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Quizzes > Mathematics > Algebra

Ready to Master Algebra? Take Our Free Algebra Quiz

Ready for Easy Algebra Questions? Dive into Simple Algebra Questions with Answers!

Editorial: Review CompletedCreated By: Nazli DasdemirUpdated Aug 27, 2025
Difficulty: Moderate
2-5mins
Learning OutcomesCheat Sheet
Paper art illustration promoting a free algebra quiz on a golden yellow background

Use this algebra quiz to practice equations, solve simple problems, and see instant answers. You'll get quick feedback on each item and an overall score, so you can spot gaps before class and build confidence - then start the free quiz to see your Algebra IQ.

Solve for x: 2x + 5 = 13.
5
3
6
4
To solve 2x + 5 = 13, subtract 5 from both sides to get 2x = 8, then divide by 2 to find x = 4. This isolates the variable and yields the solution. Linear equations follow these basic steps of inverse operations. For more practice on solving linear equations see .
Simplify the expression 3(x + 4).
3x + 4
3x + 12
x + 12
12x
By the distributive property, multiply 3 by each term inside the parentheses: 3·x = 3x and 3·4 = 12, giving 3x + 12. This rule is essential for expanding expressions. For more on distribution see .
What is the value of x if x/4 = 3?
7
12
-12
1
Multiply both sides of x/4 = 3 by 4 to isolate x: x = 3 × 4 = 12. This uses the inverse operation of division, which is multiplication. Such steps are fundamental in solving basic equations. Learn more at .
Combine like terms: 5x + 2x - 3.
7x + 3
3x - 3
7x - 3
5x - 2
Like terms have the same variable raised to the same power. Combine 5x + 2x to get 7x, then include the constant -3. The result is 7x - 3. For more on grouping like terms see .
Evaluate f(2) if f(x) = 3x².
8
10
12
6
To evaluate f(2), substitute x = 2 into 3x²: f(2) = 3·(2)² = 3·4 = 12. Function evaluation follows direct substitution of the input. See more at Math Is Fun on Functions.
Solve for x: -x + 7 = 2.
5
9
-5
-9
Subtract 7 from both sides: -x = 2 - 7 = -5, then multiply by -1 to get x = 5. This uses inverse operations to isolate x. For similar problems see .
What is the coefficient of x in the expression 4x - 9?
4
-9
-4
9
The coefficient of x is the numerical factor multiplying x. In 4x - 9, that factor is 4. Constants like -9 are not coefficients of x. Read more at .
Solve for x: 3x - 2 = x + 6.
-2
4
2
-4
Subtract x from both sides to get 2x - 2 = 6, then add 2 to both sides giving 2x = 8, and divide by 2 to find x = 4. Check the solution by plugging back in. Learn more at .
Factor the expression x² - 9.
(x - 3)(x + 3)
(x + 3)²
(x - 9)(x + 1)
(x - 1)(x + 9)
x² - 9 is a difference of squares: a² - b² = (a - b)(a + b), where a = x and b = 3. Thus it factors to (x - 3)(x + 3). For more examples see .
Simplify the product (2x³)(3x²).
6x?
6x?
5x?
5x?
Multiply coefficients: 2·3 = 6. Add exponents on x: 3 + 2 = 5. The result is 6x?. Laws of exponents are covered in .
Solve for x: 2(x - 1) + 3 = x + 4.
3
1
5
-1
Distribute 2: 2x - 2 + 3 = 2x + 1. Set equal to x + 4: 2x + 1 = x + 4, subtract x: x + 1 = 4, then x = 3. More practice at .
Solve the system: x + y = 5 and x - y = 1.
(4, 1)
(3, 2)
(2, 3)
(1, 4)
Adding the equations gives 2x = 6, so x = 3. Substitute x back into x + y = 5 to get y = 2. This yields (3, 2). See methods at .
Expand the expression (x + 2)².
x² + 2x + 4
x² + 4x + 4
x² + 2x + 2
x² + 2x + 1
Use (a + b)² = a² + 2ab + b² with a = x, b = 2. This gives x² + 2·x·2 + 4 = x² + 4x + 4. Learn more at .
Simplify the expression (x²y³)/(xy).
x²y²
x²y
xy³
xy²
Subtract exponents of like bases: x²/x¹ = x¹, y³/y¹ = y². The simplified result is xy². For exponent rules see .
Solve for x: x² - 5x + 6 = 0.
-1 and -6
-2 and -3
1 and 6
2 and 3
Factor the quadratic: x² - 5x + 6 = (x - 2)(x - 3) = 0. Setting each factor to zero gives x = 2 or x = 3. For more see .
Simplify the rational expression (x + 1)/(x² - 1).
(x + 1)/(x - 1)
1/(x - 1)
1/(x + 1)
x + 1
Factor the denominator: x² - 1 = (x - 1)(x + 1). Cancel the (x + 1) terms to get 1/(x - 1). Remember to exclude x = ±1 from the domain. See more at .
Solve for x: 2x² - 3x - 2 = 0.
2 and 0.5
2 and -0.5
-2 and 0.5
1 and -2
Use the quadratic formula: x = [3 ± ?(9 + 16)]/(4) = [3 ± 5]/4, giving solutions 2 and -0.5. Quadratic formula applies when factoring is not obvious. More at .
Simplify the expression (3x² - 12)/(3x).
x - 4/3
(3x - 12)/3
(x² - 4)/x
x - 4
Factor numerator: 3(x² - 4). Divide by 3x to get (x² - 4)/x. You can further split as x - 4/x but the factored form is standard. Learn more at .
For f(x) = x² - 4x + 3, find f(2).
3
-3
5
-1
Substitute x = 2: 2² - 4·2 + 3 = 4 - 8 + 3 = -1. This direct substitution approach works for any polynomial function. See more at Functions.
Solve for x: |x - 3| = 5.
3 and -3
2 and -8
5 and -5
8 and -2
Absolute value yields two cases: x - 3 = 5 ? x = 8, and x - 3 = -5 ? x = -2. Both satisfy the original equation since distance is positive. More at .
Perform polynomial division: divide x³ - 8 by x - 2.
x² - 2x - 4
x² - 2x + 4
x² + 2x + 4
x² + 4x + 4
Using synthetic or long division with root 2 gives quotient x² + 2x + 4. The remainder is zero since x³ - 8 = (x - 2)(x² + 2x + 4). For division steps see .
Solve for x: 2? = 16.
4
2
8
-4
Since 16 = 2?, set the exponents equal: x = 4. Understanding exponent rules helps solve such equations. For more, see .
Solve for x: x³ - 2x² - x + 2 = 0.
1, -2, 2
1, -1, -3
2, -1, -2
2, 1, -1
Factor by grouping: x²(x - 2) -1(x - 2) = (x - 2)(x² - 1) and further factor x² - 1 = (x - 1)(x + 1). The roots are x = 2, 1, -1. Detailed steps at .
Find the inverse function of f(x) = 3x/(x - 2).
2x/(x + 3)
2x/(x - 3)
3(x - 2)/x
3x/(2 - x)
Set y = 3x/(x - 2), swap x and y, then solve for y: x(y - 3) = 2y ? y = 2x/(x - 3). The inverse function is f?¹(x) = 2x/(x - 3). See the derivation at Inverse Functions.
Solve for x: 1/(x - 1) + 2/(x + 1) = 3x/(x² - 1).
x = 3
No solution
x = 1
x = -1
Combine fractions over common denominator x² - 1: (x + 1) + 2(x - 1) = 3x leads to 3x - 1 = 3x, which is a contradiction (-1 = 0). Also x ? ±1, so no solution exists. More at .
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Study Outcomes

  1. Solve Linear Equations -

    Apply step-by-step methods to solve one-variable linear equations, reinforcing key algebraic principles and boosting confidence in equation solving.

  2. Simplify Algebraic Expressions -

    Combine like terms and use the distributive property to simplify expressions, ensuring a solid grasp of foundational algebraic manipulation.

  3. Analyze Variable Relationships -

    Interpret how variables interact within equations and develop strategies to isolate and solve for unknown values accurately.

  4. Verify Solutions Instantly -

    Use instant feedback on answers to check work, understand common mistakes, and reinforce correct problem-solving approaches in real time.

  5. Apply Algebraic Strategies -

    Transfer learned techniques to a variety of simple algebra questions, enhancing overall problem-solving skills and readiness for more advanced practice.

Cheat Sheet

  1. Understanding Variables and Constants -

    Variables represent unknown values while constants are fixed numbers, for example in x + 5 = 12 you have x as the variable and 5 and 12 as constants. According to Khan Academy, clear identification of these elements is essential when tackling simple algebra questions with answers. Remember "letters for unknowns, numbers for fixed values" as a quick reference during your algebra quiz.

  2. Order of Operations (PEMDAS) -

    The PEMDAS rule (Parentheses, Exponents, Multiplication/Division, Addition/Subtraction) guides you through complex expressions step by step. The University of Texas at Austin highlights the mnemonic "Please Excuse My Dear Aunt Sally" to lock in the sequence for solving easy algebra questions. Applying PEMDAS consistently prevents errors in both algebra practise tests and quizzes.

  3. Applying the Distributive Property -

    The distributive property a(b + c) = ab + ac simplifies expressions before solving, as detailed by MIT OpenCourseWare. For instance, 3(x + 4) becomes 3x + 12, making subsequent steps in algebra questions with answers much clearer. Mastering this property speeds you through algebra quizzes with confidence.

  4. Solving One- and Two-Step Equations -

    Begin with inverse operations to isolate variables, subtracting or dividing both sides as needed; for example, to solve 2x + 3 = 11, subtract 3 then divide by 2. The University of Cambridge emphasizes practicing a variety of these problems so you'll breeze through simple algebra questions with answers on your next test. Consistent practice in algebra quizzes builds speed and accuracy.

  5. Factoring Basics and Solution Verification -

    Factoring common terms (e.g., 6x + 9 = 3(2x + 3)) is key for simplifying and solving quadratic or linear expressions, according to the National Council of Teachers of Mathematics. After finding a solution, always substitute it back into the original equation to verify accuracy - a step often overlooked in algebra practise tests. This double-check habit ensures you ace easy algebra questions every time.

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