SAT Linear Equations in One Variable: Practice Questions & Study Guide
Solving equations with a single unknown, from simple one-step problems to multi-step equations with fractions and parentheses.
Understanding Linear Equations in One Variable
A linear equation in one variable is any equation that can be written in the form ax + b = c, where x is the unknown and a, b, c are constants with a ≠ 0. The goal is to isolate x on one side by performing the same operation to both sides—a principle called the balance property of equality. On the test, these equations range from single-step problems (x + 7 = 15) to multi-step problems involving distribution, combining like terms, and fractions.
Use an ordered process: distribute first, combine like terms on each side, then move variable terms to one side and constants to the other. Writing each transformation on its own line makes a sign or arithmetic error easier to locate and correct.
Some linear equation questions are word problems where you must first write the equation, then solve it. These test whether you can translate phrases like 'three more than twice a number' into 2x + 3 or 'the total cost of n items at $4 each plus a $10 fee' into 4n + 10. Label exactly what your variable represents before writing the equation.
The test also includes equations that have no solution (a contradiction like 3 = 7) or infinitely many solutions (an identity like 2x + 4 = 2(x + 2)). When the variable drops out, inspect the remaining statement: a false statement means no solution, while a true identity means infinitely many solutions.
Key Rules & Formulas
Memorize these rules — they come up directly in practice questions.
Whatever you do to one side of an equation, do to the other side.
If 3x = 12, divide both sides by 3 to get x = 4.
Distribute before combining like terms.
2(x + 5) = 14 becomes 2x + 10 = 14, then 2x = 4, so x = 2.
Move all variable terms to one side, all constants to the other.
5x - 3 = 2x + 9 → 3x = 12 → x = 4.
Multiply through by the LCD to eliminate fractions before solving.
x/3 + 2 = 5: multiply by 3 to get x + 6 = 15, so x = 9.
If the variable cancels and the result is false (e.g., 3 = 7), the equation has no solution.
3x + 5 = 3x + 9 simplifies to 5 = 9, which is false → no solution.
If the variable cancels and the result is true (e.g., 4 = 4), the equation has infinitely many solutions.
2(x + 3) = 2x + 6 simplifies to 6 = 6 → infinitely many solutions.
Linear Equations in One Variable Practice Questions
Select an answer and click Check Answer to reveal the full explanation. Questions go from easiest to hardest.
If 4x - 7 = 17, what is the value of x?
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Correct answer: C. x = 6
Explanation
Add 7 to both sides: 4x = 24. Divide both sides by 4: x = 6. Checking: 4(6) - 7 = 24 - 7 = 17. ✓
Solving 3(x + 4) = 27 gives what value of x?
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Correct answer: D. x = 5
Explanation
Divide both sides by 3: x + 4 = 9. Subtract 4: x = 5. Alternatively, distribute: 3x + 12 = 27, so 3x = 15, x = 5.
A store sells notebooks for $3 each and pens for $1 each. Maya buys only notebooks and spends exactly $18. How many notebooks did she buy?
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Correct answer: A. 6
Explanation
Let n = number of notebooks. The equation is 3n = 18. Dividing both sides by 3 gives n = 6.
If (2x + 1)/3 = x - 1, what is the value of x?
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Correct answer: B. x = 4
Explanation
Multiply both sides by 3: 2x + 1 = 3(x - 1) = 3x - 3. Subtract 2x: 1 = x - 3. Add 3: x = 4. Check: (2(4)+1)/3 = 9/3 = 3 and 4 - 1 = 3 ✓.
If 5x + 8 = 3x - 4, what is the value of 2x + 1?
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Correct answer: C. -11
Explanation
Subtract 3x from both sides: 2x + 8 = -4. Subtract 8: 2x = -12, so x = -6. Then 2x + 1 = 2(-6) + 1 = -12 + 1 = -11.
For what value of x does 4(x - 2) - 3(x + 1) = 9?
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Correct answer: D. x = 20
Explanation
Distribute: 4x - 8 - 3x - 3 = 9. Combine like terms: x - 11 = 9. Add 11: x = 20. Check: 4(20-2) - 3(20+1) = 4(18) - 3(21) = 72 - 63 = 9 ✓.
A plumber charges a flat fee of $45 plus $60 per hour. A customer's total bill is $225. How many hours did the plumber work?
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Correct answer: A. 3
Explanation
Let h = hours worked. The equation is 45 + 60h = 225. Subtract 45: 60h = 180. Divide by 60: h = 3.
If ax + 6 = 3x + b has infinitely many solutions, which of the following must be true?
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Correct answer: B. a = 3 and b = 6
Explanation
For infinitely many solutions, the equation must be an identity (always true). This requires the coefficients of x to match (a = 3) and the constants to match (6 = b, so b = 6). Any other combination gives one solution or no solution.
The equation 3(2x + k) = 6x + 15 has infinitely many solutions. What is the value of k?
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Correct answer: C. k = 5
Explanation
Distribute: 6x + 3k = 6x + 15. Subtract 6x: 3k = 15, so k = 5. With k = 5, the equation becomes 6x + 15 = 6x + 15, which is true for all x—confirming infinitely many solutions.
If (3x + 2)/4 - (x - 1)/2 = 3, what is the value of x?
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Correct answer: A. x = 8
Explanation
Multiply every term by 4 (LCD): (3x + 2) - 2(x - 1) = 12. Distribute: 3x + 2 - 2x + 2 = 12. Combine: x + 4 = 12, so x = 8. Check: (3(8)+2)/4 - (8-1)/2 = 26/4 - 7/2 = 6.5 - 3.5 = 3 ✓.
Want more Linear Equations in One Variable practice?
Open this skill in the question bank, then filter the available practice by difficulty and score band.
Common Mistakes to Avoid
Use this checklist to catch common traps while working through Linear Equations in One Variable questions, then add your own patterns as you review missed answers.
- !Forgetting to distribute a negative sign: -(x - 4) should become -x + 4, not -x - 4.
- !Adding to one side but not the other when moving terms across the equals sign.
- !Dividing by a coefficient incorrectly when the coefficient is a fraction: if (2/3)x = 8, multiply by 3/2, not divide by 3.
- !Misidentifying 'no solution' versus 'infinitely many solutions' after the variable cancels.
- !Solving for x but reporting the wrong quantity: the problem may ask for 2x or x + 5, not x itself.
Strategy Tips: Linear Equations in One Variable
Always label your variable explicitly in word problems so you stay focused on what the question is actually asking for.
Check your answer by substituting it back into the original equation—this catches arithmetic errors in seconds.
For equations with large or messy fractions, multiply both sides by the LCD at the very first step to work with integers throughout.
When the question asks for 'no solution' or 'infinitely many solutions', set up the equation and look for when the variable terms cancel, then analyze the constant that remains.
Other Algebra Subtopics
Linear Equations in Two Variables
Interpreting and writing equations relating two quantities, including converting between equation forms and identifying key features like slope and intercepts.
Linear Functions
Understanding linear functions as rules mapping inputs to outputs, evaluating them, and interpreting function notation in real-world models.
Systems of Two Linear Equations
Finding the intersection of two lines by substitution or elimination, and analyzing when systems have no solution or infinitely many solutions.
Linear Inequalities in One or Two Variables
Solving and graphing inequalities, interpreting solution sets, and modeling real-world constraints with inequality expressions.
Master Linear Equations in One Variable
Continue with this skill in the question bank, or test how it holds up under section timing before returning to the surrounding Algebra skills.