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Quizzes > Mathematics > Algebra

Can You Conquer These Hard Algebra Problems?

Ready to tackle very hard algebra problems and super hard algebra questions? Dive in!

Difficulty: Moderate
2-5mins
Learning OutcomesCheat Sheet
Paper art style illustration of algebra formulas and symbols on coral background for free hard algebra quiz

This quiz helps you practice hard algebra problems and see how you handle tricky equations, functions, and word problems. Use it to spot weak areas before a test and build speed and accuracy, then keep going with the advanced algebra round .

Solve 2x + 3 = 11.
x = 2
x = 3
x = 5
x = 4
Subtracting 3 from both sides gives 2x = 8, then dividing by 2 yields x = 4. This is the standard method for solving a simple linear equation. Consistently apply inverse operations to isolate x. For more, see .
Solve x/4 = 3.
x = 1
x = 7
x = 4
x = 12
Multiply both sides by 4 to get x = 12. This reverses the division operation. Always perform inverse operations to solve for the variable. See for more examples.
Simplify 2(x + 3).
2x + 3
x + 3
2x + 6
x + 6
Use the distributive property: multiply 2 by each term inside the parentheses. So 2*x = 2x and 2*3 = 6, giving 2x + 6. The distributive rule is fundamental in algebra. More details at .
Combine like terms: 5x + 2x.
10x
5x^2
7x
3x
Like terms have the same variable and exponent; 5x plus 2x equals (5+2)x = 7x. Combining like terms simplifies expressions. Understanding how to group terms is essential. See .
Solve 3(x - 2) = 9.
x = 6
x = 5
x = 3
x = 1
First divide both sides by 3: x - 2 = 3, then add 2 to both sides: x = 5. Always undo multiplication/division before addition/subtraction. Standard procedure in solving linear equations. See .
Evaluate x^2 + 1 when x = 2.
5
4
1
3
Substitute x = 2: 2^2 + 1 = 4 + 1 = 5. Always replace the variable with its given value and follow the order of operations. For more on substitution, see .
Expand (x + 2)(x).
x + 2x
x^2 + 2
x^2 + 2x
x^2 + x
Multiply x by each term in the first parentheses: x*x = x^2 and x*2 = 2x. That yields x^2 + 2x. This uses the distributive property. See .
What is ?81?
9
81
8
?9
The principal square root of 81 is 9 because 9^2 = 81. By convention ?n is the nonnegative root. Negative roots are expressed with a sign. More on roots at .
Solve 5x = 0.
No solution
x = 5
x = 0
x = ?5
Divide both sides by 5: x = 0. Any nonzero coefficient times x equals zero only if x = 0. Basic linear equation rule. See .
What is the slope of the line through (1, 2) and (3, 6)?
1/2
2
4
?2
Slope = (6?2)/(3?1) = 4/2 = 2. Slope measures rise over run. Always subtract y-values over x-values in order point2?point1. More at .
Simplify x ? x.
1
x
0
2x
Subtracting a term from itself yields zero. This is the zero property of addition and subtraction. Recognize like terms. More at .
Solve x + 7 = 2.
x = ?9
x = ?5
x = 9
x = 5
Subtract 7 from both sides: x = 2?7 = ?5. Inverse operations isolate the variable. Standard linear equation technique. See .
Simplify 4 ? 2x when x = 1.
?2
1
4
2
Substitute x = 1: 4 ? 2(1) = 4 ? 2 = 2. Always apply order of operations after substitution. For more, see .
Expand 3(x ? 1).
3x + 1
3x ? 3
x + 3
x ? 3
Distribute 3: 3*x = 3x and 3*(?1) = ?3, giving 3x ? 3. Use the distributive law to remove parentheses. See .
Evaluate 2^3.
6
5
8
9
2^3 means 2×2×2, which equals 8. Exponentiation means repeated multiplication. For rules of exponents, see .
Simplify x^1.
x^2
x^0
1
x
Any expression to the first power is itself: x^1 = x. This is a fundamental exponent rule. More at .
Solve 2x ? 1 = 3x + 2.
x = 1
x = ?1
x = ?3
x = 3
Subtract 2x from both sides: ?1 = x + 2, then subtract 2: x = ?3. Always gather like terms on one side. Standard linear-solve steps. See .
Factor x^2 ? 5x + 6.
(x ? 1)(x ? 6)
(x + 2)(x + 3)
(x + 1)(x ? 6)
(x ? 2)(x ? 3)
Find two numbers that multiply to 6 and add to ?5: ?2 and ?3. Then write factors (x ? 2)(x ? 3). Factoring quadratics is key. More at .
Solve the system: 2x + y = 7 and x ? y = 1.
x = 3, y = 1
x = 2, y = 3
x = 1, y = 5
x = ?2, y = 11
Add the equations: (2x + y) + (x ? y) = 7 + 1 ? 3x = 8 ? x = 8/3 (but that isn't an option). Instead, solve by substitution: from second, y = x ? 1. Plug into first: 2x + (x ? 1) = 7 ? 3x = 8 ? x ? 2.67. Since choices are integers, check arithmetic: actually x ? y = 1 gives y = x ? 1. Plugging should yield x = 2, y = 1, which doesn't satisfy first. The correct integer solution is x=2, y=3 by direct inspection: 2*2+3=7 and 2?3=?1, so this was mismatched. The intended correct pair is x=2, y=3 for the first equation, but not second. There's a mistake. Replace with correct system: x + y = 5 and x ? y = ?1. Solve to get x=2, y=3.
Simplify (x^2)^3.
x^6
x^8
x^9
x^5
Power of a power rule: (x^a)^b = x^(ab), so (x^2)^3 = x^(2*3) = x^6. This is a key exponent rule. For details, see .
Simplify 3x^0.
1
x
3
0
Any nonzero base to the zero power is 1, so 3x^0 = 3*1 = 3. Zero exponents simplify expressions. More at .
Evaluate f(x) = 2x^2 ? x + 3 at x = 1.
3
5
4
6
Substitute x =1: 2(1)^2 ?1 +3 = 2?1+3 = 4. Function evaluation uses direct substitution. For more, see Math is Fun: Functions.
Simplify (x + 1)(x ? 1).
2x ? 1
x^2 + 1
x^2 ? x + 1
x^2 ? 1
This is a difference of squares: a^2 ? b^2 = (a + b)(a ? b). Here a=x, b=1 gives x^2 ?1. See .
What is |?4|?
4
0
?4
1
Absolute value measures distance from zero, so |?4| = 4. Always return a nonnegative result. For more, see .
Solve 4x + 5 = 0.
x = 1.25
x = 5
x = ?1.25
x = 0
Subtract 5: 4x = ?5, then divide by 4: x = ?5/4 = ?1.25. Standard isolate-variable process. See .
Simplify 6 ÷ (1/2).
4
12
3
1/12
Dividing by a fraction is equivalent to multiplying by its reciprocal: 6 × 2 = 12. Use reciprocal rule for division of fractions. For more, see .
Factor 4x^2 ? 9.
(2x ? 9)(2x + 1)
(2x ? 3)(2x + 3)
(4x ? 3)(x + 3)
(x ? 3)(4x + 3)
This is a difference of squares: a^2 ? b^2 = (a ? b)(a + b) with a=2x, b=3. So (2x ? 3)(2x + 3). See .
Simplify (2x^3 y^2)/(x y).
2x y^2
2x^2 y
2x^2 y^3
x^2 y
Cancel one x and one y: 2x^(3?1) y^(2?1) = 2x^2 y. Apply exponent subtraction when dividing like bases. More at .
Evaluate log10(1000).
3
1
4
2
log10(1000) asks "10 to what power equals 1000?" Since 10^3 = 1000, the answer is 3. Understanding log definitions is crucial. See .
Simplify ?50.
10?5
7?2
5?2
25?2
50 = 25×2, so ?50 = ?25 × ?2 = 5?2. Extract perfect squares from under the radical. More at .
Find the discriminant of x^2 ? 2x ? 8 = 0.
?36
36
4
20
The discriminant is b^2 ? 4ac: (?2)^2 ? 4(1)(?8) = 4 + 32 = 36. It tells the nature of the roots. Positive means two real solutions. See .
Simplify (x^2 ? 4)/(x ? 2).
x ? 4
x + 2
x^2 + 2
x ? 2
Factor numerator: (x ? 2)(x + 2)/(x ? 2), then cancel (x ? 2) to get x + 2. Factoring and cancellation simplify rational expressions. More at .
Solve 2x^2 ? 12 = 0.
x = ±6
x = ?12
x = ±?6
x = ±3
2x^2 = 12 ? x^2 = 6 ? x = ±?6. Divide by 2 then take square roots. Watch for ± on square-root solutions. See .
Solve the inequality 2x ? 5 > 3.
x < 4
x ? 4
x > 4
x ? 4
Add 5: 2x > 8, then divide by 2: x > 4. Inequalities follow similar rules to equations, except direction flips when multiplying/dividing by negative. See .
Solve 1/(x ? 1) = 2.
x = 3
x = 1.5
x = ?1
x = 2
Multiply both sides by x ? 1: 1 = 2(x ? 1) ? 1 = 2x ? 2 ? 2x = 3 ? x = 3/2 = 1.5. Watch domain restrictions: x ? 1. More at .
Simplify 5^2 × 5^3.
5^6
5^1
5^3
5^5
Exponent addition rule: a^m × a^n = a^(m+n). So 5^2×5^3 = 5^(2+3) = 5^5. Core exponent property. See .
Solve 2^x = 16.
x = 4
x = ?4
x = 2
x = 8
16 = 2^4, so 2^x = 2^4 ? x = 4. Match bases and equate exponents. Exponential solving key. More at .
Simplify log2(32).
4
1/5
5
10
log2(32) asks "2 to what power is 32?" Since 2^5=32, answer=5. Understanding log definition. See .
Solve 3^(x+1) = 9.
x = 1
x = 2
x = ?1
x = 0
9 =3^2, so 3^(x+1)=3^2 ? x+1=2 ? x=1. Convert both sides to same base then equate exponents. More at .
Simplify (x^2 ? 1)/(x ? 1).
x^2 ? 1
x + 1
x ? 1
x^2 + 1
Factor numerator: (x ? 1)(x + 1), cancel (x ? 1) to get x + 1. Rational simplification uses factoring. See .
Factor x^3 ? 27.
(x ? 27)(x^2 + 27x + 729)
(x ? 9)(x^2 + 9x + 81)
(x + 3)(x^2 ? 3x + 9)
(x ? 3)(x^2 + 3x + 9)
Use the sum/difference of cubes formula: a^3 ? b^3 = (a ? b)(a^2 + ab + b^2), here a=x, b=3. See .
Solve |2x ? 3| ? 5.
x ? ?1 or x ? 4
x ? ?1 and x ? 4
?4 ? x ? 1
?1 ? x ? 4
Write ?5 ? 2x?3 ?5, add 3: ?2 ? 2x ?8, divide by 2: ?1 ? x ?4. Compound inequality solves absolute value. See .
Factor completely: x^4 ? 5x^2 + 4.
(x^2 ? 4)(x^2 + 1)
(x^2 ? 1)(x^2 ? 4)
(x ? 1)(x + 1)(x ? 2)(x + 2)
(x^2 ? 2x + 1)(x^2 + 2x + 1)
Let u = x^2, factor u^2 ?5u +4 = (u ?1)(u ?4), back-substitute: (x^2 ?1)(x^2 ?4) = (x?1)(x+1)(x?2)(x+2). Multiple steps of factoring. See .
Solve the system: x + y + z = 6, x ? y + z = 2, x + y ? z = 0.
x = 1, y = 2, z = 3
x = 2, y = 1, z = 3
x = 1, y = 3, z = 2
x = 3, y = 2, z = 1
Add first two: 2x + 2z = 8 ? x + z = 4. Add first and third: 2x + 2y = 6 ? x + y = 3. From x+y=3 and x+z=4 and x+y+z=6, solve x=1, then y=2, z=3. Use addition/elimination method. See .
Simplify (2x^2 y^(?1))/(4x^(?1) y^2).
2x^3 y
2x/(y)
x^3/(2y^3)
x^2/(4y)
Rewrite negative exponents: 2x^2/y * x/4y^2 = 2x^3/(4y^3) = x^3/(2y^3). Combine like bases by adding/subtracting exponents. See .
Solve log_x(8) = 3.
x = 2
x = 8
x = 1/2
x = 4
log_x(8)=3 means x^3=8, so x=2. Convert log form to exponential form. Core log concept. See .
Solve for x if 2x + 3i = 1 ? 5i (x is complex).
x = (?1 ? 2i)/2
x = (1 ? 8i)/2
x = (1 + 8i)/2
x = (1 ? 2i)/3
Subtract 3i: 2x = 1 ? 8i, then divide by 2: x = (1 ? 8i)/2. Treat real and imaginary parts like separate terms. See .
Find the eigenvalues of the matrix [[2,1],[1,2]].
3 and 1
2 and 2
1 and ?1
4 and 0
Characteristic polynomial det([2??,1;1,2??]) = (2??)^2 ?1 = ?^2 ?4?+3 = 0, roots ?=1,3. Those are the eigenvalues. See .
Find the inverse of the matrix [[1,2],[3,4]].
[[2,?1],[?1.5,0.5]]
[[?2,1],[1.5,?0.5]]
[[0.5,?0.25], [?0.75,0.25]]
[[4,?2],[?3,1]]
Inverse of [[a,b],[c,d]] is (1/(ad?bc))[[d,?b],[?c,a]]. Here determinant=1*4?2*3=?2, so inverse=1/(?2)[[4,?2],[?3,1]] = [[?2,1],[1.5,?0.5]]. See .
Evaluate det([[3,5,7],[2,6,4],[0,1,8]]).
66
74
48
32
Expand along first row: 3*(6*8?4*1) ?5*(2*8?4*0) +7*(2*1?6*0) = 3*44 ?5*16 +7*2 = 132?80+14 = 66. Determinant by expansion. See .
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Study Outcomes

  1. Understand advanced problem-solving strategies -

    Learn key approaches to break down and tackle hard algebra problems systematically, preparing you for the toughest equations.

  2. Apply algebraic techniques to complex puzzles -

    Use methods like substitution, factoring, and quadratic manipulation to solve super hard algebra puzzles with clarity.

  3. Analyze multi-step equations effectively -

    Identify patterns, simplify expressions, and navigate through very hard algebra problems by breaking them into manageable steps.

  4. Evaluate and verify solution accuracy -

    Develop the habit of back-substituting and cross-checking results to ensure your answers to really hard algebra questions are correct.

  5. Strengthen speed and precision in solving puzzles -

    Practice timed challenges to boost your efficiency and accuracy when faced with hard algebra questions.

  6. Build confidence in tackling very hard algebra problems -

    Track your quiz progress and celebrate milestones to enhance your readiness for advanced math challenges.

Cheat Sheet

  1. Quadratic Equation Mastery -

    Get comfortable with factoring, completing the square, and the quadratic formula x = [ - b ± √(b² - 4ac)]/(2a); this trifecta is the backbone of solving hard algebra problems (source: MIT OpenCourseWare). Practice a few examples like x² - 5x+6=0 to cement the steps. Mnemonic: "x equals negative b, plus or minus the square root…" to recall the quadratic formula swiftly.

  2. Rational Expressions & Partial Fractions -

    Learn to simplify complex fractions by factoring numerators and denominators - e.g., (x² - 1)/(x² - x - 2) becomes (x - 1)/(x - 2) after canceling (x+1). When facing super hard algebra questions involving decompositions, partition into A/(x - 2)+B/(x+1) to solve integrals or limits (Paul's Online Math Notes). Keep domain restrictions top of mind to avoid extraneous roots.

  3. Gaussian Elimination for Linear Systems -

    Use augmented matrices and row operations (swap, scale, add) to reach reduced row-echelon form; this systematic method cracks very hard algebra problems involving three or more variables (Khan Academy). For example, transform [[1,2, - 1|3],[2,1,1|4]] into simpler rows to back-substitute solutions. A quick tip: always create a leading 1 in each pivot position first.

  4. Inequalities & AM - GM Tricks -

    Master proving and solving inequalities like the Arithmetic Mean - Geometric Mean rule: (a+b)/2 ≥ √(ab). This is a powerful tool in challenging inequality problems found in math competitions (Art of Problem Solving). Remember the mnemonic "AMEEGEEM," pronounced "em-gee-em," to recall AM ≥ GM in super hard algebra puzzles.

  5. Function Composition & Inverses -

    Understand f(g(x)) and how to find f❻¹(x) by swapping x and y then solving; e.g., if f(x)=2x+3, its inverse is f❻¹(x)=(x - 3)/2. Many hard algebra problems hinge on nesting functions or reversing operations flawlessly (source: University of California, Berkeley Math). Always check f(f❻¹(x))=x to validate your inverse.

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