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bg-background","children":["$","$L3a",null,{"backHref":"/a-level/maths/edexcel/questions","basePath":"/a-level/maths/edexcel/a-level-pure-mathematics-1-3-partial-fractions/questions","formulaUrl":"https://assets.mathsgenie.co.uk/board-resources/a-level-maths/edexcel/formula-book.pdf","initialIndex":2,"initialQuestionWindow":[{"id":"9d3c6e50-5295-4a06-b680-b33bd067334d","revisionId":"d8602a4c-25a3-4cfb-a708-8dfab36ff975","questionNumber":null,"questionType":"LA","specification":"","difficulty":1,"difficultyBand":"medium","totalMarks":3,"sourceTitle":null,"calculatorUse":null,"isCalculation":true,"parts":[{"id":"a66291cd-9c55-4609-bdd8-3fe0f905772d","label":"a","content":"Express $\\frac{x-1}{(x-2)(x-3)}$ in the form $\\frac{A}{(x-2)} + \\frac{B}{(x-3)}$","markscheme":"- Sets up the identity $x-1 = A(x-3) + B(x-2)$ M1\n- Finds $A = -1$ and $B = 2$ by substituting $x = 2$ and $x = 3$ (or by equating coefficients) A1\n- Final answer: $\\frac{-1}{(x-2)} + \\frac{2}{(x-3)}$ A1","marks":3,"order":0,"markingRubric":{"totalMarks":3,"points":[{"id":"p1","marks":1,"description":"Sets up a correct identity, e.g., $x-1 = A(x-3) + B(x-2)$.","code":"M"},{"id":"p2","marks":1,"description":"Determines the correct values for $A$ and $B$, typically by substitution or equating coefficients.","code":"A"},{"id":"p3","marks":1,"description":"Provides the final expression in the requested form using the correct constants.","code":"A"}],"finalAnswers":["\\frac{-1}{(x-2)} + \\frac{2}{(x-3)}","-\\frac{1}{(x-2)} + \\frac{2}{(x-3)}"],"generatedAt":"2026-05-06T14:03:29.437Z","modelVersion":"gemini-3-flash-preview"},"tableImages":[],"explanation":{"methods":[{"id":"substitution","label":"Substitution Method","steps":[{"type":"text","order":0,"title":"Set up the identity","content":"To express the algebraic fraction as partial fractions, we start by setting up an identity. We want to find constants $A$ and $B$ such that:\n\n$$\\frac{x-1}{(x-2)(x-3)} \\equiv \\frac{A}{x-2} + \\frac{B}{x-3}$$\n\nMultiply both sides of the equation by the common denominator $(x-2)(x-3)$ to clear the fractions. This gives us the key identity:\n\n$$x-1 \\equiv A(x-3) + B(x-2)$$\n\n*Awarded 1 mark (M1) for setting up this identity.*"},{"type":"hyper_latex","block":{"nodes":[{"latex":"$$$x-1 \\equiv A(x-3) + B(x-2)$$"},{"text":"Substitute the root of the second bracket, "},{"latex":"$$\\htmlData{anchor=x-val-2}{\\color{@sky}{x = 2}}$"},{"text":", into the identity to eliminate the $B$ term:"},{"latex":"$$$\\htmlData{anchor=sub-x-2}{\\color{@sky}{2}} - 1 = A(\\htmlData{anchor=sub-x-2-A}{\\color{@sky}{2}} - 3) + B(\\htmlData{anchor=sub-x-2-B}{\\color{@sky}{2}} - 2)$$"},{"latex":"$$$1 = A(-1) + B(0)$$"},{"latex":"$$$\\htmlData{anchor=A-sol}{\\color{@amber}{A = -1}}$$"}],"highlights":[{"color":"sky","style":"text-color","anchor":"x-val-2"},{"color":"amber","style":"text-color","anchor":"A-sol"}],"connections":[{"to":"sub-x-2","from":"x-val-2","color":"sky","label":null,"style":"solid"},{"to":"sub-x-2-A","from":"x-val-2","color":"sky","label":null,"style":"solid"},{"to":"sub-x-2-B","from":"x-val-2","color":"sky","label":null,"style":"solid"}]},"order":1,"title":"Find A using substitution","description":"We substitute $x = 2$ because it makes the coefficient of $B$ equal to zero, allowing us to solve directly for $A$."},{"type":"hyper_latex","block":{"nodes":[{"latex":"$$$x-1 \\equiv A(x-3) + B(x-2)$$"},{"text":"Substitute the root of the first bracket, "},{"latex":"$$\\htmlData{anchor=x-val-3}{\\color{@purple}{x = 3}}$"},{"text":", into the identity to eliminate the $A$ term:"},{"latex":"$$$\\htmlData{anchor=sub-x-3}{\\color{@purple}{3}} - 1 = A(\\htmlData{anchor=sub-x-3-A}{\\color{@purple}{3}} - 3) + B(\\htmlData{anchor=sub-x-3-B}{\\color{@purple}{3}} - 2)$$"},{"latex":"$$$2 = A(0) + B(1)$$"},{"latex":"$$$\\htmlData{anchor=B-sol}{\\color{@teal}{B = 2}}$$"}],"highlights":[{"color":"purple","style":"text-color","anchor":"x-val-3"},{"color":"teal","style":"text-color","anchor":"B-sol"}],"connections":[{"to":"sub-x-3","from":"x-val-3","color":"purple","label":null,"style":"solid"},{"to":"sub-x-3-A","from":"x-val-3","color":"purple","label":null,"style":"solid"},{"to":"sub-x-3-B","from":"x-val-3","color":"purple","label":null,"style":"solid"}]},"order":2,"title":"Find B using substitution","description":"We substitute $x = 3$ to make the coefficient of $A$ equal to zero, which lets us solve directly for $B$. Finding both $A = -1$ and $B = 2$ awards 1 mark (A1)."},{"type":"text","order":3,"title":"Write the final answer","content":"Now that we have found $A = -1$ and $B = 2$, we substitute these values back into our original partial fraction setup:\n\n$$\\frac{x-1}{(x-2)(x-3)} = \\frac{-1}{x-2} + \\frac{2}{x-3}$$\n\nAlternatively, this can be written as:\n\n$$\\frac{2}{x-3} - \\frac{1}{x-2}$$\n\n*Awarded 1 mark (A1) for the correct final expression.*"}]},{"id":"coefficients","label":"Equating Coefficients","steps":[{"type":"text","order":0,"title":"Set up the identity","content":"To express the algebraic fraction as partial fractions, we start by setting up an identity with a common denominator. We multiply both sides of the equation by the denominator $(x-2)(x-3)$:\n\n$$\\frac{x-1}{(x-2)(x-3)} = \\frac{A}{x-2} + \\frac{B}{x-3}$$\n\nMultiplying through gives us our identity:\n\n$$x-1 = A(x-3) + B(x-2)$$\n\nThis earns you the first method mark (**M1**).","highlights":[{"text":"Express $\\frac{x-1}{(x-2)(x-3)}$ in the form $\\frac{A}{(x-2)} + \\frac{B}{(x-3)}$","location":"specification"}]},{"type":"hyper_latex","block":{"nodes":[{"latex":"$$x-1 = A(x-3) + B(x-2)$"},{"latex":"$$x-1 = \\htmlData{anchor=ax}{\\color{@sky}{A}}x - 3\\htmlData{anchor=a}{\\color{@sky}{A}} + \\htmlData{anchor=bx}{\\color{@amber}{B}}x - 2\\htmlData{anchor=b}{\\color{@amber}{B}}$"},{"latex":"$$1x - 1 = (\\htmlData{anchor=eqx}{\\color{@sky}{A} + \\color{@amber}{B}})x - (\\htmlData{anchor=eqc}{3\\color{@sky}{A} + 2\\color{@amber}{B}})$"}],"highlights":[{"color":"sky","style":"text-color","anchor":"ax"},{"color":"amber","style":"text-color","anchor":"bx"}],"connections":[{"to":"eqx","from":"ax","color":"sky","label":null,"style":"solid"},{"to":"eqx","from":"bx","color":"amber","label":null,"style":"solid"}]},"order":1,"title":"Expand and group terms","description":"Now, we expand the right-hand side of the identity and group the $x$ terms and constant terms together to prepare for equating coefficients."},{"type":"text","order":2,"title":"Equate the coefficients","content":"Since this is an identity, the coefficient of $x$ and the constant term on the left-hand side must equal those on the right-hand side.\n\nBy comparing the coefficients of $x$ on both sides:\n\n$$A + B = 1 \\quad \\text{--- (Equation 1)}$$\n\nBy comparing the constant terms on both sides:\n\n$$-3A - 2B = -1 \\implies 3A + 2B = 1 \\quad \\text{--- (Equation 2)}$$"},{"type":"text","order":3,"title":"Solve the system","content":"Now we solve this system of simultaneous equations to find the values of $A$ and $B$.\n\nFrom Equation 1, we can write:\n$$B = 1 - A$$\n\nSubstitute this expression for $B$ into Equation 2:\n\n$$3A + 2(1 - A) = 1$$\n\n$$3A + 2 - 2A = 1$$\n\n$$A + 2 = 1$$\n\n$$A = -1$$\n\nNow, substitute $A = -1$ back into the expression for $B$:\n$$B = 1 - (-1) = 2$$\n\nSo we have found that $A = -1$ and $B = 2$. This earns you the second mark (**A1**).","calculator":[{"gdc":{"expression":"x = -1, y = 2","instructions":"Go to the Equation/Solf Solver menu. Choose 'Simultaneous' for 2 unknowns. Enter the coefficients matrix: [[1, 1, 1], [3, 2, 1]] and solve."},"ti84":{"expression":"x1 = -1, x2 = 2","instructions":"Press [APPS] and select 'PLYSMLT2'. Choose '2: SYSTEM SOLVER'. Set the number of equations to 2 and variables to 2. Press [NEXT] (GRAPH). Enter the coefficients: row 1 as [1, 1, 1] and row 2 as [3, 2, 1]. Press [SOLVE] (GRAPH) to find the values."},"desmos":{"expression":"(-1, 2)","instructions":"Type the equations into two separate rows: 'y + x = 1' and '3x + 2y = 1' (using x for A and y for B). Find the intersection point of the two lines on the graph."},"description":"You can quickly solve this system of linear equations on your graphing calculator."}]},{"type":"text","order":4,"title":"Write the final answer","content":"Substitute the values of $A$ and $B$ back into our original partial fractions form:\n\n$$\\frac{x-1}{(x-2)(x-3)} = \\frac{-1}{x-2} + \\frac{2}{x-3}$$\n\nThis completed expression gets the final accuracy mark (**A1**)."}]}],"generatedAt":"2026-06-09T07:17:55.738Z","modelVersion":"gemini-3.5-flash","explanationVersion":"v2"},"options":[]}],"options":[],"explanation":{"methods":[{"id":"a66291cd-9c55-4609-bdd8-3fe0f905772d:substitution","label":"Substitution Method","steps":[{"type":"text","order":0,"title":"Set up the identity","content":"To express the algebraic fraction as partial fractions, we start by setting up an identity. We want to find constants $A$ and $B$ such that:\n\n$$\\frac{x-1}{(x-2)(x-3)} \\equiv \\frac{A}{x-2} + \\frac{B}{x-3}$$\n\nMultiply both sides of the equation by the common denominator $(x-2)(x-3)$ to clear the fractions. This gives us the key identity:\n\n$$x-1 \\equiv A(x-3) + B(x-2)$$\n\n*Awarded 1 mark (M1) for setting up this identity.*"},{"type":"hyper_latex","block":{"nodes":[{"latex":"$$$x-1 \\equiv A(x-3) + B(x-2)$$"},{"text":"Substitute the root of the second bracket, "},{"latex":"$$\\htmlData{anchor=x-val-2}{\\color{@sky}{x = 2}}$"},{"text":", into the identity to eliminate the $B$ term:"},{"latex":"$$$\\htmlData{anchor=sub-x-2}{\\color{@sky}{2}} - 1 = A(\\htmlData{anchor=sub-x-2-A}{\\color{@sky}{2}} - 3) + B(\\htmlData{anchor=sub-x-2-B}{\\color{@sky}{2}} - 2)$$"},{"latex":"$$$1 = A(-1) + B(0)$$"},{"latex":"$$$\\htmlData{anchor=A-sol}{\\color{@amber}{A = -1}}$$"}],"highlights":[{"color":"sky","style":"text-color","anchor":"x-val-2"},{"color":"amber","style":"text-color","anchor":"A-sol"}],"connections":[{"to":"sub-x-2","from":"x-val-2","color":"sky","label":null,"style":"solid"},{"to":"sub-x-2-A","from":"x-val-2","color":"sky","label":null,"style":"solid"},{"to":"sub-x-2-B","from":"x-val-2","color":"sky","label":null,"style":"solid"}]},"order":1,"title":"Find A using substitution","description":"We substitute $x = 2$ because it makes the coefficient of $B$ equal to zero, allowing us to solve directly for $A$."},{"type":"hyper_latex","block":{"nodes":[{"latex":"$$$x-1 \\equiv A(x-3) + B(x-2)$$"},{"text":"Substitute the root of the first bracket, "},{"latex":"$$\\htmlData{anchor=x-val-3}{\\color{@purple}{x = 3}}$"},{"text":", into the identity to eliminate the $A$ term:"},{"latex":"$$$\\htmlData{anchor=sub-x-3}{\\color{@purple}{3}} - 1 = A(\\htmlData{anchor=sub-x-3-A}{\\color{@purple}{3}} - 3) + B(\\htmlData{anchor=sub-x-3-B}{\\color{@purple}{3}} - 2)$$"},{"latex":"$$$2 = A(0) + B(1)$$"},{"latex":"$$$\\htmlData{anchor=B-sol}{\\color{@teal}{B = 2}}$$"}],"highlights":[{"color":"purple","style":"text-color","anchor":"x-val-3"},{"color":"teal","style":"text-color","anchor":"B-sol"}],"connections":[{"to":"sub-x-3","from":"x-val-3","color":"purple","label":null,"style":"solid"},{"to":"sub-x-3-A","from":"x-val-3","color":"purple","label":null,"style":"solid"},{"to":"sub-x-3-B","from":"x-val-3","color":"purple","label":null,"style":"solid"}]},"order":2,"title":"Find B using substitution","description":"We substitute $x = 3$ to make the coefficient of $A$ equal to zero, which lets us solve directly for $B$. Finding both $A = -1$ and $B = 2$ awards 1 mark (A1)."},{"type":"text","order":3,"title":"Write the final answer","content":"Now that we have found $A = -1$ and $B = 2$, we substitute these values back into our original partial fraction setup:\n\n$$\\frac{x-1}{(x-2)(x-3)} = \\frac{-1}{x-2} + \\frac{2}{x-3}$$\n\nAlternatively, this can be written as:\n\n$$\\frac{2}{x-3} - \\frac{1}{x-2}$$\n\n*Awarded 1 mark (A1) for the correct final expression.*"}]},{"id":"a66291cd-9c55-4609-bdd8-3fe0f905772d:coefficients","label":"Equating Coefficients","steps":[{"type":"text","order":0,"title":"Set up the identity","content":"To express the algebraic fraction as partial fractions, we start by setting up an identity with a common denominator. We multiply both sides of the equation by the denominator $(x-2)(x-3)$:\n\n$$\\frac{x-1}{(x-2)(x-3)} = \\frac{A}{x-2} + \\frac{B}{x-3}$$\n\nMultiplying through gives us our identity:\n\n$$x-1 = A(x-3) + B(x-2)$$\n\nThis earns you the first method mark (**M1**).","highlights":[{"text":"Express $\\frac{x-1}{(x-2)(x-3)}$ in the form $\\frac{A}{(x-2)} + \\frac{B}{(x-3)}$","location":"specification"}]},{"type":"hyper_latex","block":{"nodes":[{"latex":"$$x-1 = A(x-3) + B(x-2)$"},{"latex":"$$x-1 = \\htmlData{anchor=ax}{\\color{@sky}{A}}x - 3\\htmlData{anchor=a}{\\color{@sky}{A}} + \\htmlData{anchor=bx}{\\color{@amber}{B}}x - 2\\htmlData{anchor=b}{\\color{@amber}{B}}$"},{"latex":"$$1x - 1 = (\\htmlData{anchor=eqx}{\\color{@sky}{A} + \\color{@amber}{B}})x - (\\htmlData{anchor=eqc}{3\\color{@sky}{A} + 2\\color{@amber}{B}})$"}],"highlights":[{"color":"sky","style":"text-color","anchor":"ax"},{"color":"amber","style":"text-color","anchor":"bx"}],"connections":[{"to":"eqx","from":"ax","color":"sky","label":null,"style":"solid"},{"to":"eqx","from":"bx","color":"amber","label":null,"style":"solid"}]},"order":1,"title":"Expand and group terms","description":"Now, we expand the right-hand side of the identity and group the $x$ terms and constant terms together to prepare for equating coefficients."},{"type":"text","order":2,"title":"Equate the coefficients","content":"Since this is an identity, the coefficient of $x$ and the constant term on the left-hand side must equal those on the right-hand side.\n\nBy comparing the coefficients of $x$ on both sides:\n\n$$A + B = 1 \\quad \\text{--- (Equation 1)}$$\n\nBy comparing the constant terms on both sides:\n\n$$-3A - 2B = -1 \\implies 3A + 2B = 1 \\quad \\text{--- (Equation 2)}$$"},{"type":"text","order":3,"title":"Solve the system","content":"Now we solve this system of simultaneous equations to find the values of $A$ and $B$.\n\nFrom Equation 1, we can write:\n$$B = 1 - A$$\n\nSubstitute this expression for $B$ into Equation 2:\n\n$$3A + 2(1 - A) = 1$$\n\n$$3A + 2 - 2A = 1$$\n\n$$A + 2 = 1$$\n\n$$A = -1$$\n\nNow, substitute $A = -1$ back into the expression for $B$:\n$$B = 1 - (-1) = 2$$\n\nSo we have found that $A = -1$ and $B = 2$. This earns you the second mark (**A1**).","calculator":[{"gdc":{"expression":"x = -1, y = 2","instructions":"Go to the Equation/Solf Solver menu. Choose 'Simultaneous' for 2 unknowns. Enter the coefficients matrix: [[1, 1, 1], [3, 2, 1]] and solve."},"ti84":{"expression":"x1 = -1, x2 = 2","instructions":"Press [APPS] and select 'PLYSMLT2'. Choose '2: SYSTEM SOLVER'. Set the number of equations to 2 and variables to 2. Press [NEXT] (GRAPH). Enter the coefficients: row 1 as [1, 1, 1] and row 2 as [3, 2, 1]. Press [SOLVE] (GRAPH) to find the values."},"desmos":{"expression":"(-1, 2)","instructions":"Type the equations into two separate rows: 'y + x = 1' and '3x + 2y = 1' (using x for A and y for B). Find the intersection point of the two lines on the graph."},"description":"You can quickly solve this system of linear equations on your graphing calculator."}]},{"type":"text","order":4,"title":"Write the final answer","content":"Substitute the values of $A$ and $B$ back into our original partial fractions form:\n\n$$\\frac{x-1}{(x-2)(x-3)} = \\frac{-1}{x-2} + \\frac{2}{x-3}$$\n\nThis completed expression gets the final accuracy mark (**A1**)."}]}],"generatedAt":"2026-06-09T07:18:58.081Z","modelVersion":"gemini-3.5-flash","explanationVersion":"v2"},"firstAttemptFullMarks":{"percent":81},"walkthroughVideo":null},{"id":"ccd01f36-c50a-4e45-ad97-44a918590c37","revisionId":"f439f861-1b8c-4996-bc35-74f23bb6c9e5","questionNumber":null,"questionType":"LA","specification":"","difficulty":1,"difficultyBand":"medium","totalMarks":3,"sourceTitle":null,"calculatorUse":null,"isCalculation":true,"parts":[{"id":"bb84d143-6ec7-457d-afa0-df137f4aed9d","label":"a","content":"Express $\\frac{3x-1}{5(x+1)(x-3)}$ in partial fractions","markscheme":"- Sets up the identity $\\frac{3x-1}{5} = A(x-3) + B(x+1)$ M1\n- Finds $A = \\frac{1}{5}$ and $B = \\frac{2}{5}$ A1\n- Final answer: $\\frac{1}{5(x+1)} + \\frac{2}{5(x-3)}$ A1","marks":3,"order":0,"markingRubric":{"totalMarks":3,"points":[{"id":"p1","marks":1,"description":"Sets up the correct partial fraction identity by clearing the common denominator.","code":"M"},{"id":"p2","marks":1,"description":"Solves for the constants $A$ and $B$ correctly.","code":"A"},{"id":"p3","marks":1,"description":"States the final expression in partial fractions correctly.","code":"A"}],"finalAnswers":["\\frac{1}{5(x+1)} + \\frac{2}{5(x-3)}","\\frac{0.2}{x+1} + \\frac{0.4}{x-3}"],"generatedAt":"2026-05-06T14:03:44.609Z","modelVersion":"gemini-3-flash-preview"},"tableImages":[],"explanation":{"methods":[{"id":"substitution","label":"Substitution Method","steps":[{"type":"text","order":0,"title":"Set up partial fractions","content":"To express the algebraic fraction in partial fractions, we first set up the standard identity.\n\nWe can factor out the constant $\\frac{1}{5}$ or include it in the denominators. Let's write the decomposition as:\n\n$$\\frac{3x-1}{5(x+1)(x-3)} \\equiv \\frac{A}{5(x+1)} + \\frac{B}{5(x-3)}$$\n\nTo make the substitution method simpler, we can multiply both sides of the identity by the entire denominator $5(x+1)(x-3)$ to eliminate the fractions:\n\n$$3x - 1 \\equiv A(x-3) + B(x+1)$$"},{"type":"text","order":1,"title":"Substitute the first root","content":"To find $A$, we can substitute the root of the other linear factor, which is $x = -1$, into our identity to eliminate $B$:\n\n$$3(-1) - 1 = A(-1 - 3) + B(-1 + 1)$$\n\n$$-4 = -4A$$\n\n$$A = 1$$"},{"type":"text","order":2,"title":"Substitute the second root","content":"Next, to find $B$, we substitute the root $x = 3$ to eliminate $A$:\n\n$$3(3) - 1 = A(3 - 3) + B(3 + 1)$$\n\n$$8 = 4B$$\n\n$$B = 2$$"},{"type":"text","order":3,"title":"Form final partial fractions","content":"Now that we have $A = 1$ and $B = 2$, we can substitute these values back into our original setup:\n\n$$\\frac{3x-1}{5(x+1)(x-3)} = \\frac{1}{5(x+1)} + \\frac{2}{5(x-3)}$$\n\nThis matches the final answer from the markscheme."}]},{"id":"equating_coefficients","label":"Equating Coefficients Method","steps":[{"type":"text","order":0,"title":"Set up the decomposition","content":"To express the expression as partial fractions, we set up an identity where each linear factor in the denominator gets its own fraction:\n\n$$\\frac{3x-1}{5(x+1)(x-3)} \\equiv \\frac{A}{x+1} + \\frac{B}{x-3}$$\n\nTo find the constants $A$ and $B$, we multiply both sides of the identity by the entire denominator $5(x+1)(x-3)$:\n\n$$3x - 1 \\equiv 5A(x-3) + 5B(x+1)$$\n\nAlternatively, we can group the factor of $\\frac{1}{5}$ with the numerators and write:\n\n$$\\frac{3x-1}{5} \\equiv A(x-3) + B(x+1)$$\n\nLet's use this simplified version to find $A$ and $B$ by equating coefficients."},{"type":"text","order":1,"title":"Expand and group terms","content":"We expand the right-hand side of the identity and collect the terms with $x$ and the constant terms together:\n\n$$\\frac{3}{5}x - \\frac{1}{5} \\equiv A(x-3) + B(x+1)$$\n\n$$\\frac{3}{5}x - \\frac{1}{5} \\equiv (A+B)x + (-3A+B)$$"},{"type":"text","order":2,"title":"Equate the coefficients","content":"By equating the coefficients of $x$ and the constant terms from both sides of the identity, we get a system of simultaneous equations:\n\n**For $x$ coefficients:**\n$$A + B = \\frac{3}{5}$$\n\n**For constant terms:**\n$$-3A + B = -\\frac{1}{5}$$"},{"type":"text","order":3,"title":"Solve the equations","content":"We can solve this system by subtracting the second equation from the first equation to eliminate $B$:\n\n$$(A + B) - (-3A + B) = \\frac{3}{5} - \\left(-\\frac{1}{5}\\right)$$\n\n$$4A = \\frac{4}{5}$$\n\n$$A = \\frac{1}{5}$$\n\nNow substitute $A = \\frac{1}{5}$ back into the first equation to find $B$:\n\n$$\\frac{1}{5} + B = \\frac{3}{5}$$\n\n$$B = \\frac{2}{5}$$"},{"type":"text","order":4,"title":"State the final fraction","content":"Substitute our values for $A$ and $B$ back into the original partial fraction setup:\n\n$$\\frac{3x-1}{5(x+1)(x-3)} \\equiv \\frac{\\frac{1}{5}}{x+1} + \\frac{\\frac{2}{5}}{x-3}$$\n\nSimplifying this gives the final answer:\n\n$$\\frac{1}{5(x+1)} + \\frac{2}{5(x-3)}$$"}]}],"generatedAt":"2026-06-09T09:06:55.897Z","modelVersion":"gemini-3.5-flash","explanationVersion":"v2"},"options":[]}],"options":[],"explanation":{"methods":[{"id":"bb84d143-6ec7-457d-afa0-df137f4aed9d:substitution","label":"Substitution Method","steps":[{"type":"text","order":0,"title":"Set up partial fractions","content":"To express the algebraic fraction in partial fractions, we first set up the standard identity.\n\nWe can factor out the constant $\\frac{1}{5}$ or include it in the denominators. Let's write the decomposition as:\n\n$$\\frac{3x-1}{5(x+1)(x-3)} \\equiv \\frac{A}{5(x+1)} + \\frac{B}{5(x-3)}$$\n\nTo make the substitution method simpler, we can multiply both sides of the identity by the entire denominator $5(x+1)(x-3)$ to eliminate the fractions:\n\n$$3x - 1 \\equiv A(x-3) + B(x+1)$$"},{"type":"text","order":1,"title":"Substitute the first root","content":"To find $A$, we can substitute the root of the other linear factor, which is $x = -1$, into our identity to eliminate $B$:\n\n$$3(-1) - 1 = A(-1 - 3) + B(-1 + 1)$$\n\n$$-4 = -4A$$\n\n$$A = 1$$"},{"type":"text","order":2,"title":"Substitute the second root","content":"Next, to find $B$, we substitute the root $x = 3$ to eliminate $A$:\n\n$$3(3) - 1 = A(3 - 3) + B(3 + 1)$$\n\n$$8 = 4B$$\n\n$$B = 2$$"},{"type":"text","order":3,"title":"Form final partial fractions","content":"Now that we have $A = 1$ and $B = 2$, we can substitute these values back into our original setup:\n\n$$\\frac{3x-1}{5(x+1)(x-3)} = \\frac{1}{5(x+1)} + \\frac{2}{5(x-3)}$$\n\nThis matches the final answer from the markscheme."}]},{"id":"bb84d143-6ec7-457d-afa0-df137f4aed9d:equating_coefficients","label":"Equating Coefficients Method","steps":[{"type":"text","order":0,"title":"Set up the decomposition","content":"To express the expression as partial fractions, we set up an identity where each linear factor in the denominator gets its own fraction:\n\n$$\\frac{3x-1}{5(x+1)(x-3)} \\equiv \\frac{A}{x+1} + \\frac{B}{x-3}$$\n\nTo find the constants $A$ and $B$, we multiply both sides of the identity by the entire denominator $5(x+1)(x-3)$:\n\n$$3x - 1 \\equiv 5A(x-3) + 5B(x+1)$$\n\nAlternatively, we can group the factor of $\\frac{1}{5}$ with the numerators and write:\n\n$$\\frac{3x-1}{5} \\equiv A(x-3) + B(x+1)$$\n\nLet's use this simplified version to find $A$ and $B$ by equating coefficients."},{"type":"text","order":1,"title":"Expand and group terms","content":"We expand the right-hand side of the identity and collect the terms with $x$ and the constant terms together:\n\n$$\\frac{3}{5}x - \\frac{1}{5} \\equiv A(x-3) + B(x+1)$$\n\n$$\\frac{3}{5}x - \\frac{1}{5} \\equiv (A+B)x + (-3A+B)$$"},{"type":"text","order":2,"title":"Equate the coefficients","content":"By equating the coefficients of $x$ and the constant terms from both sides of the identity, we get a system of simultaneous equations:\n\n**For $x$ coefficients:**\n$$A + B = \\frac{3}{5}$$\n\n**For constant terms:**\n$$-3A + B = -\\frac{1}{5}$$"},{"type":"text","order":3,"title":"Solve the equations","content":"We can solve this system by subtracting the second equation from the first equation to eliminate $B$:\n\n$$(A + B) - (-3A + B) = \\frac{3}{5} - \\left(-\\frac{1}{5}\\right)$$\n\n$$4A = \\frac{4}{5}$$\n\n$$A = \\frac{1}{5}$$\n\nNow substitute $A = \\frac{1}{5}$ back into the first equation to find $B$:\n\n$$\\frac{1}{5} + B = \\frac{3}{5}$$\n\n$$B = \\frac{2}{5}$$"},{"type":"text","order":4,"title":"State the final fraction","content":"Substitute our values for $A$ and $B$ back into the original partial fraction setup:\n\n$$\\frac{3x-1}{5(x+1)(x-3)} \\equiv \\frac{\\frac{1}{5}}{x+1} + \\frac{\\frac{2}{5}}{x-3}$$\n\nSimplifying this gives the final answer:\n\n$$\\frac{1}{5(x+1)} + \\frac{2}{5(x-3)}$$"}]}],"generatedAt":"2026-06-09T09:07:06.616Z","modelVersion":"gemini-3.5-flash","explanationVersion":"v2"},"firstAttemptFullMarks":{"percent":59},"walkthroughVideo":null},{"id":"61bf9309-865b-4108-a293-c8fc670013b0","revisionId":"e5ce1501-2cdd-4aff-88cb-b40183b2651f","questionNumber":"2","questionType":"LA","specification":"","difficulty":1,"difficultyBand":"medium","totalMarks":3,"sourceTitle":"Question Set 1: Algebraic Methods","calculatorUse":null,"isCalculation":true,"parts":[{"id":"64510fdf-2642-4f2a-817e-37f3a371759b","label":"","content":"Express $\\frac{9x-5}{2(2x-1)(x-1)}$ in partial fractions","markscheme":"- Sets up the identity $\\frac{9x-5}{2} = A(x-1) + B(2x-1)$ M1\n- Finds $A = \\frac{1}{2}$ and $B = 2$ A1\n- Final answer: $\\frac{1}{2(2x-1)} + \\frac{2}{x-1}$ A1","marks":3,"order":0,"markingRubric":{"totalMarks":3,"points":[{"id":"p1","marks":1,"description":"Sets up the correct identity $\\frac{9x-5}{2} = A(x-1) + B(2x-1)$.","code":"M"},{"id":"p2","marks":1,"description":"Correctly calculates the values of $A$ and $B$.","code":"A"},{"id":"p3","marks":1,"description":"States the final expression correctly.","code":"A"}],"finalAnswers":["\\frac{1}{2(2x-1)} + \\frac{2}{x-1}"],"generatedAt":"2026-05-06T02:36:06.047Z","modelVersion":"gemini-3-flash-preview"},"tableImages":[],"explanation":{"methods":[{"id":"substitution","label":"Substitution Method","steps":[{"type":"text","order":0,"title":"Set up partial fractions","content":"To express the algebraic fraction in partial fractions, we first write it as a sum of two simpler fractions with unknown constant numerators $A$ and $B$:\n\n$$\\frac{9x-5}{2(2x-1)(x-1)} = \\frac{A}{2x-1} + \\frac{B}{x-1}$$\n\nThis setup allows us to keep the constant $2$ on the left-hand side, leading directly to the form required by the markscheme."},{"type":"text","order":1,"title":"Clear the denominators","content":"Next, we multiply both sides of the identity by the common denominator of the right-hand side, which is $(2x-1)(x-1)$:\n\n$$\\frac{9x-5}{2} = A(x-1) + B(2x-1)$$\n\nThis gives us a linear identity where we can easily apply the **Substitution Method** to find the values of $A$ and $B$."},{"type":"text","order":2,"title":"Substitute to find B","content":"To isolate $B$, we can substitute a value of $x$ that makes the term containing $A$ equal to zero. Let's substitute $x = 1$:\n\n$$\\begin{aligned} \\frac{9(1)-5}{2} &= A(1-1) + B(2(1)-1) \\\\ \\frac{4}{2} &= A(0) + B(1) \\\\ 2 &= B \\end{aligned}$$\n\nThus, $B = 2$."},{"type":"text","order":3,"title":"Substitute to find A","content":"To isolate $A$, we choose a value of $x$ that makes the term containing $B$ equal to zero. Substituting $x = \\frac{1}{2}$ gives:\n\n$$\\begin{aligned} \\frac{9\\left(\\frac{1}{2}\\right)-5}{2} &= A\\left(\\frac{1}{2}-1\\right) + B\\left(2\\left(\\frac{1}{2}\\right)-1\\right) \\\\ \\frac{4.5-5}{2} &= A\\left(-\\frac{1}{2}\\right) + B(0) \\\\ -0.25 &= -0.5A \\\\ A &= \\frac{1}{2} \\end{aligned}$$\n\nThus, $A = \\frac{1}{2}$."},{"type":"text","order":4,"title":"Write final expression","content":"Now we substitute our values of $A = \\frac{1}{2}$ and $B = 2$ back into our initial partial fraction setup:\n\n$$\\frac{9x-5}{2(2x-1)(x-1)} = \\frac{\\frac{1}{2}}{2x-1} + \\frac{2}{x-1}$$\n\nWe simplify the first term by bringing the $2$ in the numerator down to the denominator:\n\n$$\\frac{1}{2(2x-1)} + \\frac{2}{x-1}$$"}]},{"id":"equating-coefficients","label":"Equating Coefficients","steps":[{"type":"text","order":0,"title":"Set up partial fractions","content":"To express the algebraic fraction in partial fractions, we first write it as a sum of simpler fractions with constant numerators $A$ and $B$:\n\n$$\\frac{9x-5}{2(2x-1)(x-1)} \\equiv \\frac{A}{2x-1} + \\frac{B}{x-1}$$\n\nThis is our starting identity, where we need to find the values of $A$ and $B$. This step helps earn the first method mark M1."},{"type":"text","order":1,"title":"Clear the denominators","content":"Next, we clear the denominators by multiplying both sides of our identity by $(2x-1)(x-1)$:\n\n$$\\frac{9x-5}{2} \\equiv A(x-1) + B(2x-1)$$\n\nTo make this linear equation easier to work with, we can multiply the entire equation by $2$ to clear the fraction on the left-hand side:\n\n$$9x-5 \\equiv 2A(x-1) + 2B(2x-1)$$"},{"type":"text","order":2,"title":"Expand and group terms","content":"Using the **Equating Coefficients** approach, we expand the terms on the right-hand side and group like powers of $x$:\n\n$$\\begin{aligned} 9x - 5 &\\equiv 2A(x - 1) + 2B(2x - 1) \\\\ 9x - 5 &\\equiv 2Ax - 2A + 4Bx - 2B \\\\ 9x - 5 &\\equiv (2A + 4B)x - (2A + 2B) \\end{aligned}$$"},{"type":"text","order":3,"title":"Equate the coefficients","content":"Since this is an identity, the coefficients of like powers of $x$ on both sides must be equal.\n\nBy comparing the coefficients of $x$:\n$$2A + 4B = 9 \\quad \\text{--- (Equation 1)}$$\n\nBy comparing the constant terms:\n$$-2A - 2B = -5 \\implies 2A + 2B = 5 \\quad \\text{--- (Equation 2)}$$"},{"type":"text","order":4,"title":"Solve for the constants","content":"We can solve this system of linear equations by subtracting Equation 2 from Equation 1:\n\n$$\\begin{aligned} (2A + 4B) - (2A + 2B) &= 9 - 5 \\\\ 2B &= 4 \\\\ B &= 2 \\end{aligned}$$\n\nNow, substitute $B = 2$ back into Equation 2 to find $A$:\n\n$$\\begin{aligned} 2A + 2(2) &= 5 \\\\ 2A + 4 &= 5 \\\\ 2A &= 1 \\\\ A &= \\frac{1}{2} \\end{aligned}$$\n\nThis gives us $A = \\frac{1}{2}$ and $B = 2$, which awards the second mark A1."},{"type":"text","order":5,"title":"Write the final answer","content":"Finally, we substitute the values of $A$ and $B$ back into our initial partial fraction expression:\n\n$$\\frac{\\frac{1}{2}}{2x-1} + \\frac{2}{x-1} = \\frac{1}{2(2x-1)} + \\frac{2}{x-1}$$\n\nThis is our final simplified answer, earning the final mark A1."}]}],"generatedAt":"2026-06-09T05:30:23.545Z","modelVersion":"gemini-3.5-flash","explanationVersion":"v2"},"options":[]}],"options":[],"explanation":{"methods":[{"id":"64510fdf-2642-4f2a-817e-37f3a371759b:substitution","label":"Substitution Method","steps":[{"type":"text","order":0,"title":"Set up partial fractions","content":"To express the algebraic fraction in partial fractions, we first write it as a sum of two simpler fractions with unknown constant numerators $A$ and $B$:\n\n$$\\frac{9x-5}{2(2x-1)(x-1)} = \\frac{A}{2x-1} + \\frac{B}{x-1}$$\n\nThis setup allows us to keep the constant $2$ on the left-hand side, leading directly to the form required by the markscheme."},{"type":"text","order":1,"title":"Clear the denominators","content":"Next, we multiply both sides of the identity by the common denominator of the right-hand side, which is $(2x-1)(x-1)$:\n\n$$\\frac{9x-5}{2} = A(x-1) + B(2x-1)$$\n\nThis gives us a linear identity where we can easily apply the **Substitution Method** to find the values of $A$ and $B$."},{"type":"text","order":2,"title":"Substitute to find B","content":"To isolate $B$, we can substitute a value of $x$ that makes the term containing $A$ equal to zero. Let's substitute $x = 1$:\n\n$$\\begin{aligned} \\frac{9(1)-5}{2} &= A(1-1) + B(2(1)-1) \\\\ \\frac{4}{2} &= A(0) + B(1) \\\\ 2 &= B \\end{aligned}$$\n\nThus, $B = 2$."},{"type":"text","order":3,"title":"Substitute to find A","content":"To isolate $A$, we choose a value of $x$ that makes the term containing $B$ equal to zero. Substituting $x = \\frac{1}{2}$ gives:\n\n$$\\begin{aligned} \\frac{9\\left(\\frac{1}{2}\\right)-5}{2} &= A\\left(\\frac{1}{2}-1\\right) + B\\left(2\\left(\\frac{1}{2}\\right)-1\\right) \\\\ \\frac{4.5-5}{2} &= A\\left(-\\frac{1}{2}\\right) + B(0) \\\\ -0.25 &= -0.5A \\\\ A &= \\frac{1}{2} \\end{aligned}$$\n\nThus, $A = \\frac{1}{2}$."},{"type":"text","order":4,"title":"Write final expression","content":"Now we substitute our values of $A = \\frac{1}{2}$ and $B = 2$ back into our initial partial fraction setup:\n\n$$\\frac{9x-5}{2(2x-1)(x-1)} = \\frac{\\frac{1}{2}}{2x-1} + \\frac{2}{x-1}$$\n\nWe simplify the first term by bringing the $2$ in the numerator down to the denominator:\n\n$$\\frac{1}{2(2x-1)} + \\frac{2}{x-1}$$"}]},{"id":"64510fdf-2642-4f2a-817e-37f3a371759b:equating-coefficients","label":"Equating Coefficients","steps":[{"type":"text","order":0,"title":"Set up partial fractions","content":"To express the algebraic fraction in partial fractions, we first write it as a sum of simpler fractions with constant numerators $A$ and $B$:\n\n$$\\frac{9x-5}{2(2x-1)(x-1)} \\equiv \\frac{A}{2x-1} + \\frac{B}{x-1}$$\n\nThis is our starting identity, where we need to find the values of $A$ and $B$. This step helps earn the first method mark M1."},{"type":"text","order":1,"title":"Clear the denominators","content":"Next, we clear the denominators by multiplying both sides of our identity by $(2x-1)(x-1)$:\n\n$$\\frac{9x-5}{2} \\equiv A(x-1) + B(2x-1)$$\n\nTo make this linear equation easier to work with, we can multiply the entire equation by $2$ to clear the fraction on the left-hand side:\n\n$$9x-5 \\equiv 2A(x-1) + 2B(2x-1)$$"},{"type":"text","order":2,"title":"Expand and group terms","content":"Using the **Equating Coefficients** approach, we expand the terms on the right-hand side and group like powers of $x$:\n\n$$\\begin{aligned} 9x - 5 &\\equiv 2A(x - 1) + 2B(2x - 1) \\\\ 9x - 5 &\\equiv 2Ax - 2A + 4Bx - 2B \\\\ 9x - 5 &\\equiv (2A + 4B)x - (2A + 2B) \\end{aligned}$$"},{"type":"text","order":3,"title":"Equate the coefficients","content":"Since this is an identity, the coefficients of like powers of $x$ on both sides must be equal.\n\nBy comparing the coefficients of $x$:\n$$2A + 4B = 9 \\quad \\text{--- (Equation 1)}$$\n\nBy comparing the constant terms:\n$$-2A - 2B = -5 \\implies 2A + 2B = 5 \\quad \\text{--- (Equation 2)}$$"},{"type":"text","order":4,"title":"Solve for the constants","content":"We can solve this system of linear equations by subtracting Equation 2 from Equation 1:\n\n$$\\begin{aligned} (2A + 4B) - (2A + 2B) &= 9 - 5 \\\\ 2B &= 4 \\\\ B &= 2 \\end{aligned}$$\n\nNow, substitute $B = 2$ back into Equation 2 to find $A$:\n\n$$\\begin{aligned} 2A + 2(2) &= 5 \\\\ 2A + 4 &= 5 \\\\ 2A &= 1 \\\\ A &= \\frac{1}{2} \\end{aligned}$$\n\nThis gives us $A = \\frac{1}{2}$ and $B = 2$, which awards the second mark A1."},{"type":"text","order":5,"title":"Write the final answer","content":"Finally, we substitute the values of $A$ and $B$ back into our initial partial fraction expression:\n\n$$\\frac{\\frac{1}{2}}{2x-1} + \\frac{2}{x-1} = \\frac{1}{2(2x-1)} + \\frac{2}{x-1}$$\n\nThis is our final simplified answer, earning the final mark A1."}]}],"generatedAt":"2026-06-09T05:30:48.609Z","modelVersion":"gemini-3.5-flash","explanationVersion":"v2"},"firstAttemptFullMarks":{"percent":57},"walkthroughVideo":null}],"subjectCode":"a_level_maths","summaries":[{"id":"f209f2a2-5723-43ba-82a1-fc17edfc8c67","difficultyBand":"easy","calculatorUse":null,"totalMarks":3,"questionType":"LA","isCalculation":true},{"id":"9d3c6e50-5295-4a06-b680-b33bd067334d","difficultyBand":"medium","calculatorUse":null,"totalMarks":3,"questionType":"LA","isCalculation":true},{"id":"ccd01f36-c50a-4e45-ad97-44a918590c37","difficultyBand":"medium","calculatorUse":null,"totalMarks":3,"questionType":"LA","isCalculation":true},{"id":"61bf9309-865b-4108-a293-c8fc670013b0","difficultyBand":"medium","calculatorUse":null,"totalMarks":3,"questionType":"LA","isCalculation":true},{"id":"ba7e7e9f-cffd-4764-bae7-369bfddc13e7","difficultyBand":"medium","calculatorUse":null,"totalMarks":5,"questionType":"LA","isCalculation":true},{"id":"bf5c00ea-3f06-43fb-bd0d-ec2689b1e986","difficultyBand":"easy","calculatorUse":null,"totalMarks":3,"questionType":"LA","isCalculation":true},{"id":"edbf14ba-6206-41a7-aaa9-7fc7f936d530","difficultyBand":"medium","calculatorUse":null,"totalMarks":3,"questionType":"LA","isCalculation":true},{"id":"d877c2ca-f576-439d-9a69-de51fbdd6d47","difficultyBand":"medium","calculatorUse":null,"totalMarks":3,"questionType":"LA","isCalculation":true},{"id":"f9c1abed-3a80-46e0-87b8-dba96c072800","difficultyBand":"medium","calculatorUse":null,"totalMarks":3,"questionType":"LA","isCalculation":true},{"id":"1fa2f970-a7dc-494e-8df0-cb6f72019df4","difficultyBand":"medium","calculatorUse":null,"totalMarks":4,"questionType":"LA","isCalculation":true},{"id":"5e736fe7-d52f-4446-9999-56c60bf286a7","difficultyBand":"hard","calculatorUse":null,"totalMarks":8,"questionType":"LA","isCalculation":true},{"id":"60f1d250-7fc0-4450-bfae-f94c3355b1db","difficultyBand":"medium","calculatorUse":null,"totalMarks":4,"questionType":"LA","isCalculation":false},{"id":"faa915aa-44fb-41cd-9000-18defed8f4cf","difficultyBand":"medium","calculatorUse":null,"totalMarks":9,"questionType":"LA","isCalculation":true}],"title":"1.3 Partial Fractions","topicId":"5d96c6a6-9bd0-49bd-9a4a-f7ce0af48c3d","topicSidebar":"$L3b","topicSlug":"a-level-pure-mathematics-1-3-partial-fractions","topicTitle":"1.3 Partial Fractions"}]}],"$L3c","$L3d"]}]

1.3 Partial Fractions

Question 2

1.3 Partial Fractions Questions

1.3 Partial Fractions

Question 2

1.3 Partial Fractions Questions