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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":19,"initialQuestionWindow":[{"id":"9312d566-1d8c-4634-b06e-3c683f8492d1","revisionId":"942e64cc-1e03-461d-88a7-d374f21ae726","questionNumber":null,"questionType":"LA","specification":"","difficulty":3,"difficultyBand":"medium","totalMarks":5,"sourceTitle":null,"calculatorUse":"non_calc","isCalculation":true,"parts":[{"id":"d52500d5-fe25-415f-bf48-8c84c7c7fe23","label":"a","content":"Solve the simultaneous equations\n\n$$x^2 + y^2 = 17$$\n\n$$y = x + 3$$","markscheme":"- Substitutes $y = x + 3$ into the first equation: $x^2 + (x + 3)^2 = 17$ M1\n- Expands and simplifies to a quadratic equation: $2x^2 + 6x - 8 = 0$ or $x^2 + 3x - 4 = 0$ M1\n- Factorises or uses formula to find $x$: $(x + 4)(x - 1) = 0$ so $x = -4$ or $x = 1$ M1\n- Substitutes $x$ values back to find $y$: $y = -4 + 3 = -1$ and $y = 1 + 3 = 4$ M1\n- Final answers: $x = -4, y = -1$ and $x = 1, y = 4$ A1","marks":5,"order":0,"markingRubric":{"totalMarks":5,"points":[{"id":"p1","marks":1,"description":"Substitutes the linear expression into the quadratic equation correctly.","code":"M"},{"id":"p2","marks":1,"description":"Expands the brackets and simplifies to a standard quadratic form $ax^2 + bx + c = 0$.","code":"M"},{"id":"p3","marks":1,"description":"Solves the quadratic equation to find two values for $x$.","code":"M"},{"id":"p4","marks":1,"description":"Substitutes both $x$ values back into the linear equation to find corresponding $y$ values.","code":"M"},{"id":"p5","marks":1,"description":"Provides both pairs of correct solutions clearly.","code":"A"}],"finalAnswers":["x = -4, y = -1 and x = 1, y = 4","(-4, -1), (1, 4)"],"generatedAt":"2026-05-06T14:07:13.268Z","modelVersion":"gemini-3-flash-preview"},"tableImages":[],"explanation":null,"options":[]}],"options":[],"explanation":{"methods":[{"id":"d52500d5-fe25-415f-bf48-8c84c7c7fe23:substitution_y","label":"Substitution for y","steps":[{"type":"text","order":0,"title":"Substitute the linear expression","content":"To solve these simultaneous equations, we can substitute the linear equation directly into the quadratic equation.\n\nWe are given:\n1. $$x^2 + y^2 = 17$$\n2. $$y = x + 3$$\n\nSubstitute the expression for $y$ from equation (2) into equation (1):\n\n$$x^2 + (x + 3)^2 = 17$$"},{"type":"text","order":1,"title":"Expand and simplify","content":"Next, let's expand the squared term $(x + 3)^2$ and simplify the equation:\n\n$$(x + 3)^2 = x^2 + 6x + 9$$\n\nNow substitute this back into our equation:\n\n$$x^2 + (x^2 + 6x + 9) = 17$$\n\n$$2x^2 + 6x + 9 = 17$$\n\nSubtract $17$ from both sides to set the quadratic equation to $0$:\n\n$$2x^2 + 6x - 8 = 0$$"},{"type":"text","order":2,"title":"Divide by two","content":"To make the quadratic equation easier to factorise, we can divide every term by $2$:\n\n$$x^2 + 3x - 4 = 0$$"},{"type":"text","order":3,"title":"Factorise the quadratic","content":"Now, we find two numbers that multiply to $-4$ and add to $3$. These numbers are $+4$ and $-1$.\n\nFactorising gives:\n\n$$(x + 4)(x - 1) = 0$$\n\nThis gives two possible values for $x$:\n\n$$x = -4 \\quad \\text{or} \\quad x = 1$$"},{"type":"text","order":4,"title":"Find corresponding y values","content":"Substitute each $x$ value back into the simpler linear equation $y = x + 3$ to find the corresponding $y$ values:\n\n* For $x = -4$:\n $$y = -4 + 3 = -1$$\n\n* For $x = 1$:\n $$y = 1 + 3 = 4$$\n\nSo, our two pairs of solutions are:\n\n$$x = -4, y = -1 \\quad \\text{and} \\quad x = 1, y = 4$$"}]},{"id":"d52500d5-fe25-415f-bf48-8c84c7c7fe23:substitution_x","label":"Substitution for x","steps":[{"type":"text","order":0,"title":"Rearrange for x","content":"The question asks us to solve the simultaneous equations:\n\n1) $$x^2 + y^2 = 17$$\n2) $$y = x + 3$$\n\nTo use the **Substitution for $x$** approach, we first rearrange the linear equation, equation (2), to make $x$ the subject:\n\n$$x = y - 3$$"},{"type":"hyper_latex","block":{"nodes":[{"latex":"$$(\\htmlData{anchor=x-expr}{\\color{@sky}{y - 3}})^2 + y^2 = 17$"},{"latex":"$$\\htmlData{anchor=expanded}{\\color{@sky}{y^2 - 6y + 9}} + y^2 = 17$"}],"highlights":[{"color":"sky","style":"box","anchor":"x-expr"},{"color":"sky","style":"underline","anchor":"expanded"}],"connections":[{"to":"expanded","from":"x-expr","color":"sky","label":"expand","style":"solid"}]},"order":1,"title":"Substitute into quadratic","description":"Now, substitute this expression for $x$ into the quadratic equation: $x^2 + y^2 = 17$."},{"type":"text","order":2,"title":"Simplify the equation","content":"Next, collect like terms and rearrange the equation into standard quadratic form ($ay^2 + by + c = 0$):\n\n$$\\begin{aligned} (y^2 - 6y + 9) + y^2 &= 17 \\\\ 2y^2 - 6y + 9 &= 17 \\\\ 2y^2 - 6y - 8 &= 0 \\end{aligned}$$\n\nTo make solving easier, we can divide the entire equation by $2$:\n\n$$y^2 - 3y - 4 = 0$$"},{"type":"text","order":3,"title":"Solve for y","content":"Now, we solve the simplified quadratic equation, $$y^2 - 3y - 4 = 0$$, by factorising. \n\nWe need two numbers that multiply to $-4$ and add to $-3$. These numbers are $-4$ and $+1$:\n\n$$(y - 4)(y + 1) = 0$$\n\nSetting each factor to zero gives our $y$-coordinates:\n\n$$y - 4 = 0 \\implies y = 4$$\n$$y + 1 = 0 \\implies y = -1$$"},{"type":"text","order":4,"title":"Find corresponding x values","content":"Finally, we substitute our $y$ values back into the rearranged linear equation, $$x = y - 3$$, to find the corresponding $x$ coordinates.\n\n- **If $y = 4$:**\n $$x = 4 - 3 = 1$$\n\n- **If $y = -1$:**\n $$x = -1 - 3 = -4$$\n\nSo, our two pairs of solutions are:\n\n$$\\begin{aligned} x = 1, \\quad y = 4 \\\\ x = -4, \\quad y = -1 \\end{aligned}$$"}]},{"id":"d52500d5-fe25-415f-bf48-8c84c7c7fe23:graphical","label":"Graphical Approach","steps":[{"type":"text","order":0,"title":"Identify the equations","content":"To solve simultaneous equations graphically, we need to plot both equations on a single coordinate grid. The points where the two graphs intersect (cross each other) represent the solutions to the system of equations.\n\nOur two equations are:\n1. A circle: \n\n$$x^2 + y^2 = 17$$\n\n2. A straight line:\n\n$$y = x + 3$$"},{"type":"text","order":1,"title":"Understand the circle","content":"The equation of a circle with center at the origin $(0, 0)$ is given by $$x^2 + y^2 = r^2$$, where $r$ is the radius.\n\nFor our circle equation, $$x^2 + y^2 = 17$$:\n- The center is $(0, 0)$.\n- The radius is $r = \n\\sqrt{17} \\approx 4.12$.\n\nTo draw this circle, we can plot key points on the grid. Specifically, let's look for integer coordinates that satisfy $x^2 + y^2 = 17$:\n- If $x = \\pm 1$, then $y^2 = 16 \\implies y = \\pm 4$. This gives points: $(1, 4)$, $(1, -4)$, $(-1, 4)$, and $(-1, -4)$.\n- If $x = \\pm 4$, then $y^2 = 1 \\implies y = \\pm 1$. This gives points: $(4, 1)$, $(4, -1)$, $(-4, 1)$, and $(-4, -1)$.\n- If $x = 0$, $y \\approx \\pm 4.12$.\n\nWe plot these points and draw a smooth circle of radius $\\approx 4.1$ centered at $(0,0)$."},{"type":"text","order":2,"title":"Plot the line","content":"The line equation is in gradient-intercept form, $$y = mx + c$$:\n\n$$y = x + 3$$\n\n- The $y$-intercept is $(0, 3)$ (since $c = 3$).\n- The gradient is $1$, which means for every $1$ unit we move to the right, we go up by $1$ unit.\n\nLet's calculate a few coordinate points to draw a straight line:\n- If $x = -4$, then $y = -4 + 3 = -1 \\implies (-4, -1)$\n- If $x = 0$, then $y = 0 + 3 = 3 \\implies (0, 3)$\n- If $x = 1$, then $y = 1 + 3 = 4 \\implies (1, 4)$\n\nPlot these points on the same grid and join them with a straight line."},{"type":"text","order":3,"title":"Find the intersections","content":"Now look at where the line $$y = x + 3$$ crosses the circle $$x^2 + y^2 = 17$$.\n\nBy observing your grid, the line and the circle intersect at exactly two points:\n1. In the first quadrant at the point $(1, 4)$.\n2. In the third quadrant at the point $(-4, -1)$."},{"type":"text","order":4,"title":"Use a graphing tool","content":"To verify these intersections accurately, we can use a graphing calculator or Desmos to plot both curves and locate the precise intersection points.","calculator":[{"gdc":{"instructions":"1. Go to Graph mode.\n2. Enter the circle relation or plot the two semi-circles: Y1 = \\sqrt{17 - x^2} and Y2 = -\\sqrt{17 - x^2}.\n3. Enter the line: Y3 = x + 3.\n4. Draw the graphs and use the 'Intersection' tool (G-Solve) to find where they meet."},"ti84":{"instructions":"1. Press [Y=].\n2. Since the circle cannot be entered directly as one function, solve for y: y = \\sqrt{17 - x^2} and y = -\\sqrt{17 - x^2}.\n3. Enter \\\\Y1 = \\sqrt{17 - X^2}\\\\ and \\\\Y2 = -\\sqrt{17 - X^2}\\\\.\n4. Enter \\\\Y3 = X + 3\\\\.\n5. Press [GRAPH].\n6. Press [2nd] [TRACE] and select '5: intersect' to find the intersection points at (1, 4) and (-4, -1)."},"desmos":{"expression":"x^2+y^2=17\\ny=x+3","instructions":"1. Type `x^2 + y^2 = 17` in the first input box.\n2. Type `y = x + 3` in the second input box.\n3. Click on the gray intersection points on the graph to display their coordinates."},"description":"Plot the equations to visually identify the intersections."}]},{"type":"text","order":5,"title":"Write down solutions","content":"Our graphical solution gives the coordinates of the intersection points, which correspond to the solutions of our simultaneous equations:\n\n$$x = 1, y = 4$$\n\nand \n\n$$x = -4, y = -1$$"}]}],"generatedAt":"2026-06-12T23:50:58.270Z","modelVersion":"gemini-3.5-flash","explanationVersion":"v2"},"firstAttemptFullMarks":{"percent":60},"walkthroughVideo":{"url":"https://assets.mathsgenie.co.uk/walkthroughs/videos/9312d566-1d8c-4634-b06e-3c683f8492d1.mp4","durationSeconds":186,"posterUrl":"https://assets.mathsgenie.co.uk/walkthroughs/videos/9312d566-1d8c-4634-b06e-3c683f8492d1.jpg"}},{"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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Quadratic Simultaneous Equations

Question 4

Quadratic Simultaneous Equations Questions

Quadratic Simultaneous Equations

Question 4

Quadratic Simultaneous Equations Questions