one variable linear equations

Linear equations are the heartbeat of algebra. This section focuses on one variable linear equations, which are the most common type you'll encounter on the SAT. Mastering these will not only help you solve equations quickly but also prepare you for more complex problems involving inequalities, systems of equations, and basically everything else, which we'll cover later.

variables and equations

A variable is a symbol (like $x$ or $n$ ) that stands for an unknown value. A linear equation is any equation where the variable’s highest power is 1—graphically, it forms a straight line. The goal is to rearrange the equation until you isolate the variable, checking for extraneous or no-solution scenarios as you go.

For example:

7x - 3 = 2x + 12

Step 1: Start with the original equation. The variable is present on both sides.

7x - 3 - 2x = 2x + 12 - 2x

Step 2: Subtract $2x$ from both sides to bring all the $x$ terms to the left.

5x - 3 = 12

Step 3: Combine like terms: $7x - 2x$ becomes $5x$ .

5x - 3 + 3 = 12 + 3

Step 4: Add 3 to both sides to move the constant away from the variable.

5x = 15

Step 5: Simplify both sides. Now you have the variable isolated with its coefficient.

x = \frac{15}{5} = 3

Step 6: Divide both sides by 5 to solve for $x$ .

Check: Substitute $x = 3$ back into the original equation to verify the solution.

the core moves: balance, undo, distribute, clear fractions

When solving equations, your goal is always to “undo” the operations surrounding the variable—while keeping the equation balanced. You’ll often use the distributive property to expand parentheses, and sometimes need to clear fractions so the arithmetic is easier. These core moves show up in almost every algebra problem.

Balance Principle: Whatever you do to one side, you must do to the other. This keeps the equation “fair.”

Undoing: Isolate $x$ by reversing the order of operations: undo addition/subtraction first, then multiplication/division.

Distribution: Use the distributive property to expand, for example: $a(b + c) = ab + ac$ . This helps to make it easier to solve for the variable.

Clearing Fractions: Multiply both sides by the Least Common Denominator (LCD) to clear denominators, especially in equations with messy fractions.

For example:

\frac{2x-1}{3} + \frac{x+2}{4} = 5

Step 1: Start with the original equation. Notice the fractions.

\text{Multiply both sides by 12:}

Step 2: 12 is the LCD (Least Common Denominator) of 3 and 4. Multiplying both sides by 12 will clear all denominators.

4(2x-1) + 3(x+2) = 60

Step 3: Each fraction is multiplied: $12 \div 3 = 4$ and $12 \div 4 = 3$ .

8x - 4 + 3x + 6 = 60

Step 4: Distribute to remove parentheses.

11x + 2 = 60

Step 5: Combine like terms.

11x = 58

Step 6: Subtract 2 from both sides to isolate the term with $x$ .

x = \frac{58}{11}

Step 7: Divide both sides by 11 to solve for $x$ .

Why use the LCD? Multiplying both sides by 12 clears all denominators at once, turning the equation into a simpler one with no fractions.

beyond numbers: parameters, literal equations, contradictions

Algebra isn’t just about finding numbers. Often, you work with equations involving several variables and parameters. You’ll need to solve for one variable in terms of others, determine when equations have unique, no, or infinite solutions, and recognize what happens when variables “disappear.” These skills show up frequently on standardized tests and in higher math.

Literal equations: These have more than one variable. You’ll often be asked to solve for one variable in terms of the others.

For example: Solve $A = prt$ for $t$ .

A = prt

Step 1: Start with the formula for simple interest, where $A$ is the amount, $p$ is principal, $r$ is rate, and $t$ is time.

\frac{A}{pr} = t

Step 2: Divide both sides by $pr$ to isolate $t$ . Now you have $t = \frac{A}{pr}$ .

Parameters: An equation can include a constant that’s a symbol (like $k$ or $a$ ) instead of a number. Sometimes you’re asked, “For what value(s) of $k$ does the equation have one, none, or infinite solutions?”

Contradiction: If you simplify an equation and all variables disappear, leaving a false statement (such as $4 = 7$ ), then there is no solution.

Identity: If all variables disappear and you’re left with a true statement (like $0 = 0$ ), there are infinitely many solutions.

For example:

For which value of $k$ does $2x + k = 2x + 7$ have no solution or infinite solutions?

2x + k = 2x + 7

Step 1: Start with the given equation. Both sides have $2x$ .

2x + k - 2x = 2x + 7 - 2x

Step 2: Subtract $2x$ from both sides to get all constants together.

k = 7

Step 3: Now you’re left with a statement about $k$ only.

\text{If } k \neq 7, \ 7 = k \text{ is never true } \to \text{no solution}

\text{If } k = 7, \ k = 7 \text{ is always true } \to \text{infinitely many solutions}

Once the variable terms cancel, the equation will only be true if the constants are equal. If not, it’s impossible!

modeling: real-life word problems

SAT word problems test your ability to translate real-world situations into mathematical equations. Success comes from clear variable definitions, careful reading, and always making sure you actually answer the question being asked!

Assign variables clearly (Let $x$ be...)
Translate sentences into equations—pay close attention to words like “less than”, “more than”, “is”, and “per”.
Always answer the actual question, not just $x$ !

For example:

Julia buys 5 notebooks and 3 pens. Each notebook costs $2.50 and each pen costs $1.20. If she pays with a $20 bill, how much change does she get?

\text{Let } x = \text{change Julia receives (in dollars)}

Step 1: Assign the variable. We want to find the amount of change Julia receives.

\text{Total cost} = 5 \times 2.50 + 3 \times 1.20

Step 2: Write an expression for the total cost of the notebooks and pens.

= 12.50 + 3.60 = 16.10

Step 3: Calculate the total cost.

x = 20 - 16.10

Step 4: Subtract the total cost from the amount paid to find the change.

x = 3.90

Answer: Julia receives $3.90 in change.

Always define your variable, carefully model each part of the situation, and double-check you’re answering what the question asks!

Let's practice with a few College Board questions!

Question 1

Hector used a tool called an auger to remove corn from a storage bin at a constant rate. The bin contained 24,000 bushels of corn when Hector began to use the auger. After 5 hours of using the auger, 19,350 bushels of corn remained in the bin. If the auger continues to remove corn at this rate, what is the total number of hours Hector will have been using the auger when 12,840 bushels of corn remain in the bin?

Question 2

What value of $p$ satisfies the equation $5p + 180 = 250$ ?

Question 4

medium SAT math

John paid a total of $165 for a microscope by making a down payment of $37 plus $p$ monthly payments of $16 each. Which of the following equations represents this situation?

✓

linear equations: expert mode unlocked

You now understand not only how to solve any linear equation, but also how to recognize identities, contradictions, word problems, and parameter-based scenarios—just like a top SAT scorer.

Next Up

22: linear functions

Learn how to interpret and analyze linear functions.

one variable linear equations

variables and equations

the core moves: balance, undo, distribute, clear fractions

beyond numbers: parameters, literal equations, contradictions

modeling: real-life word problems

Question 1

Question 2

What value of ppp satisfies the equation 5p+180=2505p + 180 = 2505p+180=250?

Question 4

John paid a total of $165 for a microscope by making a down payment of $37 plus ppp monthly payments of $16 each. Which of the following equations represents this situation?

linear equations: expert mode unlocked

22: linear functions

What value of $p$ satisfies the equation $5p + 180 = 250$ ?

John paid a total of $165 for a microscope by making a down payment of $37 plus $p$ monthly payments of $16 each. Which of the following equations represents this situation?