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bg-background","children":["$","$L3b",null,{"backHref":"/gcse/maths/aqa/questions","basePath":"/gcse/maths/aqa/grade-8-9-8-1-quadratic-simultaneous-equations/questions","formulaUrl":"https://www.aqa.org.uk/files/resources.mathematics.AQA-8300H-FS-INS-2025_PDF/c87a0d4512612ebfb053370b84e9339e546e9e27.pdf","initialIndex":24,"initialQuestionWindow":[{"id":"9d0ab159-7531-4fc5-b8a8-094cb630657a","revisionId":"01561f85-fbd3-44da-afb7-b16dd65fb07b","questionNumber":null,"questionType":"LA","specification":"","difficulty":3,"difficultyBand":"medium","totalMarks":5,"sourceTitle":null,"calculatorUse":"non_calc","isCalculation":true,"parts":[{"id":"066f7be6-5ce5-4619-9041-8f2bb551320c","label":"a","content":"Solve the simultaneous equations\n\n$$x^2 + y^2 = 34$$\n\n$$y - x = 2$$","markscheme":"- Rearranges linear equation: $y = x + 2$ (or $x = y - 2$) M1\n- Substitutes into quadratic: $x^2 + (x + 2)^2 = 34$ M1\n- Simplifies to $2x^2 + 4x - 30 = 0$ or $x^2 + 2x - 15 = 0$ M1\n- Solves for $x$: $(x + 5)(x - 3) = 0$ so $x = -5$ or $x = 3$ M1\n- Finds corresponding $y$ values: $y = -3$ and $y = 5$ A1","marks":5,"order":0,"markingRubric":{"totalMarks":5,"points":[{"id":"p1","marks":1,"description":"Rearranges the linear equation to make one variable the subject.","code":"M"},{"id":"p2","marks":1,"description":"Substitutes the linear expression into the quadratic equation.","code":"M"},{"id":"p3","marks":1,"description":"Simplifies the resulting equation into a standard quadratic form $ax^2 + bx + c = 0$.","code":"M"},{"id":"p4","marks":1,"description":"Solves the quadratic equation to find two values for the first variable.","code":"M"},{"id":"p5","marks":1,"description":"Finds the corresponding values for the second variable to form two correct pairs.","code":"A"}],"finalAnswers":["x = -5, y = -3","x = 3, y = 5"],"generatedAt":"2026-05-06T14:07:17.363Z","modelVersion":"gemini-3-flash-preview"},"tableImages":[],"explanation":{"methods":[{"id":"substitution","label":"Substitution","steps":[{"type":"hyper_latex","block":{"nodes":[{"latex":"$$y-x=2$"},{"latex":"$$y=\\color{@sky}{x+2}$"}]},"order":0,"title":"Rearrange the line","description":"Add x to both sides to make y the subject. This earns the first M1 mark."},{"type":"hyper_latex","block":{"nodes":[{"latex":"$$x^2+y^2=34$"},{"latex":"$$x^2+(x+2)^2=34$"}]},"order":1,"title":"Substitute for y","description":"Replace y in the quadratic equation with x + 2. Writing this substitution correctly earns the second M1 mark."},{"type":"hyper_latex","block":{"nodes":[{"latex":"$$x^2+x^2+4x+4=34$"},{"latex":"$$2x^2+4x+4=34$"},{"latex":"$$2x^2+4x-30=0$"},{"latex":"$$x^2+2x-15=0$"}]},"order":2,"title":"Simplify the quadratic","description":"Expand the bracket, collect like terms, subtract 34 from both sides, then divide every term by 2. Reaching either quadratic form earns the third M1 mark."},{"type":"hyper_latex","block":{"nodes":[{"latex":"$$x^2+2x-15=0$"},{"latex":"$$(x+5)(x-3)=0$"},{"latex":"$$x+5=0$"},{"latex":"$$x=-5$"},{"latex":"$$x-3=0$"},{"latex":"$$x=3$"}]},"order":3,"title":"Factorise and solve","description":"Factorise the quadratic, then solve the two separate factor equations. Correctly factorising and obtaining both x-values earns the fourth M1 mark."},{"type":"hyper_latex","block":{"nodes":[{"latex":"$$x=-5$"},{"latex":"$$y=-5+2=-3$"},{"latex":"$$(x,y)=(-5,-3)$"},{"latex":"$$x=3$"},{"latex":"$$y=3+2=5$"},{"latex":"$$(x,y)=(3,5)$"}]},"order":4,"title":"Find both y-values","description":"Substitute each x-value into y = x + 2. Matching each x-value with its corresponding y-value earns the final A1 mark."}]}],"generatedAt":"2026-07-20T11:13:19.621Z","modelVersion":"gpt-5.6-sol","explanationVersion":"v3"},"options":[]}],"options":[],"explanation":{"methods":[{"id":"066f7be6-5ce5-4619-9041-8f2bb551320c:substitution_y","label":"Substitution for y","steps":[{"type":"text","order":0,"title":"Rearrange the linear equation","content":"To solve these simultaneous equations using the substitution method, we start by rearranging the linear equation to express $y$ in terms of $x$.\n\nGiven the linear equation:\n$$y - x = 2$$\n\nAdd $x$ to both sides to solve for $y$:\n$$y = x + 2$$\n\nThis earns the first method mark **(M1)**.","highlights":[{"text":"y - x = 2","location":"specification"}]},{"type":"hyper_latex","block":{"nodes":[{"text":"Substitute our expression for $y$ into the quadratic equation:"},{"latex":"$$$x^2 + \\htmlData{anchor=y-var}{\\color{@sky}{y}}^2 = 34$$"},{"latex":"$$$x^2 + (\\htmlData{anchor=y-expr}{\\color{@sky}{x + 2}})^2 = 34$$"}],"highlights":[{"color":"sky","style":"text-color","anchor":"y-var"},{"color":"sky","style":"text-color","anchor":"y-expr"}],"connections":[{"to":"y-expr","from":"y-var","color":"sky","label":"substitute","style":"solid"}]},"order":1,"title":"Substitute into quadratic","description":"We replace $y$ in the quadratic equation with $(x + 2)$ to get an equation in terms of $x$ only. This earns the second method mark (M1)."},{"type":"text","order":2,"title":"Expand and simplify","content":"Now we expand the bracket $(x + 2)^2$ and simplify the equation:\n\n$$(x + 2)^2 = (x + 2)(x + 2) = x^2 + 4x + 4$$\n\nSubstitute this back into our equation:\n$$x^2 + (x^2 + 4x + 4) = 34$$\n\nCombine like terms:\n$$2x^2 + 4x + 4 = 34$$\n\nSubtract $34$ from both sides to set the quadratic to $0$:\n$$2x^2 + 4x - 30 = 0$$\n\nTo make factorising easier, we can divide the entire equation by $2$:\n$$x^2 + 2x - 15 = 0$$\n\nThis simplified quadratic equation earns the third method mark **(M1)**."},{"type":"text","order":3,"title":"Solve for x","content":"Next, we solve the quadratic equation $x^2 + 2x - 15 = 0$ by factorising.\n\nWe need two numbers that multiply to $-15$ and add to $+2$. These numbers are $+5$ and $-3$:\n$$(x + 5)(x - 3) = 0$$\n\nSet each bracket to $0$ to find the solutions for $x$:\n- $x + 5 = 0 \\implies x = -5$\n- $x - 3 = 0 \\implies x = 3$\n\nThis earns the fourth method mark **(M1)**."},{"type":"text","order":4,"title":"Find corresponding y values","content":"Finally, substitute each $x$ value back into our rearranged linear equation $y = x + 2$ to find the corresponding values for $y$.\n\n**When $x = -5$:**\n$$y = -5 + 2 = -3$$\n\n**When $x = 3$:**\n$$y = 3 + 2 = 5$$\n\nSo, the two pairs of solutions are:\n$$(x = -5, y = -3) \\quad \\text{and} \\quad (x = 3, y = 5)$$\n\nFinding both correct pairs of values completes the solution and earns the accuracy mark **(A1)**."}]},{"id":"066f7be6-5ce5-4619-9041-8f2bb551320c:substitution_x","label":"Substitution for x","steps":[{"type":"text","order":0,"title":"Rearrange for x","content":"To solve this system using the substitution method for $x$, we first rearrange the linear equation to express $x$ in terms of $y$.\n\nStarting with the linear equation:\n$$y - x = 2$$\n\nAdd $x$ and subtract $2$ from both sides to get:\n$$x = y - 2$$","highlights":[{"text":"y - x = 2","location":"specification"}]},{"type":"hyper_latex","block":{"nodes":[{"latex":"$$\\{ \n\\begin{aligned} \n&x^2 + y^2 = 34 \\\\\n&\\htmlData{anchor=x-val}{\\color{@sky}{x}} = \\htmlData{anchor=y-expr}{\\color{@sky}{y - 2}} \n\\end{aligned}\n$"},{"latex":"$$\\left(\\htmlData{anchor=sub-x}{\\color{@sky}{y - 2}}\\right)^2 + y^2 = 34$"}],"highlights":[{"color":"sky","style":"text-color","anchor":"x-val"},{"color":"sky","style":"text-color","anchor":"y-expr"},{"color":"sky","style":"box","anchor":"sub-x"}],"connections":[{"to":"sub-x","from":"y-expr","color":"sky","label":"substitute x","style":"solid"}]},"order":1,"title":"Substitute into quadratic","description":"Substitute our expression for $x$ into the quadratic equation."},{"type":"text","order":2,"title":"Expand and simplify","content":"Now expand the bracket and simplify the quadratic equation:\n\n$$(y - 2)^2 + y^2 = 34$$\n$$(y^2 - 4y + 4) + y^2 = 34$$\n$$2y^2 - 4y + 4 = 34$$\n\nSubtract $34$ from both sides to set the quadratic to $0$:\n$$2y^2 - 4y - 30 = 0$$"},{"type":"text","order":3,"title":"Solve for y","content":"To make solving easier, divide every term by $2$:\n$$y^2 - 2y - 15 = 0$$\n\nNow, factorise the quadratic. We need two numbers that multiply to $-15$ and add up to $-2$. These numbers are $-5$ and $+3$:\n$$(y - 5)(y + 3) = 0$$\n\nThis gives us two possible values for $y$:\n$$y = 5 \\quad \\text{or} \\quad y = -3$$"},{"type":"text","order":4,"title":"Find corresponding x","content":"Substitute each $y$-value back into the rearranged linear formula $$x = y - 2$$ to find the corresponding values of $x$.\n\n- **Case 1:** When $y = 5$\n $$x = 5 - 2 = 3$$\n\n- **Case 2:** When $y = -3$\n $$x = -3 - 2 = -5$$\n\nThus, the solutions are:\n$$x = 3, y = 5 \\quad \\text{and} \\quad x = -5, y = -3$$"}]}],"generatedAt":"2026-06-13T01:00:25.651Z","modelVersion":"gemini-3.5-flash","explanationVersion":"v2"},"firstAttemptFullMarks":{"percent":48},"walkthroughVideo":{"url":"https://assets.mathsgenie.co.uk/walkthroughs/videos/9d0ab159-7531-4fc5-b8a8-094cb630657a.mp4","durationSeconds":209,"posterUrl":"https://assets.mathsgenie.co.uk/walkthroughs/videos/9d0ab159-7531-4fc5-b8a8-094cb630657a.jpg"}},{"id":"ba64323a-522d-444c-bb02-5c99ef808894","revisionId":"61c97560-f355-4148-a321-e67004a3f3bb","questionNumber":null,"questionType":"LA","specification":"","difficulty":3,"difficultyBand":"medium","totalMarks":5,"sourceTitle":null,"calculatorUse":"non_calc","isCalculation":true,"parts":[{"id":"54588627-dab8-410e-bdf2-06765bb28ccc","label":"a","content":"Solve the simultaneous equations\n\n$$x^2 + y^2 = 26$$\n\n$$x + y = 6$$","markscheme":"- Rearranges linear equation: $x = 6 - y$ (or $y = 6 - x$) M1\n- Substitutes into quadratic: $(6 - y)^2 + y^2 = 26$ M1\n- Simplifies to $2y^2 - 12y + 10 = 0$ or $y^2 - 6y + 5 = 0$ M1\n- Solves for $y$: $(y - 5)(y - 1) = 0$ so $y = 5$ or $y = 1$ M1\n- Finds $x$: $x = 1$ or $x = 5$ A1","marks":5,"order":0,"markingRubric":{"totalMarks":5,"points":[{"id":"p1","marks":1,"description":"Rearranges the linear equation to make $x$ or $y$ the subject.","code":"M","acceptable":["$$x = 6 - y$","$$y = 6 - x$"]},{"id":"p2","marks":1,"description":"Substitutes the linear expression into the quadratic equation.","code":"M","acceptable":["$$(6 - y)^2 + y^2 = 26$","$$x^2 + (6 - x)^2 = 26$"]},{"id":"p3","marks":1,"description":"Simplifies the resulting equation into a standard quadratic form equal to zero.","code":"M","acceptable":["$$2y^2 - 12y + 10 = 0$","$$y^2 - 6y + 5 = 0$","$$2x^2 - 12x + 10 = 0$","$$x^2 - 6x + 5 = 0$"]},{"id":"p4","marks":1,"description":"Solves the quadratic equation to find the values of one variable.","code":"M","acceptable":["$$(y-5)(y-1)=0$","$$(x-5)(x-1)=0$"]},{"id":"p5","marks":1,"description":"Finds the corresponding values for the second variable to produce two pairs of solutions.","code":"A"}],"finalAnswers":["x = 1, y = 5 and x = 5, y = 1"],"generatedAt":"2026-05-06T14:07:41.061Z","modelVersion":"gemini-3-flash-preview"},"tableImages":[],"explanation":{"methods":[{"id":"substitution","label":"Substitution","steps":[{"type":"hyper_latex","block":{"nodes":[{"latex":"$$\\htmlData{anchor=line}{x+y=6}$"},{"latex":"$$\\htmlData{anchor=rearranged}{\\color{@sky}{x=6-y}}$"}],"connections":[{"to":"rearranged","from":"line","color":"sky","label":"subtract y"}]},"order":0,"title":"Rearrange the line","description":"Make x the subject of the linear equation. This earns the first M1 mark."},{"type":"hyper_latex","block":{"nodes":[{"latex":"$$x=\\color{@orange}{6-y}$"},{"latex":"$$26=y^2+\\htmlData{anchor=original_x_squared}{\\color{@orange}{x^2}}$"},{"latex":"$$26=y^2+\\htmlData{anchor=replacement_squared}{\\color{@orange}{(6-y)^2}}$"}],"connections":[{"to":"replacement_squared","from":"original_x_squared","color":"orange","label":"substitute"}]},"order":1,"title":"Substitute for x","description":"Replace x in the quadratic equation with 6 − y. Writing this substitution earns the second M1 mark."},{"type":"hyper_latex","block":{"nodes":[{"latex":"$$\\htmlData{anchor=substituted_eq}{(6-y)^2+y^2=26}$"},{"latex":"$$\\htmlData{anchor=expanded_eq}{36-12y+y^2+y^2=26}$"},{"latex":"$$\\htmlData{anchor=rearranged_eq}{2y^2-12y+10=0}$"},{"latex":"$$\\htmlData{anchor=simpler_eq}{y^2-6y+5=0}$"}],"connections":[{"to":"expanded_eq","from":"substituted_eq","color":"orange","label":"expand"},{"to":"rearranged_eq","from":"expanded_eq","color":"sky","label":"collect and rearrange"},{"to":"simpler_eq","from":"rearranged_eq","color":"teal","label":"divide by 2"}]},"order":2,"title":"Form the quadratic","description":"Expand the bracket, then collect like terms and rearrange. Dividing throughout by 2 gives an equivalent, simpler quadratic; reaching either final form earns the third M1 mark."},{"type":"hyper_latex","block":{"nodes":[{"latex":"$$y^2-6y+5=0$"},{"latex":"$$\\htmlData{anchor=factors}{\\color{@green}{(y-5)(y-1)=0}}$"},{"latex":"$$\\htmlData{anchor=roots}{\\color{@green}{y=5\\text{ or }y=1}}$"}],"connections":[{"to":"roots","from":"factors","color":"green","label":"either factor is zero"}]},"order":3,"title":"Factorise and solve","description":"Find two numbers that multiply to 5 and add to −6. Setting each factor equal to zero gives both possible values of y and earns the fourth M1 mark."},{"type":"hyper_latex","block":{"nodes":[{"latex":"$$\\htmlData{anchor=x_from_five}{x=6-5}$"},{"latex":"$$\\htmlData{anchor=x_one}{\\color{@purple}{x=1}}$"},{"latex":"$$\\htmlData{anchor=x_from_one}{x=6-1}$"},{"latex":"$$\\htmlData{anchor=x_five}{\\color{@purple}{x=5}}$"},{"latex":"$$\\color{@purple}{(x,y)=(1,5)\\text{ or }(5,1)}$"}],"connections":[{"to":"x_one","from":"x_from_five","color":"purple","label":"simplify"},{"to":"x_five","from":"x_from_one","color":"purple","label":"simplify"}]},"order":4,"title":"Find matching x-values","description":"Substitute each y-value into x = 6 − y. Keep the values paired correctly; finding both matching x-values earns A1."}]}],"generatedAt":"2026-07-19T16:45:03.427Z","modelVersion":"gpt-5.6-sol","explanationVersion":"v3"},"options":[]}],"options":[],"explanation":{"methods":[{"id":"54588627-dab8-410e-bdf2-06765bb28ccc:substitution","label":"Substitution Method","steps":[{"type":"text","order":0,"title":"Rearrange the linear equation","content":"To use the substitution method, we start by rearranging the simpler linear equation to express one variable in terms of the other.\n\nLet's rearrange $x + y = 6$ to make $x$ the subject:\n\n$$x = 6 - y$$","highlights":[{"text":"x + y = 6","location":"specification"}]},{"type":"hyper_latex","block":{"nodes":[{"latex":"$$\\{\\htmlData{anchor=x-val}{\\color{@sky}{x}} = \\htmlData{anchor=sub-val}{\\color{@sky}{6 - y}}\\}$"},{"latex":"$$\\htmlData{anchor=quad-eq}{\\color{@sky}{x}}^2 + y^2 = 26$"},{"latex":"$$\\htmlData{anchor=substituted-eq}{\\color{@sky}{(6 - y)}}^2 + y^2 = 26$"}],"highlights":[{"color":"sky","style":"text-color","anchor":"x-val"},{"color":"sky","style":"text-color","anchor":"sub-val"},{"color":"sky","style":"text-color","anchor":"quad-eq"},{"color":"sky","style":"box","anchor":"substituted-eq"}],"connections":[{"to":"substituted-eq","from":"sub-val","color":"sky","label":"substitute","style":"solid"}]},"order":1,"title":"Substitute into quadratic","description":"Now, we substitute this expression for $x$ into the quadratic equation $x^2 + y^2 = 26$."},{"type":"text","order":2,"title":"Expand and simplify","content":"Next, expand the bracket and collect like terms to form a standard quadratic equation.\n\nFirst, expand $(6 - y)^2$:\n\n$$(6 - y)^2 = 36 - 12y + y^2$$\n\nNow, substitute this back into our equation:\n\n$$\\begin{aligned}\n(36 - 12y + y^2) + y^2 &= 26 \\\\\n2y^2 - 12y + 36 &= 26\n\\end{aligned}$$\n\nSubtract $26$ from both sides to set the quadratic to $0$:\n\n$$2y^2 - 12y + 10 = 0$$"},{"type":"text","order":3,"title":"Solve the quadratic equation","content":"To make solving easier, divide the entire equation by $2$:\n\n$$y^2 - 6y + 5 = 0$$\n\nWe can factorise this quadratic by finding two numbers that multiply to $+5$ and add up to $-6$. These numbers are $-5$ and $-1$:\n\n$$(y - 5)(y - 1) = 0$$\n\nThis gives two possible solutions for $y$:\n\n$$y = 5 \\quad \\text{or} \\quad y = 1$$"},{"type":"text","order":4,"title":"Find corresponding x-values","content":"Now, substitute each $y$-value back into our rearranged linear equation $x = 6 - y$ to find the corresponding values of $x$.\n\n- If $y = 5$:\n $$x = 6 - 5 = 1$$\n\n- If $y = 1$:\n $$x = 6 - 1 = 5$$\n\nSo, the solutions are:\n\n$$x = 1, y = 5 \\quad \\text{and} \\quad x = 5, y = 1$$"}]},{"id":"54588627-dab8-410e-bdf2-06765bb28ccc:algebraic_identities","label":"Algebraic Identities Method","steps":[{"type":"text","order":0,"title":"Recall the algebraic identity","content":"We are given the system of equations:\n\n1) $$x^2 + y^2 = 26$$\n2) $$x + y = 6$$\n\nInstead of substituting one variable into another, we can use the algebraic identity for the square of a sum:\n\n$$(x + y)^2 = x^2 + y^2 + 2xy$$"},{"type":"hyper_latex","block":{"nodes":[{"latex":"$$\\{ \\htmlData{anchor=sum-val}{\\color{@sky}{x + y = 6}}, \\quad \\htmlData{anchor=squares-val}{\\color{@amber}{x^2 + y^2 = 26}} \\}$"},{"latex":"$$(\\htmlData{anchor=sum-sub}{\\color{@sky}{x + y}})^2 = (\\htmlData{anchor=squares-sub}{\\color{@amber}{x^2 + y^2}}) + 2xy$"},{"latex":"$$(\\color{@sky}{6})^2 = \\color{@amber}{26} + 2xy$"}],"highlights":[{"color":"sky","style":"text-color","anchor":"sum-val"},{"color":"amber","style":"text-color","anchor":"squares-val"}],"connections":[{"to":"sum-sub","from":"sum-val","color":"sky","label":"substitute sum","style":"solid"},{"to":"squares-sub","from":"squares-val","color":"amber","label":"substitute sum of squares","style":"solid"}]},"order":1,"title":"Substitute known values","description":"We substitute our known values for $(x + y)$ and $(x^2 + y^2)$ directly into the algebraic identity."},{"type":"text","order":2,"title":"Solve for the product xy","content":"Now we simplify the equation to find the value of the product $xy$:\n\n$$\\begin{aligned} 36 &= 26 + 2xy \\\\ 10 &= 2xy \\\\ xy &= 5 \\end{aligned}$$"},{"type":"text","order":3,"title":"Form a quadratic equation","content":"We now have a system of sum and product equations:\n\n* **Sum:** $x + y = 6$\n* **Product:** $xy = 5$\n\nAccording to the properties of quadratic equations, if two numbers have a sum $S$ and a product $P$, they are the roots of the quadratic equation:\n\n$$t^2 - St + P = 0$$\n\nSubstituting $S = 6$ and $P = 5$ gives:\n\n$$t^2 - 6t + 5 = 0$$"},{"type":"text","order":4,"title":"Factorise and solve","content":"We can factorise this quadratic equation to find the possible values for $t$ (which represent our variables $x$ and $y$):\n\n$$(t - 5)(t - 1) = 0$$\n\nThis gives two solutions for $t$:\n\n$$t = 5 \\quad \\text{or} \\quad t = 1$$"},{"type":"text","order":5,"title":"State the final solutions","content":"Since $x$ and $y$ are the roots, one must be $5$ and the other must be $1$:\n\n* If $x = 1$, then $y = 5$\n* If $x = 5$, then $y = 1$\n\nThus, the solutions are:\n\n$$(1, 5) \\quad \\text{and} \\quad (5, 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Simultaneous Equations","topicId":"04a1cdc8-5e08-44e9-aeb1-9c07bd3bf814","topicSidebar":"$L3c","topicSlug":"grade-8-9-8-1-quadratic-simultaneous-equations","topicTitle":"Quadratic Simultaneous Equations"}]}],"$L3d","$L3e"]}]

Quadratic Simultaneous Equations

Question 5

Quadratic Simultaneous Equations Questions

Quadratic Simultaneous Equations

Question 5

Quadratic Simultaneous Equations Questions