2(4)+3=8+3=11$$

This is true, so \(x=4\) is the solution.

This is called a conditional equation because it is true on the condition that \(x=4\).

2. Identities

An identity is an equation that stays true no matter what value you choose for the variable.

When you simplify an identity, the variable terms usually disappear and you are left with a true statement.

For example, you might end with:

$$5=5$$

Since \(5=5\) is always true, the original equation is true for all values of the variable.

Here is a simple identity:

\(3(x+2)=3x+6\)

Simplify the left side:

$$3x+6=3x+6$$

Subtract \(3x\) from both sides:

$$6=6$$

This final statement is always true. That means the equation is an identity.

Solution: all real numbers.

3. Contradictions

A contradiction is an equation that is never true, no matter what value is used for the variable.

When you simplify a contradiction, the variable terms usually disappear and you are left with a false statement.

For example, you might end with:

$$4=9$$

Since \(4=9\) is never true, the original equation has no solution.

Here is a simple contradiction:

\(2(x+1)=2x+5\)

Distribute on the left side:

$$2x+2=2x+5$$

Subtract \(2x\) from both sides:

$$2=5$$

This is false, so the equation is a contradiction.

Solution: no solution.

4. How to Classify an Equation

To decide whether an equation is conditional, an identity, or a contradiction, solve it as usual.

1. Simplify both sides if needed. This may include distributing and combining like terms.
2. Move variable terms to one side and constants to the other side.
3. Look at the result.

There are three possible outcomes:

• If you get something like \(x=7\), the equation is conditional.
• If the variable disappears and you get a true statement like \(0=0\) or \(3=3\), the equation is an identity.
• If the variable disappears and you get a false statement like \(0=5\) or \(8=-2\), the equation is a contradiction.

Important idea: The final simplified statement tells you the type of equation.

5. Worked Examples

Example 1: A conditional equation

Solve and classify: \(5x-7=13\)

Step 1: Add 7 to both sides.

$$5x=20$$

Step 2: Divide by 5.

$$x=4$$

The equation has exactly one solution, so it is a conditional equation.

Answer: \(x=4\)

Example 2: An identity

Solve and classify: \(4(x-1)+8=4x+4\)

Step 1: Distribute on the left.

$$4x-4+8=4x+4$$

Step 2: Combine like terms.

$$4x+4=4x+4$$

Step 3: Subtract \(4x\) from both sides.

$$4=4$$

This is always true, so the equation is an identity.

Answer: infinitely many solutions, or all real numbers.

Example 3: A contradiction

Solve and classify: \(3(2x-1)=6x+4\)

Step 1: Distribute on the left.

$$6x-3=6x+4$$

Step 2: Subtract \(6x\) from both sides.

$$-3=4$$

This is false, so the equation is a contradiction.

Answer: no solution.

Example 4: A more detailed example

Solve and classify: \(7-2(x+3)=1-2x\)

Step 1: Distribute \(-2\).

$$7-2x-6=1-2x$$

Step 2: Combine like terms on the left.

$$1-2x=1-2x$$

Both sides are exactly the same.

Step 3: Add \(2x\) to both sides.

$$1=1$$

This is always true, so the equation is an identity.

Answer: all real numbers.

6. A Quick Comparison

• Conditional equation: one solution
  Example result: \(x=2\)
• Identity: infinitely many solutions
  Example result: \(0=0\)
• Contradiction: no solution
  Example result: \(5=1\)

7. Common Mistakes to Avoid

• Stopping too early: Do not classify the equation before fully simplifying both sides.
• Forgetting to distribute: In equations like \(2(x+3)\), multiply the 2 by both terms inside the parentheses.
• Misreading the final statement: A true statement means identity. A false statement means contradiction.
• Thinking every equation has one answer: Some equations have no solution, and some have infinitely many.

8. How to Check Yourself

After solving, ask these questions:

• Did I simplify both sides correctly?
• Did the variable disappear?
• Is the final statement true or false?
• If I got a value for \(x\), does it make the original equation true?

If you can answer these questions, you can confidently classify the equation.

9. Summary

Equations can be grouped by how many solutions they have.

• A conditional equation has one solution.
• An identity is true for every value of the variable.
• A contradiction is never true, so it has no solution.

When solving, keep simplifying until you get either a variable answer like \(x=3\), a true statement like \(2=2\), or a false statement like \(2=7\). That final result tells you how to classify the equation.

Absolute Value Equations are equations that include absolute value symbols, like \(|x|\) or \(|2x-5|\).

To solve them, it helps to understand what absolute value means. Absolute value tells how far a number is from 0 on a number line. Since distance cannot be negative, absolute value is always 0 or positive.

For example:

• \(|4| = 4\)
• \(|-4| = 4\)
• \(|0| = 0\)

This is why absolute value equations often have two solutions. If a number is 6 units from 0, that number could be \(6\) or \(-6\).

For example, if

$$|x| = 6$$

then the solutions are

$$x = 6 \quad \text{or} \quad x = -6$$

Main Idea: When you see an equation of the form

$$|A| = b$$

where \(b\) is a positive number, you solve it by writing two equations:

$$A = b \quad \text{or} \quad A = -b$$

This works because the expression inside the absolute value can be positive or negative, and both give the same distance from 0.

There is also an important special case. If you have

$$|A| = 0$$

then there is only one solution:

$$A = 0$$

And if you have

$$|A| = \text{a negative number}$$

then there is no solution, because an absolute value can never be negative.

Steps for Solving Absolute Value Equations

1. Get the absolute value by itself on one side of the equation.
2. Check the number on the other side.
3. If it is positive, write two equations.
4. If it is 0, write one equation.
5. If it is negative, say there is no solution.
6. Solve each equation.
7. Check your answers in the original equation.

Worked Example 1: A basic absolute value equation

Solve:

$$|x| = 9$$

Since 9 is positive, write two equations:

$$x = 9 \quad \text{or} \quad x = -9$$

So the solutions are:

$$x = 9, -9$$

Worked Example 2: Absolute value with an expression inside

Solve:

$$|x-3| = 5$$

Write two equations:

$$x - 3 = 5 \quad \text{or} \quad x - 3 = -5$$

Now solve each one.

First equation:

$$x - 3 = 5$$

$$x = 8$$

Second equation:

$$x - 3 = -5$$

$$x = -2$$

So the solutions are:

$$x = 8, -2$$

Let us check one answer:

For \(x=8\),

$$|8-3| = |5| = 5$$

For \(x=-2\),

$$|-2-3| = |-5| = 5$$

Both are correct.

Worked Example 3: Isolate the absolute value first

Solve:

$$|2x+1| - 4 = 10$$

First, isolate the absolute value by adding 4 to both sides:

$$|2x+1| = 14$$

Now write two equations:

$$2x+1 = 14 \quad \text{or} \quad 2x+1 = -14$$

Solve the first equation:

$$2x+1 = 14$$

$$2x = 13$$

$$x = \frac{13}{2}$$

Solve the second equation:

$$2x+1 = -14$$

$$2x = -15$$

$$x = -\frac{15}{2}$$

So the solutions are:

$$x = \frac{13}{2}, -\frac{15}{2}$$

Worked Example 4: A case with no solution

Solve:

$$|3x-7| = -2$$

This equation has no solution.

Why? Because absolute value is never negative. No matter what value of \(x\) you choose, \(|3x-7|\) cannot equal \(-2\).

One-Solution Case

Now look at this equation:

$$|5x-10| = 0$$

An absolute value equals 0 only when the expression inside is 0.

So:

$$5x - 10 = 0$$

$$5x = 10$$

$$x = 2$$

This equation has only one solution:

$$x = 2$$

Why Two Equations?

Suppose you have

$$|x-1| = 4$$

This means the expression \(x-1\) is 4 units away from 0. There are two ways that can happen:

• \(x-1 = 4\)
• \(x-1 = -4\)

That is the key idea in solving these equations. The inside expression can be positive or negative, but the absolute value makes both distances positive.

Common Mistakes to Avoid

• Forgetting the second equation: If \(|A| = b\) and \(b>0\), always write both \(A=b\) and \(A=-b\).
• Not isolating the absolute value first: Solve equations like \(|2x-1|+3=8\) by first getting \(|2x-1|\) alone.
• Trying to solve when the other side is negative: If the equation is like \(|A|=-5\), there is no solution.
• Not checking answers: Substituting back helps catch mistakes.

Quick Practice Thinking

Use the rules above to think about these:

• \(|x+2| = 7\) has two solutions.
• \(|4x-1| = 0\) has one solution.
• \(|2x+9| = -3\) has no solution.

Summary

To solve an absolute value equation, first get the absolute value alone. Then look at the number on the other side.

• If it is positive, write two equations: one positive and one negative.
• If it is zero, write one equation.
• If it is negative, there is no solution.

Absolute value equations are really about distance. Once you remember that distance can be on either side of 0, the idea of two possible solutions makes sense.

Linear Inequalities and Sign Reversal is about solving statements that compare values using symbols like \(<\), \(>\), \(\le\), and \(\ge\).

A linear inequality in one variable looks a lot like a linear equation, but instead of an equals sign, it uses an inequality sign. For example, \(2x+3>11\) is a linear inequality.

When solving inequalities, we try to isolate the variable, just as we do with equations. But there is one very important extra rule: if you multiply or divide both sides by a negative number, you must reverse the inequality sign.

This lesson will explain why that sign reversal happens, how to solve simple and chained inequalities, and how to check your answers.

1. Inequality symbols

• \(<\) means less than
• \(>\) means greater than
• \(\le\) means less than or equal to
• \(\ge\) means greater than or equal to

For example:

• \(x<5\) means any number smaller than 5
• \(x\ge -2\) means \(-2\) and every number greater than \(-2\)

2. Solving inequalities: same idea as equations

You can do most of the same operations to inequalities that you do to equations:

• Add the same number to both sides
• Subtract the same number from both sides
• Multiply both sides by the same positive number
• Divide both sides by the same positive number

These operations keep the inequality true and do not change the direction of the sign.

Example:

Start with

$$x+4<9$$

Subtract 4 from both sides:

$$x<5$$

The sign stayed the same because we only subtracted a number.

3. The special rule: sign reversal

The one operation that changes the direction of an inequality is multiplying or dividing by a negative number.

For example, we know that

$$3<7$$

If we multiply both sides by \(-1\), we get

$$-3\quad ? \quad -7$$

Now \(-3\) is actually greater than \(-7\), so the true statement is

$$-3>-7$$

The inequality sign had to reverse. This is why:

$$3<7 \quad \Rightarrow \quad -3>-7$$

The same thing happens when dividing by a negative number.

Rule:

If you multiply or divide both sides of an inequality by a negative number, reverse the inequality sign.

4. Why does the sign reverse?

Think about a number line. Numbers to the right are greater, and numbers to the left are smaller.

When you multiply by \(-1\), positive numbers become negative and negative numbers become positive. This reflects the numbers across 0 on the number line.

So the order flips. A number that was smaller becomes larger after multiplying by a negative, compared with the other number.

That is why the sign must reverse.

5. Worked Example 1: no sign reversal needed

Solve:

$$2x+5<13$$

Step 1: Subtract 5 from both sides.

$$2x<8$$

Step 2: Divide both sides by 2.

$$x<4$$

Answer: \(x<4\)

We did not reverse the sign because we divided by a positive number.

Check: Try \(x=3\).

$$2(3)+5=11$$

Since \(11<13\), the solution makes sense.

6. Worked Example 2: sign reversal when dividing by a negative

Solve:

$$-3x\ge 12$$

Step 1: Divide both sides by \(-3\).

Because we are dividing by a negative number, we must reverse \(\ge\) to \(\le\).

$$x\le -4$$

Answer: \(x\le -4\)

Check: Try \(x=-5\).

$$-3(-5)=15$$

Since \(15\ge 12\), the answer works.

If we had forgotten to reverse the sign, we would have gotten the wrong solution.

7. Worked Example 3: variables on both sides

Solve:

$$4x-7>2x+5$$

Step 1: Subtract \(2x\) from both sides.

$$2x-7>5$$

Step 2: Add 7 to both sides.

$$2x>12$$

Step 3: Divide both sides by 2.

$$x>6$$

Answer: \(x>6\)

No sign reversal happened because we divided by a positive number.

8. Worked Example 4: a chained inequality

Solve:

$$-2<3x+1\le 10$$

A chained inequality means one expression is between two values. We solve all parts at the same time, doing the same thing to each part.

Step 1: Subtract 1 from all three parts.

$$-3<3x\le 9$$

Step 2: Divide all three parts by 3.

$$-1

Answer: \(-1

This means \(x\) is greater than \(-1\) and at the same time less than or equal to 3.

Now look at a chained inequality that includes sign reversal:

Solve:

$$6\ge -2x> -8$$

Step 1: Divide every part by \(-2\).

Because we divided by a negative number, both inequality signs reverse.

$$-3\le x<4$$

Answer: \(-3\le x<4\)

This is a very important idea: in a chained inequality, if you multiply or divide all parts by a negative number, every inequality sign must reverse.

9. Common mistakes to avoid

• Forgetting to reverse the sign when multiplying or dividing by a negative number
• Reversing the sign when you should not; adding or subtracting does not reverse it
• Only reversing one sign in a chained inequality; if you divide or multiply all parts by a negative, reverse all the inequality signs
• Not checking the final answer by testing a value that should work

10. Quick comparison: equation vs inequality

• An equation usually has one answer or a set of exact answers, like \(x=4\)
• An inequality usually has many answers, like \(x<4\)

For example, if \(x<4\), then \(3\), \(0\), and \(-10\) are all solutions.

11. How to check an inequality solution

After solving, pick a number that fits your answer and substitute it into the original inequality.

Example: if your answer is \(x\le -4\), test \(x=-4\) or \(x=-5\).

If the original inequality is true, your solution is likely correct.

12. Steps for solving linear inequalities

1. Simplify each side if needed.
2. Move variable terms to one side and number terms to the other side.
3. Use addition or subtraction as needed.
4. Multiply or divide to isolate the variable.
5. If you multiply or divide by a negative number, reverse the inequality sign.
6. Check your answer with a test value.

Brief Summary

Linear inequalities are solved much like linear equations, but they have one special rule. When you multiply or divide both sides by a negative number, the inequality sign must reverse.

This rule also applies to chained inequalities, where every inequality sign must reverse if all parts are multiplied or divided by a negative. Remembering this rule helps you solve inequalities correctly and avoid one of the most common mistakes in algebra.

Compound Inequalities are two inequalities put together into one statement. They are used when a value must satisfy more than one condition.

In 9th Grade maths, you will usually see compound inequalities joined by the words and or or. Understanding the difference between these two words is the key to solving them correctly.

This lesson will show you how to solve compound inequalities, how to write the solution, and how to graph the answer on a number line.

1. What is a compound inequality?

A compound inequality combines two simple inequalities. For example:

• AND: \(x > 2\) and \(x < 7\)
• OR: \(x < -1\) or \(x \ge 4\)

Each part gives a condition for the variable. The word connecting them tells you how the conditions work together.

2. AND inequalities

An AND compound inequality means the solution must satisfy both inequalities at the same time.

This means you are looking for the values that are in the overlap of the two solution sets. Another word for this is intersection.

For example, in

$$x > 2 \text{ and } x < 7$$

the number must be greater than 2 and also less than 7. So the solution is all numbers between 2 and 7:

$$2 < x < 7$$

On a number line, this would be graphed with open circles at 2 and 7, because 2 and 7 are not included, and shading between them.

3. OR inequalities

An OR compound inequality means the solution must satisfy at least one of the inequalities.

This means you combine both solution sets. Another word for this is union.

For example, in

$$x < -1 \text{ or } x \ge 4$$

the solution includes any number less than \(-1\), together with any number greater than or equal to 4.

On a number line, this would be graphed with:

• an open circle at \(-1\) and shading to the left
• a closed circle at 4 and shading to the right

4. Open circles and closed circles

When graphing inequalities, the endpoint matters.

• Use an open circle for \(<\) or \(>\), because the endpoint is not included.
• Use a closed circle for \(\le\) or \(\ge\), because the endpoint is included.

Examples:

• \(x < 5\): open circle at 5
• \(x \le 5\): closed circle at 5

5. How to solve compound inequalities

The steps depend on whether the inequality uses AND or OR.

For AND inequalities:

1. Solve each inequality, or solve the whole chain together if possible.
2. Find the values that make both parts true.
3. Write the solution as one combined inequality when possible.
4. Graph the overlap on the number line.

For OR inequalities:

1. Solve each inequality separately.
2. Keep all values that make either part true.
3. Write the answer as two parts if needed.
4. Graph both rays or regions on the number line.

Important rule: If you multiply or divide both sides of an inequality by a negative number, you must reverse the inequality sign.

For example:

$$-2x > 6$$

Divide both sides by \(-2\):

$$x < -3$$

The sign changes from \(>\) to \(<\).

Worked Example 1: Simple AND inequality

Solve:

$$3 < x + 1 < 8$$

This is an AND inequality written as one chain. We can solve all three parts together by subtracting 1 from each part:

$$3 - 1 < x + 1 - 1 < 8 - 1$$

$$2 < x < 7$$

Answer: \(2 < x < 7\)

This means \(x\) is between 2 and 7, but does not include 2 or 7.

Graph:

• open circle at 2
• open circle at 7
• shade between them

Worked Example 2: AND inequality with included endpoints

Solve:

$$-4 \le 2x \le 10$$

Divide all three parts by 2:

$$\frac{-4}{2} \le \frac{2x}{2} \le \frac{10}{2}$$

$$-2 \le x \le 5$$

Answer: \(-2 \le x \le 5\)

This means \(x\) can be any number from \(-2\) to 5, including both endpoints.

Graph:

• closed circle at \(-2\)
• closed circle at 5
• shade between them

Worked Example 3: OR inequality

Solve:

$$2x - 1 < 5 \text{ or } 3x + 2 \ge 11$$

Solve each part separately.

First inequality:

$$2x - 1 < 5$$

Add 1 to both sides:

$$2x < 6$$

Divide by 2:

$$x < 3$$

Second inequality:

$$3x + 2 \ge 11$$

Subtract 2 from both sides:

$$3x \ge 9$$

Divide by 3:

$$x \ge 3$$

So the compound inequality becomes:

$$x < 3 \text{ or } x \ge 3$$

This includes every real number, because any number is either less than 3 or greater than or equal to 3.

Answer: all real numbers

Graph:

• open circle at 3 with shading to the left
• closed circle at 3 with shading to the right

Together, the whole number line is shaded.

Worked Example 4: OR inequality with a negative coefficient

Solve:

$$-2x > 8 \text{ or } x - 5 \le -1$$

Solve each inequality separately.

First inequality:

$$-2x > 8$$

Divide both sides by \(-2\). Because we divide by a negative, reverse the inequality sign:

$$x < -4$$

Second inequality:

$$x - 5 \le -1$$

Add 5 to both sides:

$$x \le 4$$

So we have:

$$x < -4 \text{ or } x \le 4$$

Now think carefully. Every number less than \(-4\) is already included in \(x \le 4\). That means the larger set is just:

$$x \le 4$$

Answer: \(x \le 4\)

Graph:

• closed circle at 4
• shade to the left

6. Checking whether a value is a solution

A good way to understand compound inequalities is to test values.

Suppose the inequality is

$$1 < x \le 6$$

Test some values:

• \(x = 1\): not a solution, because \(1 < x\) is false
• \(x = 4\): yes, because \(1 < 4\) and \(4 \le 6\) are both true
• \(x = 6\): yes, because \(1 < 6\) and \(6 \le 6\) are both true
• \(x = 8\): not a solution, because \(8 \le 6\) is false

For an AND inequality, both parts must be true.

For an OR inequality, only one part needs to be true.

7. Common mistakes to avoid

• Mixing up AND and OR: AND means overlap; OR means combine both parts.
• Forgetting to reverse the sign: If you multiply or divide by a negative, flip the inequality sign.
• Using the wrong circle: Open for \(<\) or \(>\), closed for \(\le\) or \(\ge\).
• Graphing AND like OR: AND usually shades a section between numbers, while OR often gives two rays.

8. Quick guide: AND vs OR

• AND = values in the middle that satisfy both conditions
• OR = values in either part of the graph

Examples:

• $$x > 1 \text{ and } x \le 5 \quad \Rightarrow \quad 1 < x \le 5$$
• $$x < -2 \text{ or } x > 4$$

The first is one continuous section. The second is two separate parts.

Summary

Compound inequalities contain two inequalities joined by and or or. With and, the solution is the overlap of the two parts. With or, the solution is everything in either part.

To solve them, isolate the variable just like in a regular inequality. Then write the solution carefully and graph it using open or closed circles depending on whether the endpoint is included.

If you remember the meanings of AND = overlap and OR = combine, you will be able to solve and graph compound inequalities correctly.

Absolute Value Inequalities tell us how far a number is from another number on the number line, and whether that distance is less than or greater than a certain amount.

In this lesson, you will learn what absolute value inequalities mean, how to solve them, and how to write and graph their solution sets.

Remember: absolute value means distance from 0. For example, \(|3|=3\) and \(|-3|=3\). Distance is always nonnegative.

This idea helps us understand expressions like \(|x-4|<2\) or \(|x+1|\ge 5\). These are not just equations—they describe all the numbers that are close to or far from a certain value.

Key idea: \(|x-a|\) means the distance between \(x\) and \(a\) on the number line.

So:

• \(|x-a|<b\) means the distance from \(x\) to \(a\) is less than \(b\)
• \(|x-a|\le b\) means the distance from \(x\) to \(a\) is at most \(b\)
• \(|x-a|>b\) means the distance from \(x\) to \(a\) is greater than \(b\)
• \(|x-a|\ge b\) means the distance from \(x\) to \(a\) is at least \(b\)

These ideas lead to two very important patterns.

Pattern 1: “Less than” absolute value inequalities

If

$$|x-a|<b$$

then

$$-b<x-a<b$$

And if

$$|x-a|\le b$$

then

$$-b\le x-a\le b$$

This creates a solution between two numbers. That is why these are called compound inequalities with “and”.

Pattern 2: “Greater than” absolute value inequalities

If

$$|x-a|>b$$

then

$$x-a<-b \quad \text{or} \quad x-a>b$$

And if

$$|x-a|\ge b$$

then

$$x-a\le -b \quad \text{or} \quad x-a\ge b$$

This creates a solution outside two numbers. These are compound inequalities with “or”.

Why? If a number is more than 3 units away from 0, it could be to the left of \(-3\) or to the right of \(3\). That is why “greater than” becomes an “or” statement.

Important note: the number on the right side, such as \(b\), should be nonnegative when using these patterns. Since distance cannot be negative, some inequalities have special outcomes.

• If \(|x-a|<-2\), there is no solution, because a distance cannot be less than a negative number.
• If \(|x-a|\ge -2\), the solution is all real numbers, because every distance is greater than or equal to \(-2\).

Now let’s solve some inequalities step by step.

Example 1: Solve \(|x-5|<3\)

This is a “less than” absolute value inequality, so we rewrite it as a compound inequality:

$$-3<x-5<3$$

Now add 5 to all three parts:

$$-3+5<x<3+5$$ $$2<x<8$$

Solution: \(2<x<8\)

This means \(x\) is within 3 units of 5, so it must lie between 2 and 8.

Example 2: Solve \(|2x+1|\le 7\)

This is also a “less than or equal to” inequality, so write it as:

$$-7\le 2x+1\le 7$$

Subtract 1 from all three parts:

$$-8\le 2x\le 6$$

Now divide all three parts by 2:

$$-4\le x\le 3$$

Solution: \(-4\le x\le 3\)

Because the symbol was \(\le\), the endpoints are included.

Example 3: Solve \(|x+4|>6\)

This is a “greater than” inequality, so we split it into two parts using or:

$$x+4<-6 \quad \text{or} \quad x+4>6$$

Now solve each inequality:

$$x<-10 \quad \text{or} \quad x>2$$

Solution: \(x<-10\) or \(x>2\)

This means \(x\) is more than 6 units away from \(-4\), so it must be far to the left or far to the right.

Example 4: Solve \(3|x-2|\ge 12\)

First isolate the absolute value by dividing both sides by 3:

$$|x-2|\ge 4$$

Now use the “greater than or equal to” pattern:

$$x-2\le -4 \quad \text{or} \quad x-2\ge 4$$

Add 2 to both sides of each inequality:

$$x\le -2 \quad \text{or} \quad x\ge 6$$

Solution: \(x\le -2\) or \(x\ge 6\)

How to graph solutions

After solving, it helps to picture the answer on a number line.

• If the solution is between two numbers, shade in the middle.
• If the solution is outside two numbers, shade to the left and right.
• Use an open circle for \(<\) or \(>\).
• Use a closed circle for \(\le\) or \(\ge\).

For example:

• \(2<x<8\) means open circles at 2 and 8, shaded between them.
• \(x\le -2\) or \(x\ge 6\) means closed circles at \(-2\) and 6, shaded outward.

Real-world meaning

Absolute value inequalities are useful when talking about tolerance or error.

Suppose a machine should cut a piece to 10 cm, but it can be off by at most 0.2 cm. If \(x\) is the actual length, then:

$$|x-10|\le 0.2$$

This means the distance between the actual length and 10 is no more than 0.2.

Rewrite as a compound inequality:

$$-0.2\le x-10\le 0.2$$ $$9.8\le x\le 10.2$$

So the acceptable lengths are from 9.8 cm to 10.2 cm.

Common mistakes to avoid

• Mixing up “and” and “or”: Less than gives an and inequality. Greater than gives an or inequality.
• Forgetting to isolate the absolute value first: In \(3|x-2|\ge 12\), divide by 3 before splitting.
• Not solving both parts completely: After rewriting, finish solving each inequality.
• Using the wrong endpoint circles on a graph: Open for strict inequalities, closed for inclusive inequalities.

Steps for solving absolute value inequalities

1. Isolate the absolute value expression if needed.
2. Decide whether it is a less than type or a greater than type.
3. Rewrite it as a compound inequality.
4. Solve the resulting inequality or inequalities.
5. Write the final answer clearly and graph it if needed.

Quick reference

$$|x-a|<b \iff -b<x-a<b$$ $$|x-a|\le b \iff -b\le x-a\le b$$ $$|x-a|>b \iff x-a<-b \text{ or } x-a>b$$ $$|x-a|\ge b \iff x-a\le -b \text{ or } x-a\ge b$$

Summary

Absolute value inequalities describe distance on the number line. When the inequality is “less than,” the solution is between two values and uses and. When the inequality is “greater than,” the solution is outside two values and uses or. If you remember that absolute value means distance, these problems become much easier to understand.