Algebra Questions with Worked Solutions
Practice original algebra questions with answers and step-by-step solutions, from linear equations to identities and algebraic fractions.
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1. Linear equation: 3x + 5 = 20
x = 5
- Subtract 5 from both sides: 3x = 15.
- Divide by 3: x = 5.
- Check: 3 × 5 + 5 = 20.
2. Equation with brackets: 4(x − 2) = 20
x = 7
- Divide both sides by 4: x − 2 = 5.
- Add 2 to both sides: x = 7.
- Check: 4(7 − 2) = 20.
3. Variables on both sides: 5x + 2 = 2x + 17
x = 5
- Subtract 2x: 3x + 2 = 17.
- Subtract 2 and divide by 3: x = 5.
- Check: both original sides equal 27.
4. Fraction coefficient: x/3 + 2 = 5
x = 9
- Subtract 2: x/3 = 3.
- Multiply both sides by 3: x = 9.
- Check: 9/3 + 2 = 5.
5. Identity equation: 2(x + 3) = 2x + 6
All real x
- Expand the left side to 2x + 6.
- Subtract the right side: 0 = 0.
- Every real x satisfies the equation.
6. Contradictory equation: 2x + 3 = 2x + 8
No solution
- Subtract 2x from both sides: 3 = 8.
- This is false regardless of x.
- There is no value that satisfies the equation.
7. Combine like terms: 3x + 2x − 7 + 4
5x − 3
- Combine 3x and 2x to get 5x.
- Combine −7 and 4 to get −3.
- The simplified expression is 5x − 3; it is not a solved value of x.
8. Expand brackets: (x + 3)(x − 2)
x² + x − 6
- Multiply each pair of terms: x² − 2x + 3x − 6.
- Combine the x terms: x² + x − 6.
- At x = 4, both forms equal 14.
9. Square of a sum: (x + 5)^2
x² + 10x + 25
- Write the square as (x + 5)(x + 5).
- Distribute: x² + 5x + 5x + 25.
- Combine the middle terms to get 10x.
10. Factor a quadratic: x^2 − 7x + 12
(x − 3)(x − 4)
- Find two numbers with product 12 and sum −7: −3 and −4.
- Use the factors (x − 3)(x − 4).
- Expand to check: x² − 4x − 3x + 12 = x² − 7x + 12.
11. Quadratic roots: x^2 − 5x + 6 = 0
x = 2 or x = 3
- Factor the expression as (x − 2)(x − 3).
- A product is zero when at least one factor is zero.
- Set x − 2 = 0 or x − 3 = 0; both roots pass substitution.
12. Repeated quadratic root: x^2 − 6x + 9 = 0
x = 3
- Factor as (x − 3)² = 0.
- The discriminant is 36 − 36 = 0.
- There is one distinct root, repeated twice: x = 3.
13. Irrational roots: x^2 − 2 = 0
x = ±√2
- Move −2 to the other side: x² = 2.
- Take both square-root signs: x = √2 or x = −√2.
- The decimal approximations are ±1.41421356; the radical form is exact.
14. Complex roots: x^2 + 4 = 0
x = ±2i
- Move 4 to the other side: x² = −4.
- There is no real number whose square is −4.
- Using i² = −1 gives the two complex roots 2i and −2i.
15. Non-unit quadratic coefficient: 2x^2 + 7x + 3
(2x + 1)(x + 3)
- Find factors of 2x² and 3 that yield a middle coefficient of 7.
- Expand (2x + 1)(x + 3) to get 2x² + 6x + x + 3.
- Combine 6x + x to verify 2x² + 7x + 3.
16. Solve a two-equation system: 2x + y = 7 and x − y = 2
x = 3, y = 1
- Add the equations to eliminate y: 3x = 9.
- Divide by 3 to get x = 3.
- Substitute into x − y = 2: 3 − y = 2, so y = 1.
- Check: 2 × 3 + 1 = 7, and 3 − 1 = 2.
17. Simplify (x² − 9)/(x − 3)
x + 3, with x ≠ 3
- The original denominator is zero at x = 3, so exclude that value first.
- Factor x² − 9 as (x − 3)(x + 3).
- Cancel the common factor x − 3. Keep the exclusion x ≠ 3.
18. Is (a + b)² = a² + b² an identity?
No; the middle term 2ab is missing
- Expand the left side: a² + 2ab + b².
- Subtract the proposed right side: the difference is 2ab, which is not always zero.
- At a = b = 1, the left side is 4 and the right side is 2. This counterexample disproves the identity.
19. Evaluate 2x² − 3x + 4 when x = −2
18
- Substitute the signed value with brackets: 2(−2)² − 3(−2) + 4.
- Square first: 2 × 4 + 6 + 4.
- Add the terms: 8 + 6 + 4 = 18.
20. A service costs $12 plus $7 per hour. How many hours cost $47?
5 hours
- Let h be the number of hours and write 12 + 7h = 47.
- Subtract the fixed charge: 7h = 35.
- Divide by the hourly charge: h = 5. Check: 12 + 7 × 5 = 47.
How to check your algebra answer
For an equation, substitute the proposed solution into both original sides. If they have different values, review the step where a term moved, a sign changed or a division occurred. Quadratics can have two roots, one repeated root or complex roots, so avoid assuming every problem has exactly one real answer.
For simplification or factoring, expand or evaluate both forms to see whether the transformation is consistent. A few numerical checks help detect an error, but coefficient comparison is what verifies a polynomial identity for all values. For fractions, a matching simplified value does not remove the original domain restrictions.
Keep exact fractions and radicals while working. A rounded answer can look close without satisfying an equation exactly. Round only when the problem requests an approximation, and mark that result with ≈.
What to practice next
If equations with negative signs are difficult, revisit arithmetic with signed numbers before increasing the number of steps. If bracket expansion is the sticking point, write every product separately and combine like terms afterward. If you forget exclusions in fractions, start each problem by writing which denominator values are forbidden.
The worksheet generator can create focused sets for these topics and mixed practice for review. Use the formula reference to check the conditions of a rule, then test a fresh example yourself.