l,"questionType":"LA","specification":"","difficulty":3,"difficultyBand":null,"totalMarks":4,"sourceTitle":null,"calculatorUse":null,"parts":[{"id":"27cbcf5a-54cb-417f-ae38-9e7de21a0c64","label":"a","content":"Solve $\\frac{1}{x - 1} + \\frac{12}{2x + 3} = 3$","markscheme":"- Multiply by common denominator: $1(2x + 3) + 12(x - 1) = 3(x - 1)(2x + 3)$ M1\n- Expand and simplify: $14x - 9 = 6x^2 + 3x - 9$ M1\n- Rearrange to $6x^2 - 11x = 0$ M1\n- Factorise $x(6x - 11) = 0$ to get $x = 0$ or $x = \\frac{11}{6}$ A1","marks":4,"order":0,"markingRubric":{"totalMarks":4,"points":[{"id":"p1","marks":1,"description":"Multiply the entire equation by the common denominator $(x-1)(2x+3)$ or equate the numerators over a common denominator.","code":"M"},{"id":"p2","marks":1,"description":"Expand brackets and simplify both sides of the equation.","code":"M"},{"id":"p3","marks":1,"description":"Rearrange the equation into a standard quadratic form $ax^2 + bx + c = 0$.","code":"M"},{"id":"p4","marks":1,"description":"Solve the quadratic equation to find both values of $x$.","code":"A"}],"finalAnswers":["x = 0, x = \\frac{11}{6}","0, 1.83"],"generatedAt":"2026-08-10T10:36:33.482Z","modelVersion":"gemini-3-flash-preview"},"explanation":null,"options":[]}],"options":[],"explanation":null,"firstAttemptFullMarks":null},{"id":"63f39f22-9af6-42e5-8505-b7faa2f277a8","revisionId":"ffe7009b-8f80-4817-8c4c-293e569dafbd","questionNumber":null,"questionType":"LA","specification":"","difficulty":3,"difficultyBand":null,"totalMarks":4,"sourceTitle":null,"calculatorUse":null,"parts":[{"id":"bd0090e3-cdd0-4ba8-816e-0caff76644df","label":"a","content":"Solve $\\frac{1}{x - 2} + \\frac{4}{x + 1} = 1$","markscheme":"- Multiply by common denominator: $1(x + 1) + 4(x - 2) = (x - 2)(x + 1)$ M1\n- Expand and simplify to quadratic: $5x - 7 = x^2 - x - 2$ M1\n- Rearrange to $x^2 - 6x + 5 = 0$ M1\n- Solve to get $x = 1$ or $x = 5$ A1","marks":4,"order":0,"markingRubric":{"totalMarks":4,"points":[{"id":"p1","marks":1,"description":"Multiply both sides by the common denominator $(x - 2)(x + 1)$ correctly.","code":"M"},{"id":"p2","marks":1,"description":"Expand brackets and simplify both sides of the equation to reach a quadratic stage.","code":"M"},{"id":"p3","marks":1,"description":"Rearrange the equation into the standard quadratic form $x^2 - 6x + 5 = 0$.","code":"M"},{"id":"p4","marks":1,"description":"Solve the quadratic equation to find $x = 1$ and $x = 5$.","code":"A"}],"finalAnswers":["x = 1, x = 5","x = 5, x = 1"],"generatedAt":"2026-08-10T10:37:13.033Z","modelVersion":"gemini-3-flash-preview"},"explanation":null,"options":[]}],"options":[],"explanation":null,"firstAttemptFullMarks":null},{"id":"65038096-a2f3-4ac6-a5d9-40c46647dce0","revisionId":"999a6d5f-ffaf-427d-9c90-c9f9f5e25ec4","questionNumber":null,"questionType":"LA","specification":"","difficulty":2,"difficultyBand":null,"totalMarks":3,"sourceTitle":null,"calculatorUse":null,"parts":[{"id":"ba13d3d6-4b7e-42e9-929f-411b684cba9f","label":"a","content":"Write $\\frac{x^2 + 5x - 14}{3x^2 - x - 10}$ in the form $\\frac{x + a}{bx + c}$ where $a$, $b$, and $c$ are integers.","markscheme":"- Factorise numerator: $(x + 7)(x - 2)$ M1\n- Factorise denominator: $(3x + 5)(x - 2)$ M1\n- Cancel common factor $(x - 2)$ to get $\\frac{x + 7}{3x + 5}$ A1","marks":3,"order":0,"markingRubric":{"totalMarks":3,"points":[{"id":"p1","marks":1,"description":"Attempt to factorise the quadratic numerator into two linear brackets.","code":"M"},{"id":"p2","marks":1,"description":"Attempt to factorise the quadratic denominator into two linear brackets.","code":"M"},{"id":"p3","marks":1,"description":"Correct simplification by cancelling the common factor $(x - 2)$.","code":"A"}],"finalAnswers":["\\frac{x+7}{3x+5}"],"generatedAt":"2026-08-10T10:37:47.610Z","modelVersion":"gemini-3-flash-preview"},"explanation":null,"options":[]}],"options":[],"explanation":null,"firstA

Algebraic Fractions Questions

Algebraic Fractions Questions

Algebraic Fractions Questions

Algebraic Fractions Questions