Linear Equations in One Variable

Linear Equations in One Variable

SAT Math | Algebra | Foundation to Advanced

Linear equations are an essential part of SAT Math. This lesson explains how to solve equations, simplify expressions, work with fractions and unknown constants, identify different types of solutions, and apply algebra to real-world problems.

1. Understanding Linear Equations

A linear equation in one variable is an equation that contains one unknown variable, usually x, and can be simplified into the form:

ax + b = c

where a, b, and c are constants and a ≠ 0.

A linear equation has an equals sign (=), which tells us that the expressions on both sides have the same value.

Consider:

3x + 5 = 20

In this equation:

• x is the variable (unknown value).

• 3 is the coefficient (number multiplying x).

• 5 is a constant.

• 20 is the value on the right side.

The goal is to determine which value of x makes the equation true.

Example

3x + 5 = 20

Step 1: Subtract 5 from both sides.

3x = 15

Step 2: Divide both sides by 3.

x = 5

Checking the answer

Substitute x = 5 into the original equation.

3(5) + 5 = 20

15 + 5 = 20

20 = 20

Since both sides are equal, x = 5 is correct.

Key idea: Solving an equation means finding the value of the variable that makes the equation true.

The balance principle

An equation works like a balanced scale. Both sides must remain equal.

You may add, subtract, multiply, or divide both sides by the same number, provided you do not divide by zero.

For example:

x + 7 = 15

Subtracting 7 from both sides gives:

x = 8

If you subtract 7 from only one side, the equality is no longer preserved.

On the SAT, understanding this principle is more important than memorizing shortcuts such as "move the number to the other side."

2. Solving Linear Equations Step by Step

The main goal when solving a linear equation is to isolate the variable.

This means rewriting the equation so that the variable appears by itself on one side.

Inverse operations

Inverse operations are mathematical operations that undo each other.

Addition is undone by subtraction.

Subtraction is undone by addition.

Multiplication is undone by division.

Division is undone by multiplication.

For example, if x is multiplied by 6, dividing by 6 reverses that multiplication.

General method

Step 1: Simplify expressions on both sides.

Step 2: Remove parentheses if necessary.

Step 3: Combine like terms.

Step 4: Collect variable terms on one side.

Step 5: Collect constants on the other side.

Step 6: Divide by the coefficient of the variable.

Step 7: Check the answer when appropriate.

Worked Example 1 — Two-step equation

Difficulty: Easy

What value of x satisfies the equation?

7x − 9 = 26

Step 1: Add 9 to both sides.

7x = 35

Step 2: Divide both sides by 7.

x = 5

Final answer: 5

Worked Example 2 — Negative coefficient

Difficulty: Medium

Solve:

9 − 2x = −5

Step 1: Subtract 9 from both sides.

−2x = −14

Step 2: Divide both sides by −2.

x = 7

Check:

9 − 2(7) = −5

9 − 14 = −5

Final answer: 7

SAT tip: A negative number divided by another negative number gives a positive result. Pay careful attention to signs.

3. Simplifying Equations with Parentheses, Fractions, and Decimals

SAT questions often contain expressions that must be simplified before the variable can be isolated.

A. Distributive property

The distributive property allows a number outside parentheses to multiply every term inside.

a(b + c) = ab + ac

a(b − c) = ab − ac

For example:

4(x + 3) = 4x + 12

−2(x − 5) = −2x + 10

Be especially careful when distributing a negative number.

Worked Example 3 — Parentheses

Difficulty: Medium

Solve:

3(2x − 5) + 4 = 2x + 17

Step 1: Distribute 3.

6x − 15 + 4 = 2x + 17

Step 2: Combine like terms.

6x − 11 = 2x + 17

Step 3: Subtract 2x from both sides.

4x − 11 = 17

Step 4: Add 11.

4x = 28

Step 5: Divide by 4.

x = 7

Final answer: 7

B. Equations containing fractions

To simplify equations with fractions, multiply every term by the least common denominator (LCD).

The LCD is the smallest positive number that is divisible by all denominators.

For example, the LCD of 2 and 3 is 6.

Worked Example 4 — Fractions

Difficulty: Medium

Solve:

(x − 2)/3 + (x + 4)/2 = 8

Step 1: Identify the LCD.

The denominators are 3 and 2.

LCD = 6

Step 2: Multiply every term by 6.

2(x − 2) + 3(x + 4) = 48

Step 3: Expand the parentheses.

2x − 4 + 3x + 12 = 48

Step 4: Combine like terms.

5x + 8 = 48

Step 5: Subtract 8.

5x = 40

Step 6: Divide by 5.

x = 8

Final answer: 8

Common mistake: Multiplying only the fractions by the LCD while forgetting the other terms in the equation.

C. Equations containing decimals

Decimals can be handled directly or removed by multiplying the entire equation by an appropriate power of 10.

Worked Example 5 — Decimals

Difficulty: Medium

Solve:

0.4x − 1.2 = 2.8

Step 1: Add 1.2 to both sides.

0.4x = 4

Step 2: Divide both sides by 0.4.

x = 10

Final answer: 10

An alternative approach is to multiply the original equation by 10.

4x − 12 = 28

4x = 40

x = 10

Both methods produce the same answer.

4. Variables on Both Sides and Types of Solutions

Some SAT questions contain variables on both sides of the equals sign.

For example:

5x + 18 = 2x + 39

Subtract 2x from both sides.

3x + 18 = 39

Subtract 18.

3x = 21

Divide by 3.

x = 7

However, not every linear equation has exactly one solution.

There are three possible cases.

Case 1 — One solution

An equation has one solution when exactly one value of x makes it true.

Example:

2x + 3 = 11

2x = 8

x = 4

Conclusion: One solution, x = 4.

Case 2 — No solution

An equation has no solution when no value of x can make it true.

Example:

5x + 4 = 5x + 9

Subtract 5x from both sides.

4 = 9

This is impossible.

Conclusion: No solution.

Case 3 — Infinitely many solutions

An equation has infinitely many solutions when every possible value of x makes it true.

Example:

4(x + 2) = 4x + 8

Expand:

4x + 8 = 4x + 8

Subtract 4x.

8 = 8

This is always true.

Conclusion: Infinitely many solutions.

SAT shortcut — Comparing coefficients

Consider the general equation:

ax + b = cx + d

If a ≠ c, the equation has exactly one solution.

If a = c but b ≠ d, the equation has no solution.

If a = c and b = d, the equation has infinitely many solutions.

This shortcut is especially useful for questions involving unknown constants.

5. Unknown Constants and Algebraic Expressions

More advanced SAT questions may include unknown constants such as a, b, or k.

Instead of simply solving for x, students may need to determine which value of a constant causes a certain condition.

Worked Example 6 — Unknown constant

Difficulty: Hard

For what value of k does the equation have infinitely many solutions?

4(2x − 3) = 8x + k

Step 1: Distribute 4.

8x − 12 = 8x + k

Step 2: Subtract 8x from both sides.

−12 = k

Therefore:

k = −12

Final answer: −12

When k = −12, the equation becomes:

8x − 12 = 8x − 12

Both sides are identical, so every value of x is a solution.

Worked Example 7 — No solution with a parameter

Difficulty: Hard

For what value of k does the following equation have no solution?

(k + 1)x + 4 = 5x + 9

Step 1: Recall that an equation has no solution when the coefficients of x are equal but the constants are different.

Step 2: Set the coefficients equal.

k + 1 = 5

Step 3: Subtract 1.

k = 4

Step 4: Check the equation.

5x + 4 = 5x + 9

Subtract 5x.

4 = 9

This is false.

Final answer: 4

Finding an expression instead of x

Sometimes the SAT asks for a value such as 3x + 2 rather than the value of x.

Consider:

5(3x + 2) − 4 = 41

What is the value of 3x + 2?

Step 1: Add 4 to both sides.

5(3x + 2) = 45

Step 2: Divide both sides by 5.

3x + 2 = 9

Final answer: 9

You do not need to solve for x.

SAT tip: Always identify the exact quantity requested. Avoid unnecessary calculations.

6. Creating Equations from Real-World Problems

Not all SAT Algebra questions present an equation directly.

Some questions describe a situation and require students to create an equation.

A common structure is:

Total = Fixed amount + (Rate × Quantity)

The fixed amount stays the same.

The rate describes how much is added or removed for each unit.

The quantity represents the number of units.

Worked Example 8 — Tutoring sessions

Difficulty: Medium

A tutoring center charges a one-time registration fee of $35 and $12 per tutoring session.

A student pays $131 in total.

How many tutoring sessions did the student attend?

Step 1: Define the variable.

Let s represent the number of sessions.

Step 2: Write the equation.

35 + 12s = 131

Step 3: Subtract 35.

12s = 96

Step 4: Divide by 12.

s = 8

Final answer: 8 sessions

Interpreting the equation

In the equation:

35 + 12s = 131

35 represents the one-time registration fee.

12 represents the cost per session.

s represents the number of sessions.

131 represents the total amount paid.

The SAT may ask students what one of these values represents instead of asking them to calculate s.

Worked Example 9 — Comparing two plans

Difficulty: Hard

Company A charges a fixed fee of $10 and $4 per hour.

Company B charges a fixed fee of $34 and $2 per hour.

After how many hours will both companies charge the same amount?

Step 1: Let h represent the number of hours.

Step 2: Write the cost of Company A.

10 + 4h

Step 3: Write the cost of Company B.

34 + 2h

Step 4: Set the expressions equal.

10 + 4h = 34 + 2h

Step 5: Subtract 2h.

10 + 2h = 34

Step 6: Subtract 10.

2h = 24

Step 7: Divide by 2.

h = 12

Final answer: 12 hours

Checking:

Company A: 10 + 4(12) = 58

Company B: 34 + 2(12) = 58

Both companies charge $58 after 12 hours.

7. SAT Strategies and Common Mistakes

Strategy 1 — Simplify before solving

Distribute and combine like terms before isolating the variable when needed.

Example:

3(x + 2) + 2x = 21

3x + 6 + 2x = 21

5x + 6 = 21

5x = 15

x = 3

Strategy 2 — Recognize special solution cases

When the variable disappears during simplification, inspect the remaining statement.

A true statement indicates infinitely many solutions.

A false statement indicates no solution.

Strategy 3 — Look for shortcuts

If a question asks for an expression such as 2x + 5, try isolating that expression directly.

You may not need to find x.

Strategy 4 — Check by substitution

After solving an equation, substitute the answer into the original equation when useful.

This is especially helpful when the answer choices contain similar numbers.

Strategy 5 — Interpret context carefully

In a word problem, identify:

The unknown quantity.

The starting or fixed amount.

The rate per unit.

The total amount.

Then construct the equation.

Strategy 6 — Use your calculator wisely

The Digital SAT allows calculator use throughout the Math section, including the built-in Desmos calculator.

However, straightforward equations can often be solved more quickly using algebra.

A calculator is useful for checking decimal calculations or testing a possible solution, but it does not replace understanding the mathematical relationship.

Common mistakes to avoid

Mistake 1 — Incorrect distribution

Incorrect:

3(x + 4) = 3x + 4

Correct:

3(x + 4) = 3x + 12

Mistake 2 — Losing negative signs

Incorrect:

−(2x − 5) = −2x − 5

Correct:

−(2x − 5) = −2x + 5

Mistake 3 — Performing an operation on only one side

When solving an equation, apply the same valid operation to both sides.

Mistake 4 — Confusing no solution and infinitely many solutions

5 = 9 is false, so there is no solution.

5 = 5 is true, so the equation has infinitely many solutions when the variables have canceled.

Mistake 5 — Answering the wrong quantity

If the question asks for 2x + 3, do not provide x as the final answer.

Mistake 6 — Misinterpreting fixed fees

In 20 + 5x = 70, the number 20 is the fixed amount, while 5 represents the rate per unit.

Mistake 7 — Forgetting to multiply every term when clearing fractions

For an equation containing fractions, multiply every term on both sides by the LCD.

Mistake 8 — Rushing through arithmetic

Simple sign and arithmetic errors can cause incorrect answers even when the solving method is correct.

8. Check Your Understanding

Use these questions to review the main ideas before starting Baloo Prep's full practice bank.

Question 1 — Solving an equation

What is the value of x?

6x + 11 = 47

A. 4

B. 5

C. 6

D. 8

Answer: C — 6

Subtract 11.

6x = 36

Divide by 6.

x = 6

Question 2 — Distribution

Solve:

2(3x − 4) = 4x + 10

A. 7

B. 8

C. 9

D. 11

Answer: C — 9

Distribute.

6x − 8 = 4x + 10

Subtract 4x.

2x − 8 = 10

Add 8.

2x = 18

Divide by 2.

x = 9

Question 3 — Infinitely many solutions

For what value of k does the equation have infinitely many solutions?

3(4x + 5) = 12x + k

A. 5

B. 12

C. 15

D. 20

Answer: C — 15

Distribute.

12x + 15 = 12x + k

The expressions must be identical.

k = 15

Question 4 — No solution

Which equation has no solution?

A. 2x + 4 = 2x + 4

B. 3x + 7 = 3x + 10

C. 4x − 5 = 11

D. 5x + 2 = 2x + 8

Answer: B

Subtract 3x.

7 = 10

This is impossible, so the equation has no solution.

Question 5 — Real-world application

A membership costs $24 plus $8 per month. The total cost is $88.

How many months were purchased?

A. 6

B. 7

C. 8

D. 11

Answer: C — 8

Create the equation.

24 + 8m = 88

Subtract 24.

8m = 64

Divide by 8.

m = 8

Question 6 — Finding an expression

If:

4(2x + 3) = 44

what is the value of 2x + 3?

A. 8

B. 11

C. 14

D. 17

Answer: B — 11

Divide both sides by 4.

2x + 3 = 11

The requested expression is already isolated.

9. Lesson Summary

A linear equation in one variable contains one unknown variable and can be simplified to a first-degree equation.

To solve an equation, use valid operations that preserve equality and isolate the variable.

The distributive property is important for equations containing parentheses.

Fractions can often be removed by multiplying every term by the least common denominator.

Some linear equations have one solution, no solution, or infinitely many solutions.

SAT questions may involve unknown constants, equations with variables on both sides, and expressions that can be evaluated without solving for x.

Real-world problems require identifying variables, fixed amounts, rates, and totals before constructing an equation.

Always read the question carefully, show logical steps, and verify answers when appropriate.

You are now ready to practice Linear Equations in One Variable in the Baloo Prep question bank.