Linear Equations and Inequalities in One Variable

Multi-Step Linear Equations are equations that take more than one step to solve. You may need to distribute, combine like terms, and then use inverse operations to get the variable by itself.

In this lesson, you will learn how to solve these equations in a clear, organized way. You will also learn how to check your answer to make sure it is correct.

A linear equation is an equation where the variable has an exponent of 1, such as \(x\) or \(3y\). A multi-step linear equation might look like this:

$$3(x+4)-2x=11$$

This equation cannot be solved in just one step. You must work through several steps in the correct order.

Main idea: Your goal is always to get the variable alone on one side of the equation.

Here is a helpful step-by-step process:

1. Distribute if there are parentheses.
2. Combine like terms on each side if possible.
3. Move variable terms to one side.
4. Move constants to the other side.
5. Solve for the variable.
6. Check your answer by substituting it back into the original equation.

Let’s review two important skills used in these problems.

1. Distributive Property

When a number is outside parentheses, multiply it by every term inside.

For example:

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

$$-3(y-4)=-3y+12$$

Be careful with negative signs. A negative multiplied by a negative becomes positive.

2. Combining Like Terms

Like terms have the same variable part. For example, \(3x\) and \(-5x\) are like terms, but \(3x\) and \(3y\) are not.

Examples:

$$4x+2x=6x$$

$$7-3=4$$

$$5x+2-3x+6=2x+8$$

Now let’s solve some equations step by step.

Worked Example 1: Combine like terms first

Solve:

$$3x+5+2x=20$$

Step 1: Combine like terms on the left side.

$$5x+5=20$$

Step 2: Subtract 5 from both sides.

$$5x=15$$

Step 3: Divide both sides by 5.

$$x=3$$

Check: Substitute \(x=3\) into the original equation.

$$3(3)+5+2(3)=9+5+6=20$$

The solution is correct.

Worked Example 2: Use the distributive property

Solve:

$$4(x+2)=20$$

Step 1: Distribute the 4.

$$4x+8=20$$

Step 2: Subtract 8 from both sides.

$$4x=12$$

Step 3: Divide both sides by 4.

$$x=3$$

Check:

$$4(3+2)=4(5)=20$$

The solution is correct.

Worked Example 3: Distribute and combine like terms

Solve:

$$2(x+3)+4x=18$$

Step 1: Distribute the 2.

$$2x+6+4x=18$$

Step 2: Combine like terms.

$$6x+6=18$$

Step 3: Subtract 6 from both sides.

$$6x=12$$

Step 4: Divide both sides by 6.

$$x=2$$

Check:

$$2(2+3)+4(2)=2(5)+8=10+8=18$$

The solution is correct.

Worked Example 4: Variables on both sides

Solve:

$$5x+7=2x+19$$

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

$$3x+7=19$$

Step 2: Move constants to the other side. Subtract 7 from both sides.

$$3x=12$$

Step 3: Divide both sides by 3.

$$x=4$$

Check:

Left side: \(5(4)+7=20+7=27\)

Right side: \(2(4)+19=8+19=27\)

Both sides are equal, so the solution is correct.

What if there are parentheses and variables on both sides?

You follow the same process: distribute, combine like terms, move variable terms, move constants, and solve.

For example:

$$3(x+2)+x=18$$

Distribute:

$$3x+6+x=18$$

Combine like terms:

$$4x+6=18$$

Subtract 6:

$$4x=12$$

Divide by 4:

$$x=3$$

Common mistakes to avoid

• Forgetting to distribute to every term inside parentheses.
• Making sign mistakes, especially with negative numbers.
• Combining unlike terms, such as adding \(3x\) and \(4\).
• Doing different operations on each side of the equation. Whatever you do to one side, do to the other side.
• Not checking your answer in the original equation.

Helpful tips

• Write each step on a new line so your work stays organized.
• If there are parentheses, distribute first.
• If you can combine like terms, do that before solving.
• Try to keep the variable on the left side, although either side works.
• Always check by substitution at the end.

Practice thinking

When you look at a multi-step equation, ask yourself:

• Are there parentheses to distribute?
• Are there like terms to combine?
• Are there variable terms on both sides?
• What should I undo first to isolate the variable?

Summary

To solve a multi-step linear equation, work in a clear order: distribute, combine like terms, move variable terms, move constants, and solve. The goal is to get the variable alone. If you stay organized and check your answer, you can solve even more complicated equations with confidence.

Variables on Both Sides means solving equations where the variable appears on the left side and the right side of the equals sign.

For example, in the equation \(3x + 5 = x + 13\), the variable \(x\) is on both sides. To solve it, we collect all the variable terms on one side and all the number terms on the other side.

This skill is important because many real equations are not already “cleaned up.” Learning how to move terms carefully helps you solve more complex linear equations with confidence.

Goal: Get the variable alone on one side of the equation.

Main Idea

When solving equations with variables on both sides, use opposite operations to keep the equation balanced. Whatever you do to one side, you must do to the other side.

A helpful plan is:

1. Move all variable terms to one side.
2. Move all constant terms to the other side.
3. Simplify.
4. Check your answer by substituting it back into the original equation.

How to Move Terms

• To move a variable term, add or subtract that term from both sides.
• To move a constant, add or subtract the number from both sides.
• If the variable has a coefficient, divide both sides by that coefficient at the end.

Example 1: A basic equation

Solve:

$$3x + 5 = x + 13$$

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

$$3x + 5 - x = x + 13 - x$$

$$2x + 5 = 13$$

Step 2: Move the constant terms to the other side. Subtract \(5\) from both sides.

$$2x + 5 - 5 = 13 - 5$$

$$2x = 8$$

Step 3: Divide both sides by \(2\).

$$x = 4$$

Check:

Left side: \(3(4) + 5 = 12 + 5 = 17\)

Right side: \(4 + 13 = 17\)

Both sides are equal, so \(x = 4\) is correct.

Example 2: Variables with negative terms

Solve:

$$5x - 7 = 2x + 8$$

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

$$5x - 7 - 2x = 2x + 8 - 2x$$

$$3x - 7 = 8$$

Step 2: Move the constant terms. Add \(7\) to both sides.

$$3x - 7 + 7 = 8 + 7$$

$$3x = 15$$

Step 3: Divide by \(3\).

$$x = 5$$

Check:

Left side: \(5(5) - 7 = 25 - 7 = 18\)

Right side: \(2(5) + 8 = 10 + 8 = 18\)

The answer is correct.

Example 3: Variables and constants on both sides

Solve:

$$4x + 9 = 2x - 3$$

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

$$4x + 9 - 2x = 2x - 3 - 2x$$

$$2x + 9 = -3$$

Step 2: Move the constants. Subtract \(9\) from both sides.

$$2x + 9 - 9 = -3 - 9$$

$$2x = -12$$

Step 3: Divide by \(2\).

$$x = -6$$

Check:

Left side: \(4(-6) + 9 = -24 + 9 = -15\)

Right side: \(2(-6) - 3 = -12 - 3 = -15\)

So \(x = -6\).

Example 4: Distribute first, then solve

Sometimes you must simplify each side before moving terms.

Solve:

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

Step 1: Distribute.

$$2x + 6 = x + 11$$

Step 2: Move the variable terms. Subtract \(x\) from both sides.

$$2x + 6 - x = x + 11 - x$$

$$x + 6 = 11$$

Step 3: Subtract \(6\) from both sides.

$$x = 5$$

Check:

Original equation: \(2(5 + 3) = 5 + 11\)

Left side: \(2(8) = 16\)

Right side: \(16\)

The answer is correct.

Choosing Which Side Gets the Variable

You can move the variables to either side of the equation. Most students choose the side that keeps the coefficient positive.

For example, in \(6x + 4 = 2x + 20\), subtracting \(2x\) from both sides gives \(4x + 4 = 20\). This is often easier than subtracting \(6x\), which would create a negative coefficient.

Common Mistakes to Avoid

• Forgetting to do the same thing to both sides. An equation stays balanced only if both sides are treated equally.
• Moving a term without changing it correctly. If you subtract \(x\) from one side, you must subtract \(x\) from the other side too.
• Combining unlike terms. For example, \(3x + 5\) cannot become \(8x\). Variable terms and constants are different kinds of terms.
• Not distributing first. In an equation like \(3(x + 2) = x + 10\), distribute before collecting variable terms.
• Skipping the check. Checking helps catch mistakes.

Special Cases

Sometimes solving an equation with variables on both sides leads to a surprising result.

Case 1: No solution

Example:

$$2x + 5 = 2x + 1$$

Subtract \(2x\) from both sides:

$$5 = 1$$

This is false, so there is no solution.

Case 2: Infinitely many solutions

Example:

$$3x + 4 = 3x + 4$$

Subtract \(3x\) from both sides:

$$4 = 4$$

This is always true, so there are infinitely many solutions.

Quick Strategy Reminder

1. Simplify both sides if needed.
2. Move variable terms to one side.
3. Move constants to the other side.
4. Divide to get the variable alone.
5. Check the solution.

Try Thinking Through These

• \(7x + 2 = 4x + 14\)
• \(9 + 2x = 5x - 6\)
• \(4(x - 1) = 2x + 6\)

As you practice, remember that solving equations is like keeping a scale balanced. Every step should keep both sides equal.

Summary

To solve equations with variables on both sides, collect the variable terms on one side and the constants on the other side. Use addition or subtraction to move terms, then divide if needed to isolate the variable. Always simplify carefully and check your answer in the original equation.

Clearing Fractions and Decimals is a helpful strategy for solving linear equations in one variable. Sometimes an equation has fractions or decimals, and that can make the work feel messy. Instead of solving it right away, we can first rewrite the equation so it uses whole numbers.

The main idea is simple: multiply every term in the equation by the same number. This keeps the equation balanced while removing fractions or decimals.

When we clear fractions, we usually multiply by the least common denominator (LCD). When we clear decimals, we usually multiply by a power of 10, such as 10, 100, or 1000.

After the fractions or decimals are gone, the equation is usually much easier to solve.

Why does this work?

An equation is like a balanced scale. If you multiply one side by a number, you must multiply the other side by the same number. As long as you multiply every term on both sides by the same value, the two sides stay equal.

For example, if

we can multiply every term by 2:

This becomes

which is much easier to solve.

How to clear fractions

1. Find the denominators in the equation.
2. Find the least common denominator (LCD).
3. Multiply every term in the equation by the LCD.
4. Simplify to remove the fractions.
5. Solve the equation.
6. Check your answer if possible.

Important reminder: Multiply every term, not just the fractions. Terms without fractions must also be multiplied.

Example of terms in an equation:

The terms are \(\frac{1}{3}x\), \(5\), and \(\frac{2}{3}\). If we multiply by 3, all three terms must be multiplied by 3.

How to clear decimals

1. Look at how many decimal places appear in the equation.
2. Choose a power of 10 that moves all decimals to the right enough places.
3. Multiply every term by that power of 10.
4. Simplify and solve.

For example, if the greatest number of decimal places is 2, then multiplying by 100 will clear the decimals.

Worked Example 1: Clearing fractions with one denominator

Solve:

Step 1: Find the denominator.

The only denominator is 4, so the LCD is 4.

Step 2: Multiply every term by 4.

Step 3: Simplify.

Step 4: Solve.

Check:

The solution is correct.

Worked Example 2: Clearing fractions with different denominators

Solve:

Step 1: Find the denominators.

The denominators are 2, 3, and 6. The LCD is 6.

Step 2: Multiply every term by 6.

Step 3: Simplify.

Step 4: Solve.

It is okay if the answer is still a fraction. Clearing fractions helps simplify the equation, but the solution itself can still be a fraction.

Worked Example 3: Clearing decimals

Solve:

Step 1: Count decimal places.

Each decimal has 1 decimal place, so multiply every term by 10.

Step 2: Multiply every term by 10.

Step 3: Simplify.

Step 4: Solve.

Check:

The solution is correct.

Worked Example 4: Clearing decimals and using distribution

Solve:

Step 1: Choose a power of 10.

The greatest number of decimal places is 2, so multiply every term by 100.

Step 2: Simplify.

Step 3: Divide both sides by 25.

Step 4: Solve.

Check:

The solution is correct.

Tips for success

• Find the LCD carefully. A good LCD makes the equation easier.
• Multiply every term. Do not forget whole numbers or terms in parentheses.
• Use parentheses when needed. If a term has more than one part, like \((x+4)\), keep it together when multiplying.
• Simplify before solving. Once the fractions or decimals are gone, solve like a regular linear equation.
• Check your answer. Plug it back into the original equation.

Common mistakes to avoid

• Multiplying only one side of the equation.
• Multiplying only the fractions and forgetting the other terms.
• Choosing a number that does not clear all denominators.
• Forgetting to distribute when parentheses are involved.
• Moving decimals incorrectly when multiplying by 10, 100, or 1000.

When should you use this strategy?

Use clearing fractions or decimals when an equation looks harder because of fractional or decimal coefficients. Turning the equation into one with whole numbers often makes the solving steps quicker and less confusing.

Quick comparison

• If the equation has fractions like \(\frac{1}{2}\) or \(\frac{3}{4}\), multiply by the LCD.
• If the equation has decimals like 0.6 or 1.25, multiply by a power of 10.

Practice thinking

Before solving, ask yourself:

• What is making this equation hard to read?
• Are there fractions? Then what is the LCD?
• Are there decimals? Then what power of 10 should I use?
• Did I multiply every term on both sides?

Summary

Clearing fractions and decimals is a smart way to simplify linear equations before solving them. To clear fractions, multiply every term by the least common denominator. To clear decimals, multiply every term by a power of 10. After that, solve the simpler equation and check your answer in the original problem.

Identities and Contradictions are special kinds of equations you may see when solving linear equations in one variable.

Usually, when you solve an equation, you expect to find one value for the variable. But sometimes something different happens:

• the equation is true for every value of the variable, or
• the equation is true for no value of the variable.

Learning to recognize these cases helps you understand what an equation is really saying, not just how to move terms around.

In this lesson, you will learn:

• what an identity is,
• what a contradiction is,
• how they appear when solving equations, and
• how to explain why an equation has infinitely many solutions or no solution.

1. Review: What usually happens when solving an equation?

An equation says two expressions are equal. For example, in

$$2x + 3 = 11$$

there is one value of \(x\) that makes the equation true.

We solve:

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

So this equation has one solution.

But not all equations work this way.

2. What is an identity?

An identity is an equation that is true for all values of the variable.

That means it has infinitely many solutions.

When solving, an identity usually ends with a statement that is always true, such as:

• \(0 = 0\)
• \(5 = 5\)
• \(-2 = -2\)

These are true no matter what value the variable has, because the variable has disappeared and both sides match.

Why does this happen?

It happens when both sides of the equation are really just different ways of writing the same expression.

For example,

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

If we simplify the left side, we get

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

That is true for every value of \(x\), so it is an identity.

3. What is a contradiction?

A contradiction is an equation that is true for no value of the variable.

That means it has no solution.

When solving, a contradiction usually ends with a statement that is always false, such as:

• \(0 = 5\)
• \(3 = -1\)
• \(7 = 2\)

These statements can never be true, so the equation has no value that works.

Why does this happen?

It happens when the variable terms on both sides match, but the constant numbers do not match.

For example,

$$2x + 1 = 2x + 4$$

If we subtract \(2x\) from both sides, we get

$$1 = 4$$

That is false, so this equation is a contradiction.

4. How to tell what kind of equation you have

When solving a linear equation, keep going until the variable is isolated or disappears.

Then decide which of these three results happened:

• One solution: you get something like \(x = 3\)
• Identity: you get a true statement like \(0 = 0\)
• Contradiction: you get a false statement like \(0 = 7\)

You can think of it like this:

• If everything balances perfectly, it is an identity.
• If the variable disappears but the numbers do not match, it is a contradiction.

5. Worked Examples

Example 1: Identity

Solve:

$$4x + 8 = 2(2x + 4)$$

Step 1: Simplify the right side.

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

Step 2: Subtract \(4x\) from both sides.

$$8 = 8$$

This is always true.

Conclusion: The equation is an identity, so it has infinitely many solutions.

Example 2: Contradiction

Solve:

$$5(x - 1) = 5x + 10$$

Step 1: Distribute.

$$5x - 5 = 5x + 10$$

Step 2: Subtract \(5x\) from both sides.

$$-5 = 10$$

This is never true.

Conclusion: The equation is a contradiction, so it has no solution.

Example 3: One solution

Solve:

$$3x + 7 = x + 15$$

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

$$2x + 7 = 15$$

Step 2: Subtract 7 from both sides.

$$2x = 8$$

Step 3: Divide by 2.

$$x = 4$$

Conclusion: This equation has one solution, \(x = 4\).

This example is helpful because it shows that not every equation with variables on both sides is an identity or contradiction. Sometimes you still get one answer.

Example 4: A more complex identity

Solve:

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

Step 1: Distribute on both sides.

$$6x + 6 - 2x = 4x + 8 - 2$$

Step 2: Combine like terms.

$$4x + 6 = 4x + 6$$

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

$$6 = 6$$

This is always true.

Conclusion: The equation is an identity with infinitely many solutions.

6. Looking at structure

It is useful to notice the structure of an equation before you even finish solving.

Suppose an equation simplifies to this form:

$$ax + b = ax + b$$

Both sides are exactly the same. This will be an identity.

Now suppose an equation simplifies to this form:

$$ax + b = ax + c$$

where \(b\) and \(c\) are different numbers.

If you subtract \(ax\) from both sides, you get

$$b = c$$

If \(b\) and \(c\) are different, this is false. So it is a contradiction.

This is the structural reason behind these special cases:

• same variable part and same number part  identity
• same variable part but different number part  contradiction

7. Common mistakes to avoid

• Stopping too early: Always simplify both sides fully before deciding.
• Thinking \(0 = 0\) means no solution: It actually means infinitely many solutions.
• Thinking \(0 = 5\) means one solution: It actually means no solution.
• Forgetting to distribute: Expressions like \(2(x + 3)\) must become \(2x + 6\).
• Not combining like terms: Simplify carefully so you can see the true result.

8. Quick check questions

Try to decide what each equation has: one solution, infinitely many solutions, or no solution.

1. $$2x + 6 = 2(x + 3)$$
2. $$7x - 4 = 7x + 1$$
3. $$4x + 9 = 3x + 11$$

Answers:

1. \(2x + 6 = 2x + 6\)  identity  infinitely many solutions
2. Subtract \(7x\): \(-4 = 1\)  contradiction  no solution
3. Subtract \(3x\): \(x + 9 = 11\), so \(x = 2\)  one solution

9. Summary

When solving linear equations, sometimes the variable disappears.

• If you end with a true statement like \(0 = 0\), the equation is an identity and has infinitely many solutions.
• If you end with a false statement like \(0 = 4\), the equation is a contradiction and has no solution.
• If you end with something like \(x = 5\), the equation has one solution.

The key idea is to look at what the equation becomes after simplifying. That tells you whether both sides always match, never match, or match for just one value.

Solving Literal Equations means rearranging a formula so that a different variable is by itself.

A literal equation is an equation with two or more variables. For example, the formula for distance is \(d = rt\), where \(d\) is distance, \(r\) is rate, and \(t\) is time.

Sometimes a formula is already solved for one variable, but you may need to solve it for a different one. This is done using the same idea as solving regular equations: use inverse operations to undo what is happening to the variable you want.

For example, if you know \(d = rt\) and want to solve for \(r\), you ask: what is happening to \(r\)? It is being multiplied by \(t\). To undo that, divide both sides by \(t\).

Main idea: Treat the variable you want like the “answer” you are trying to isolate, and treat the other letters like numbers.

Here are the basic steps for solving a literal equation:

1. Choose the variable you want to isolate.
2. Undo addition or subtraction first.
3. Undo multiplication or division next.
4. Do the same thing to both sides of the equation.
5. Check that the variable is alone.

It helps to remember that letters other than the one you want are not special. They act like fixed values while you rearrange the equation.

Example 1: Solve \(A = lw\) for \(l\)

We want \(l\) by itself.

In \(A = lw\), the variable \(l\) is multiplied by \(w\). To undo multiplication by \(w\), divide both sides by \(w\).

So the equation solved for \(l\) is:

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

We want \(x\) alone.

First, undo the \(+ b\) by subtracting \(b\) from both sides.

Now \(x\) is being multiplied by \(m\). Divide both sides by \(m\).

So the equation solved for \(x\) is:

Notice the order: subtract first, then divide. This matches the order of operations in reverse.

Example 3: Solve \(P = 2l + 2w\) for \(w\)

We want \(w\) by itself.

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

Now divide both sides by 2.

So the equation solved for \(w\) is:

You may also see the answer written as:

Both forms mean the same thing.

Example 4: Solve \(v = u + at\) for \(a\)

This example has more than one operation, but the steps are still the same.

We want \(a\) by itself. First subtract \(u\) from both sides.

Now divide both sides by \(t\).

So the equation solved for \(a\) is:

Tips for Success

• Keep the equation balanced. Whatever you do to one side, do to the other side.
• Move terms in a careful order. Undo addition and subtraction before multiplication and division.
• Watch parentheses. If a group of terms stays together, keep them grouped.
• Do not plug in numbers unless asked. Literal equations usually stay in variable form.
• Rewrite neatly. A clean layout makes it easier to see each step.

Common Mistakes

• Forgetting to do the same operation to both sides.
• Dividing only part of an expression instead of the entire side.
• Changing the order of terms incorrectly.
• Stopping before the chosen variable is fully isolated.

For example, in solving \(y = mx + b\) for \(x\), it would be incorrect to write \(x = y - \frac{b}{m}\). You must subtract \(b\) first, then divide the entire expression:

How to Check Your Answer

After solving, ask yourself these questions:

• Is the variable I wanted completely alone?
• Did I use inverse operations correctly?
• Did I apply each operation to both sides?
• Does the final equation still make sense as a formula?

Summary

Solving literal equations is just like solving regular equations, except the equation has more than one variable. The goal is to rearrange the formula so one chosen variable is isolated.

Use inverse operations, keep the equation balanced, and treat the other variables like numbers. With practice, you will be able to rewrite formulas in the form you need.

Multi-Step Inequalities are a lot like multi-step equations, but there is one important difference: instead of finding one exact value, you are finding a set of values that make the inequality true.

An inequality uses symbols like:

•  less than
•  greater than
• = less than or equal to
• = greater than or equal to

For example, in the inequality \(x + 3  7\), the solution is not just one number. Any number greater than 4 will make the inequality true.

In this lesson, you will learn how to solve multi-step inequalities, including the very important rule about reversing the inequality sign when you multiply or divide by a negative number.

Why inequalities are different from equations

An equation, such as \(2x + 1 = 9\), has a specific answer. But an inequality, such as \(2x + 1  9\), has many possible answers.

When you solve an inequality, your goal is to get the variable by itself, just like with equations. You use inverse operations to undo addition, subtraction, multiplication, and division.

Rules for solving inequalities

• Add the same number to both sides: the inequality sign stays the same.
• Subtract the same number from both sides: the inequality sign stays the same.
• Multiply both sides by the same positive number: the inequality sign stays the same.
• Divide both sides by the same positive number: the inequality sign stays the same.
• Multiply or divide both sides by a negative number: reverse the inequality sign.

This last rule is the most important one to remember.

For example:

$$-2x  10$$

To solve, divide both sides by \(-2\). Since you are dividing by a negative number, reverse the inequality sign:

$$x  -5$$

Steps for solving multi-step inequalities

1. Simplify each side if needed. Combine like terms or use the distributive property.
2. Move variable terms to one side and number terms to the other side.
3. Undo multiplication or division to isolate the variable.
4. If you multiply or divide by a negative number, reverse the inequality sign.
5. Check your answer by testing a value that should work.

Worked Example 1: Two-step inequality

Solve:

$$x + 5  12$$

Subtract 5 from both sides:

$$x + 5 - 5  12 - 5$$

$$x  7$$

Answer: \(x  7\)

This means 7 is a solution, and any number greater than 7 is also a solution.

Worked Example 2: Multiplying or dividing by a negative

Solve:

$$-3x + 4  19$$

First subtract 4 from both sides:

$$-3x  15$$

Now divide both sides by \(-3\). Because you are dividing by a negative, reverse the inequality sign:

$$x  -5$$

Answer: \(x  -5\)

Check with \(x = 0\):

$$-3(0) + 4  19$$

$$4  19$$

This is true, and 0 is greater than \(-5\), so the answer makes sense.

Worked Example 3: Inequality with distribution

Solve:

$$2(x - 3) + 5  11$$

First use the distributive property:

$$2x - 6 + 5  11$$

Combine like terms:

$$2x - 1  11$$

Add 1 to both sides:

$$2x  12$$

Divide both sides by 2:

$$x  6$$

Answer: \(x  6\)

Because we divided by a positive number, the inequality sign stayed the same.

Worked Example 4: Variables on both sides

Solve:

$$5x - 2  2x + 7$$

Subtract \(2x\) from both sides:

$$3x - 2  7$$

Add 2 to both sides:

$$3x  9$$

Divide both sides by 3:

$$x  3$$

Answer: \(x  3\)

This means 3 and all numbers greater than 3 are solutions.

How to write and understand the solution

When you solve an inequality, the answer tells you a whole group of numbers.

• \(x  4\) means all numbers less than 4
• \(x  -2\) means all numbers greater than -2
• \(x = 6\) means 6 and all numbers less than 6
• \(x = 1\) means 1 and all numbers greater than 1

If you draw the solution on a number line:

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

Common mistakes to avoid

• Forgetting to reverse the sign when dividing or multiplying by a negative number.
• Not distributing correctly when there are parentheses.
• Combining unlike terms. Only like terms can be combined.
• Treating an inequality like an equation. Remember, the answer is usually many numbers, not just one.

Quick reminder about reversing the sign

If you see something like:

$$-4x  20$$

Divide by \(-4\):

$$x  -5$$

The sign changes from \(\) to \(\) because of the negative division.

Try these on your own

1. \(3x + 2  14\)
2. \(-2x - 1  7\)
3. \(4(x + 1) - 3 = 13\)
4. \(6x - 5  2x + 11\)

Summary

To solve a multi-step inequality, use the same basic process as solving equations: simplify, move terms, and isolate the variable.

The most important rule is this: if you multiply or divide both sides by a negative number, reverse the inequality sign.

Always check your work by substituting a value from your solution set back into the original inequality.

Compound Inequalities are two inequalities joined together in one statement. They help describe a range of possible values for a variable.

For example, if a number is greater than 2 and less than 7, we can write that as a compound inequality:

$$2 < x < 7$$

In this lesson, you will learn what compound inequalities mean, how to solve them, and how to graph their solutions on a number line.

There are two main types of compound inequalities:

• And inequalities
• Or inequalities

1. And Inequalities

An and inequality means a value must satisfy both parts at the same time.

Example:

$$x > 3 \text{ and } x < 8$$

This means the number must be bigger than 3 and also smaller than 8. So the solution is all numbers between 3 and 8.

We can also write it as:

$$3 < x < 8$$

On a number line, this would have:

• an open circle at 3 because 3 is not included
• an open circle at 8 because 8 is not included
• a shaded line between 3 and 8

If the inequality uses \(\leq\) or \(\geq\), then the endpoint is included, so we use a closed circle.

For example:

$$-2 \leq x < 4$$

This means:

• \(x\) can be -2, so use a closed circle at -2
• \(x\) cannot be 4, so use an open circle at 4
• shade between them

2. Or Inequalities

An or inequality means a value can satisfy either part. It does not have to satisfy both.

Example:

$$x < -1 \text{ or } x > 5$$

This means the number can be less than -1, or greater than 5.

On a number line, this would have:

• an open circle at -1 and shading to the left
• an open circle at 5 and shading to the right

The graph has two separate parts because the solutions are in two different places.

Important idea:

• And means the overlap, or the values that work for both inequalities.
• Or means the combination, or all values that work for either inequality.

How to Solve Compound Inequalities

The steps depend on how the inequality is written.

For an "and" inequality written in one line, do the same operation to all three parts.

Example:

$$4 < x + 1 \leq 9$$

Subtract 1 from all three parts:

$$4 - 1 < x + 1 - 1 \leq 9 - 1$$

$$3 < x \leq 8$$

This means:

• open circle at 3
• closed circle at 8
• shade between them

For inequalities joined by “and” or “or”, solve each part separately.

Example:

$$2x - 1 < 7 \text{ and } 3x + 2 \geq 5$$

Solve the first part:

$$2x - 1 < 7$$

$$2x < 8$$

$$x < 4$$

Solve the second part:

$$3x + 2 \geq 5$$

$$3x \geq 3$$

$$x \geq 1$$

Because this is an and inequality, we need values that satisfy both:

$$x < 4 \text{ and } x \geq 1$$

So the solution is:

$$1 \leq x < 4$$

A very important rule: If you multiply or divide by a negative number, flip the inequality sign.

For example:

$$-2x > 6$$

Divide by -2:

$$x < -3$$

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

Worked Example 1: Basic And Inequality

Solve and graph:

$$5 < x - 2 \leq 10$$

Add 2 to all three parts:

$$5 + 2 < x - 2 + 2 \leq 10 + 2$$

$$7 < x \leq 12$$

Answer: $$7 < x \leq 12$$

Graph:

• open circle at 7
• closed circle at 12
• shade between 7 and 12

Worked Example 2: And Inequality Solved Separately

Solve:

$$3x + 1 \leq 13 \text{ and } x - 4 > -2$$

First part:

$$3x + 1 \leq 13$$

$$3x \leq 12$$

$$x \leq 4$$

Second part:

$$x - 4 > -2$$

$$x > 2$$

Now combine them using and:

$$x > 2 \text{ and } x \leq 4$$

So the solution is:

$$2 < x \leq 4$$

Worked Example 3: Or Inequality

Solve:

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

First part:

$$2x + 3 < 1$$

$$2x < -2$$

$$x < -1$$

Second part:

$$x - 5 \geq 2$$

$$x \geq 7$$

Because this is an or inequality, the solution is all values from either part:

$$x < -1 \text{ or } x \geq 7$$

Graph:

• open circle at -1, shade left
• closed circle at 7, shade right

Worked Example 4: Inequality with a Negative Coefficient

Solve:

$$-3x + 6 \geq 12 \text{ or } 2x - 1 < 3$$

First part:

$$-3x + 6 \geq 12$$

$$-3x \geq 6$$

Divide by -3 and flip the sign:

$$x \leq -2$$

Second part:

$$2x - 1 < 3$$

$$2x < 4$$

$$x < 2$$

Now combine with or:

$$x \leq -2 \text{ or } x < 2$$

Since every number less than or equal to -2 is already included in \(x < 2\), the full solution is simply:

$$x < 2$$

This shows that sometimes one part of an or inequality is already covered by the other part.

How to Check Your Answer

• Solve each inequality carefully.
• If it is and, look for the overlap.
• If it is or, include all values from both parts.
• Check whether endpoints should be open or closed.
• If you divide or multiply by a negative, flip the inequality sign.

Common Mistakes to Avoid

• Mixing up and and or
• Forgetting to flip the sign when dividing by a negative
• Using a closed circle when the endpoint is not included
• Using an open circle when the endpoint is included
• Shading the wrong part of the number line

Real-World Meaning

Compound inequalities are useful when values must stay within a range or can fall into different allowed groups.

For example, if a ride at an amusement park is for children who are at least 48 inches tall and less than 72 inches tall, the height rule can be written as:

$$48 \leq h < 72$$

If a store gives a discount to shoppers age 12 or younger or age 65 and older, that could be written as:

$$a \leq 12 \text{ or } a \geq 65$$

Summary

A compound inequality combines two inequalities using and or or.

Use and when both conditions must be true. This usually gives one connected section on a number line.

Use or when either condition can be true. This often gives two separate rays on a number line.

When solving, work carefully, graph the solution, and remember to flip the inequality sign if you multiply or divide by a negative number.

Modeling with Linear Equations means taking a real-world situation, turning it into an equation, solving the equation, and then explaining what the answer means in the situation.

This is an important skill because many word problems are really asking you to find an unknown number. A linear equation helps you represent that unknown clearly.

In this lesson, you will learn how to:

• identify the unknown quantity,
• define a variable,
• write a linear equation from a word problem,
• solve the equation, and
• check and interpret the answer in context.

1. What is a linear equation?

A linear equation is an equation where the variable has an exponent of 1. It can look like:

$$x + 7 = 19$$

or

$$3x - 5 = 16$$

When we model a situation with a linear equation, we are saying, “The relationship in this problem can be described with a straight-line pattern.” In 8th grade word problems, this often means there is a constant rate, a fixed fee, or a simple total.

2. Steps for modeling a word problem

When you read a word problem, do not rush to calculate right away. Follow these steps.

1. Read carefully. Figure out what the problem is about.
2. Choose a variable. Let the unknown be a letter, such as \(x\).
3. Translate words into math. Write an equation that matches the situation.
4. Solve the equation.
5. Check your answer. Substitute it back into the situation.
6. Interpret the answer. Write what the number means in the real world.

3. Common word phrases and what they mean

These phrases often appear in word problems:

• sum means add
• difference means subtract
• product means multiply
• quotient means divide
• is usually means equals \((=)\)
• more than means add
• less than means subtract, but be careful with order
• twice means multiply by 2
• three times means multiply by 3

For example:

• “5 more than a number” becomes \(x + 5\)
• “12 less than a number” becomes \(x - 12\)
• “twice a number” becomes \(2x\)
• “7 less than twice a number” becomes \(2x - 7\)

4. Look for the structure of the problem

Many real-world problems fit one of these patterns:

• Starting amount + change = total
• Rate × number of units + fixed amount = total
• One quantity compared to another

These patterns help you decide how to write the equation.

Worked Example 1: Simple total

A number increased by 9 is 23. What is the number?

Step 1: Define the variable.

Let \(x\) be the number.

Step 2: Write the equation.

“Increased by 9” means add 9, and “is 23” means equals 23.

$$x + 9 = 23$$

Step 3: Solve.

Subtract 9 from both sides:

$$x = 14$$

Step 4: Interpret.

The number is 14.

Check: \(14 + 9 = 23\), so the answer makes sense.

Worked Example 2: Cost with a fixed fee

A gym charges a $15 sign-up fee and $8 per class. If Maya spent $63 in total, how many classes did she attend?

Step 1: Define the variable.

Let \(x\) be the number of classes.

Step 2: Write the equation.

The total cost is:

$$\text{sign-up fee} + \text{cost per class} = \text{total}$$

So:

$$15 + 8x = 63$$

Step 3: Solve.

Subtract 15 from both sides:

$$8x = 48$$

Divide by 8:

$$x = 6$$

Step 4: Interpret.

Maya attended 6 classes.

Check: \(15 + 8(6) = 15 + 48 = 63\).

Worked Example 3: Comparing two amounts

Jalen has 5 fewer stickers than Priya. Together they have 37 stickers. How many stickers does each person have?

Step 1: Define the variable.

Let \(x\) be the number of stickers Priya has.

Then Jalen has \(x - 5\) stickers.

Step 2: Write the equation.

Together they have 37:

$$x + (x - 5) = 37$$

Step 3: Solve.

Combine like terms:

$$2x - 5 = 37$$

Add 5:

$$2x = 42$$

Divide by 2:

$$x = 21$$

Step 4: Interpret.

Priya has \(21\) stickers.

Jalen has \(21 - 5 = 16\) stickers.

Check: \(21 + 16 = 37\).

Worked Example 4: A more complex situation

A taxi ride costs a $4 starting fee plus $2.50 per mile. The total fare was $26.50. How many miles was the ride?

Step 1: Define the variable.

Let \(m\) be the number of miles.

Step 2: Write the equation.

$$4 + 2.5m = 26.5$$

Step 3: Solve.

Subtract 4 from both sides:

$$2.5m = 22.5$$

Divide by 2.5:

$$m = 9$$

Step 4: Interpret.

The taxi ride was 9 miles.

Check: \(4 + 2.5(9) = 4 + 22.5 = 26.5\).

5. How to know what the variable should represent

The variable should represent the quantity you are asked to find, or a quantity that makes the equation easy to write.

For example, if a problem asks for the number of tickets sold, let \(t\) represent the number of tickets. If the problem compares two people’s ages, choose one age as the variable, then write the other age using that variable.

6. Interpreting the answer in context

Solving the equation gives you a number, but the problem is not finished until you explain what that number means.

For example, if \(x = 6\), say “The student bought 6 notebooks” instead of only writing 6.

Also think about whether the answer makes sense in real life.

• If you get a negative number for the number of tickets, something is wrong.
• If you get \(4.5\) people, that does not make sense unless the problem allows parts.
• If the problem is about money or distance, decimals may make sense.

7. Watch out for contextual constraints

A constraint is a rule from the real-world situation.

For example:

• The number of students must be a whole number.
• The number of items sold cannot be negative.
• A distance or time usually cannot be negative.

Even if an equation has a solution, you must decide whether that solution makes sense in the problem.

Example of checking a constraint

A school bus holds 48 students. There are 12 already on the bus. If each stop adds the same number of students and after 3 stops the bus is full, how many students got on at each stop?

Let \(x\) be the number of students getting on at each stop.

$$12 + 3x = 48$$

Subtract 12:

$$3x = 36$$

Divide by 3:

$$x = 12$$

This makes sense because 12 is a whole number and cannot be negative.

8. Special cases to understand

Sometimes, after simplifying an equation, you may get a special result.

No solution

Example:

$$2x + 3 = 2x - 5$$

Subtract \(2x\) from both sides:

$$3 = -5$$

This is false, so there is no solution.

In a real-world problem, this means the information does not fit together.

One solution

Most word problems have one solution.

Example:

$$3x + 4 = 19$$

This gives one value of \(x\).

When modeling, most realistic 8th grade problems will have one solution, but it is still useful to know that sometimes the numbers in a problem may not make sense together.

9. Tips for success

• Underline important information as you read.
• Name the variable clearly.
• Write an equation before solving.
• Use parentheses when combining expressions, like \(x + (x - 5)\).
• Check your answer in the original problem, not just in the equation.
• Include units when needed, such as dollars, miles, or students.

10. Quick practice thinking

Here are some situations and the equations they model:

• A number decreased by 7 is 18: \(x - 7 = 18\)
• $12 plus $3 per notebook totals $27: \(12 + 3x = 27\)
• A rectangle’s length is 4 more than its width, and the length is 15: \(w + 4 = 15\)
• Two identical shirts and a $5 fee cost $41: \(2x + 5 = 41\)

Brief Summary

To model with a linear equation, first decide what the unknown is and choose a variable. Next, translate the words into an equation, solve it, and check the answer. Finally, explain the answer in the context of the problem and make sure it fits any real-world constraints.