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quadratic form: $2x^2 - 6x - 8 = 0$ or $x^2 - 3x - 4 = 0$ M1\n- Solves the quadratic equation: $(x - 4)(x + 1) = 0$ so $x = 4$ or $x = -1$ M1\n- Finds corresponding $y$ values: $y = 4 - 3 = 1$ and $y = -1 - 3 = -4$ M1\n- Final answers stated as pairs or coordinates: $(4, 1)$ and $(-1, -4)$ A1","marks":5,"order":0,"markingRubric":{"totalMarks":5,"points":[{"id":"p1","marks":1,"description":"Substitution of linear equation into quadratic equation.","code":"M"},{"id":"p2","marks":1,"description":"Expansion of brackets and simplification to a three-term quadratic equal to zero.","code":"M"},{"id":"p3","marks":1,"description":"Correct method to solve their quadratic equation (factorisation, formula, or completing the square).","code":"M"},{"id":"p4","marks":1,"description":"Correct method to find both corresponding values of the second variable.","code":"M"},{"id":"p5","marks":1,"description":"Both pairs of solutions correctly identified.","code":"A"}],"finalAnswers":["(4, 1) and (-1, -4)","x = 4, y = 1; x = -1, y = -4"],"generatedAt":"2026-05-06T14:07:18.379Z","modelVersion":"gemini-3-flash-preview"},"tableImages":[],"explanation":{"methods":[{"id":"substitution_factorising","label":"Substitution and factorising","steps":[{"type":"text","order":0,"title":"Link the equations","content":"At an intersection, the same $x$- and $y$-values satisfy **both** equations. Since the line already gives $y$ in terms of $x$, substitute $x-3$ for $y$ in the circle equation.","highlights":[{"text":"points of intersection of the circle and the straight line","location":"specification"}]},{"type":"hyper_latex","block":{"nodes":[{"latex":"$$x^2+\\color{@sky}{y^2}=17$"},{"latex":"$$x^2+\\color{@sky}{(x-3)^2}=17$"}]},"order":1,"title":"Substitute for y","description":"Replace y in the circle equation with the equal expression x − 3. Writing this substitution earns the first method mark, M1."},{"type":"hyper_latex","block":{"nodes":[{"latex":"$$x^2+\\color{@orange}{(x-3)^2}=17$"},{"latex":"$$x^2+\\color{@orange}{x^2-6x+9}=17$"},{"latex":"$$\\color{@teal}{2x^2-6x-8}=0$"},{"latex":"$$\\color{@teal}{x^2-3x-4}=0$"}]},"order":2,"title":"Form the quadratic","description":"Expand the squared bracket, collect like terms, then divide every term by 2. Reaching either quadratic form earns the second M1."},{"type":"hyper_latex","block":{"nodes":[{"latex":"$$x^2-3x-4=0$"},{"latex":"$$\\color{@purple}{(x-4)(x+1)}=0$"},{"latex":"$$x-4=0\\quad\\text{or}\\quad x+1=0$"},{"latex":"$$\\color{@green}{x=4}\\quad\\text{or}\\quad\\color{@orange}{x=-1}$"}]},"order":3,"title":"Solve by factorising","description":"Choose two numbers with product −4 and sum −3: −4 and 1. Solving the factorised quadratic earns the third M1."},{"type":"hyper_latex","block":{"nodes":[{"latex":"$$x=4:\\quad y=4-3$"},{"latex":"$$\\color{@green}{y=1}$"},{"latex":"$$x=-1:\\quad y=-1-3$"},{"latex":"$$\\color{@orange}{y=-4}$"}]},"order":4,"title":"Find both y-values","description":"Substitute each x-value into the line equation. Finding both corresponding y-values earns the fourth M1."},{"type":"text","order":5,"title":"State the coordinates","content":"Pair each $x$-value with its corresponding $y$-value. The points of intersection are\n\n$$(4,1)\\quad\\text{and}\\quad(-1,-4).$$\n\nStating both answers as coordinate pairs earns the final A1 mark. Keep the matching pairs together: $x=4$ goes with $y=1$, while $x=-1$ goes with $y=-4$.","highlights":[{"text":"Find the coordinates","location":"specification"}]}]}],"generatedAt":"2026-07-20T11:00:55.720Z","modelVersion":"gpt-5.6-sol","explanationVersion":"v3"},"options":[]}],"options":[],"explanation":{"methods":[{"id":"7c45f242-9697-4d60-94a9-a0b4b7dcbc30:substitution_y","label":"Substitution for y","steps":[{"type":"hyper_latex","block":{"nodes":[{"latex":"$$$x^2 + \\htmlData{anchor=y-var}{\\color{@sky}{y}}^2 = 17$$"},{"latex":"$$$\\htmlData{anchor=y-expr}{\\color{@sky}{y}} = x - 3$$"},{"latex":"$$$x^2 + (\\htmlData{anchor=y-sub}{\\color{@sky}{x - 3}})^2 = 17$$"}],"highlights":[{"color":"sky","style":"box","anchor":"y-expr"},{"color":"sky","style":"box","anchor":"y-sub"}],"connections":[{"to":"y-sub","from":"y-expr","color":"sky","label":"substitute","style":"solid"}]},"order":0,"title":"Substitute y into circle equation","description":"We are given the equation of a circle $x^2 + y^2 = 17$ and the equation of a line $y = x - 3$. Since the line equation is already solved for $y$, we can substitute this expression directly into the circle equation. This will give us a single quadratic equation containing only the variable $x$."},{"type":"text","order":1,"title":"Expand and simplify","content":"Now we expand the squared bracket and simplify the equation into a standard quadratic form $ax^2 + bx + c = 0$:\n\n$$\\begin{aligned} x^2 + (x - 3)^2 &= 17 \\\\ x^2 + (x^2 - 6x + 9) &= 17 \\\\ 2x^2 - 6x + 9 &= 17 \\end{aligned}$$\n\nSubtract $17$ from both sides to set the equation to $0$:\n\n$$2x^2 - 6x - 8 = 0$$\n\nTo make the equation easier to solve, we divide all terms by $2$:\n\n$$x^2 - 3x - 4 = 0$$"},{"type":"text","order":2,"title":"Solve the quadratic equation","content":"We can solve the quadratic equation $x^2 - 3x - 4 = 0$ by factorising. We need to find two numbers that multiply to $-4$ and add up to $-3$. These numbers are $-4$ and $+1$.\n\n$$\\begin{aligned} (x - 4)(x + 1) &= 0 \\end{aligned}$$\n\nThis gives us two possible solutions for $x$:\n\n$$x = 4 \\quad \\text{or} \\quad x = -1$$","calculator":[{"gdc":{"expression":"x^2 - 3x - 4 = 0","instructions":"Go to the Equation Solver menu. Select 'Polynomial' and set the degree to 2. Input the coefficients: a = 1, b = -3, c = -4. Solve to obtain the roots."},"ti84":{"expression":"x^2 - 3x - 4 = 0","instructions":"Press [apps] and select 'PlySmlt2'. Choose '1: Polynomial Root Finder'. Set order to 2 and press [NEXT]. Enter a = 1, b = -3, and c = -4. Press [SOLVE] to see the roots."},"desmos":{"expression":"y = x^2 - 3x - 4","instructions":"Type y = x^2 - 3x - 4 into the expression bar. Click on the x-intercepts of the plotted curve to find the solutions."},"description":"You can find the roots of the quadratic equation using a graphing calculator."}]},{"type":"hyper_latex","block":{"nodes":[{"latex":"$$$\\text{When } \\htmlData{anchor=x1}{\\color{@pink}{x = 4}}:$$"},{"latex":"$$$y = \\htmlData{anchor=x1-sub}{\\color{@pink}{4}} - 3 = 1$$"},{"latex":"$$$\\text{When } \\htmlData{anchor=x2}{\\color{@teal}{x = -1}}:$$"},{"latex":"$$$y = \\htmlData{anchor=x2-sub}{\\color{@teal}{-1}} - 3 = -4$$"}],"highlights":[{"color":"pink","style":"text-color","anchor":"x1"},{"color":"teal","style":"text-color","anchor":"x2"}],"connections":[{"to":"x1-sub","from":"x1","color":"pink","label":"substitute x","style":"solid"},{"to":"x2-sub","from":"x2","color":"teal","label":"substitute x","style":"solid"}]},"order":3,"title":"Find the corresponding y-coordinates","description":"Substitute each of our $x$-values back into the linear equation $y = x - 3$ to find their corresponding $y$-coordinates."},{"type":"text","order":4,"title":"State the final coordinates","content":"Combine the matching $x$ and $y$ values to state the final points of intersection:\n\n$$(4, 1) \\quad \\text{and} \\quad (-1, -4)$$"}]},{"id":"7c45f242-9697-4d60-94a9-a0b4b7dcbc30:substitution_x","label":"Substitution for x","steps":[{"type":"text","order":0,"title":"Rearrange linear equation","content":"To use the **Substitution for $x$** method, we first rearrange our linear equation to express $x$ in terms of $y$.\n\nGiven the linear equation:\n$$y = x - 3$$\n\nAdd $3$ to both sides to solve for $x$:\n$$x = y + 3$$","highlights":[{"text":"y = x - 3","location":"specification"}]},{"type":"hyper_latex","block":{"nodes":[{"latex":"$$\\{ \nx^2 + y^2 = 17, \\quad \n\\htmlData{anchor=x-expr}{\\color{@sky}{x = y + 3}}\n\\}$"},{"latex":"$$(\\htmlData{anchor=x-sub}{\\color{@sky}{y + 3}})^2 + y^2 = 17$"}],"highlights":[{"color":"sky","style":"text-color","anchor":"x-expr"},{"color":"sky","style":"text-color","anchor":"x-sub"}],"connections":[{"to":"x-sub","from":"x-expr","color":"sky","label":"substitute","style":"solid"}]},"order":1,"title":"Substitute into circle equation","description":"Now substitute our expression for $x$ into the equation of the circle, $x^2 + y^2 = 17$."},{"type":"text","order":2,"title":"Expand and simplify","content":"Next, we expand the brackets and simplify the equation to find a standard quadratic equation in terms of $y$:\n\n$$\\begin{aligned}\n(y + 3)^2 + y^2 &= 17 \\\\\n(y^2 + 6y + 9) + y^2 &= 17 \\\\\n2y^2 + 6y + 9 &= 17\n\\end{aligned}$$\n\nSubtract $17$ from both sides to set the quadratic equal to zero:\n$$2y^2 + 6y - 8 = 0$$\n\nWe can make the equation simpler by dividing every term by $2$:\n$$y^2 + 3y - 4 = 0$$"},{"type":"text","order":3,"title":"Solve for y","content":"Now we solve the 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$$(y + 4)(y - 1) = 0$$\n\nSetting each bracket to zero gives our $y$-coordinates:\n$$y = -4 \\quad \\text{or} \\quad y = 1$$"},{"type":"hyper_latex","block":{"nodes":[{"latex":"$$x = \\htmlData{anchor=y-var}{\\color{@pink}{y}} + 3$"},{"latex":"$$y = \\htmlData{anchor=val1}{\\color{@pink}{-4}} \\implies x = \\htmlData{anchor=sub1}{\\color{@pink}{-4}} + 3 = -1$"},{"latex":"$$y = \\htmlData{anchor=val2}{\\color{@pink}{1}} \\implies x = \\htmlData{anchor=sub2}{\\color{@pink}{1}} + 3 = 4$"}],"highlights":[{"color":"pink","style":"text-color","anchor":"y-var"},{"color":"pink","style":"text-color","anchor":"val1"},{"color":"pink","style":"text-color","anchor":"val2"}],"connections":[{"to":"val1","from":"y-var","color":"pink","label":null,"style":"dashed"},{"to":"val2","from":"y-var","color":"pink","label":null,"style":"dashed"}]},"order":4,"title":"Find corresponding x-coordinates","description":"Substitute each of the $y$-values back into our linear expression $x = y + 3$ to find their corresponding $x$-coordinates."},{"type":"text","order":5,"title":"Write coordinates of intersection","content":"Write down the final coordinates as coordinate pairs $(x, y)$:\n\n* For $x = -1$, the corresponding $y$-value is $-4$, giving the point **$(-1, -4)$**.\n* For $x = 4$, the corresponding $y$-value is $1$, giving the point **$(4, 1)$**."}]}],"generatedAt":"2026-06-12T17:53:10.296Z","modelVersion":"gemini-3.5-flash","explanationVersion":"v2"},"firstAttemptFullMarks":{"percent":42},"walkthroughVideo":{"url":"https://assets.mathsgenie.co.uk/walkthroughs/videos/5e9e88ab-b405-442a-b038-baf7a7438a0c.mp4","durationSeconds":169,"posterUrl":"https://assets.mathsgenie.co.uk/walkthroughs/videos/5e9e88ab-b405-442a-b038-baf7a7438a0c.jpg"}},{"id":"7a6b6985-38bc-41be-818e-2673e2f2d760","revisionId":"d834bbd2-636b-4a07-812e-c508aa8be0e5","questionNumber":null,"questionType":"LA","specification":"","difficulty":3,"difficultyBand":"medium","totalMarks":5,"sourceTitle":null,"calculatorUse":"non_calc","isCalculation":true,"parts":[{"id":"7c6a363c-a64d-4c9a-a369-dda3fdb28814","label":"","content":"Solve the simultaneous equations\n\n$$x^2 + y^2 = 13$$\n\n$$y = x + 1$$","markscheme":"- Substitutes $y = x + 1$ into the first equation: $x^2 + (x + 1)^2 = 13$ M1\n- Expands and simplifies: $2x^2 + 2x - 12 = 0$ or $x^2 + x - 6 = 0$ M1\n- Solves quadratic: $(x + 3)(x - 2) = 0$ so $x = -3$ or $x = 2$ M1\n- Finds $y$ values: $y = -3 + 1 = -2$ and $y = 2 + 1 = 3$ M1\n- Final answers: $x = -3, y = -2$ and $x = 2, y = 3$ A1","marks":5,"order":0,"markingRubric":{"totalMarks":5,"points":[{"id":"p1","marks":1,"description":"Substitutes the linear equation into the quadratic equation correctly.","code":"M"},{"id":"p2","marks":1,"description":"Expands the bracket and simplifies to a standard quadratic form $ax^2 + bx + c = 0$.","code":"M"},{"id":"p3","marks":1,"description":"Solves the quadratic equation to find two values for $x$.","code":"M"},{"id":"p4","marks":1,"description":"Substitutes $x$ values back into the linear equation to find corresponding $y$ values.","code":"M"},{"id":"p5","marks":1,"description":"Provides both pairs of solutions clearly and correctly.","code":"A","acceptable":["(-3, -2) and (2, 3)","x = -3, y = -2; x = 2, y = 3"]}],"finalAnswers":["x = -3, y = -2 and x = 2, y = 3"],"generatedAt":"2026-05-06T11:36:33.352Z","modelVersion":"gemini-3-flash-preview"},"tableImages":[],"explanation":null,"options":[]}],"options":[],"explanation":{"methods":[{"id":"7c6a363c-a64d-4c9a-a369-dda3fdb28814:substitution_y","label":"Substitution for y","steps":[{"type":"text","order":0,"title":"Substitute the linear equation","content":"We start with our two simultaneous equations:\n\n1) $x^2 + y^2 = 13$\n2) $y = x + 1$\n\nSince the second equation is already written in terms of $y$, we can substitute the expression for $y$ directly into the first equation:\n\n$$x^2 + (x + 1)^2 = 13$$","highlights":[{"text":"x^2 + y^2 = 13","location":"specification"},{"text":"y = x + 1","location":"specification"}]},{"type":"text","order":1,"title":"Expand and simplify","content":"Next, we expand the brackets and simplify the equation to form a quadratic in the standard form $ax^2 + bx + c = 0$.\n\nFirst, expand $(x+1)^2$:\n$$(x + 1)^2 = x^2 + 2x + 1$$\n\nNow substitute this back into the equation:\n$$x^2 + (x^2 + 2x + 1) = 13$$\n\nCombine like terms:\n$$2x^2 + 2x + 1 = 13$$\n\nSubtract $13$ from both sides to set the quadratic to $0$:\n$$2x^2 + 2x - 12 = 0$$"},{"type":"text","order":2,"title":"Solve the quadratic equation","content":"To make factorising easier, we can divide the entire equation by $2$:\n$$x^2 + x - 6 = 0$$\n\nWe need to find two numbers that multiply to $-6$ and add up to $1$. These numbers are $3$ and $-2$. We factorise the expression:\n$$(x + 3)(x - 2) = 0$$\n\nThis gives us two solutions for $x$:\n$$x = -3 \\quad \\text{or} \\quad x = 2$$"},{"type":"text","order":3,"title":"Find the corresponding y-values","content":"Now substitute each value of $x$ back into the linear equation $y = x + 1$ to find the corresponding values of $y$.\n\n* **For $x = -3$:**\n $$y = -3 + 1 = -2$$\n\n* **For $x = 2$:**\n $$y = 2 + 1 = 3$$"},{"type":"text","order":4,"title":"State the final solutions","content":"The final simultaneous solutions are the coordinate pairs:\n\n$$x = -3, \\; y = -2$$\n$$\\text{and}$$\n$$x = 2, \\; y = 3$$"}]},{"id":"7c6a363c-a64d-4c9a-a369-dda3fdb28814:substitution_x","label":"Substitution for x","steps":[{"type":"text","order":0,"title":"Rearrange for x","content":"To use the **Substitution for x** approach, we first rearrange the linear equation to make $x$ the subject.\n\nWe are given the linear equation:\n$$y = x + 1$$\n\nSubtract $1$ from both sides to isolate $x$:\n$$x = y - 1$$"},{"type":"hyper_latex","block":{"nodes":[{"latex":"$$x^2 + y^2 = 13$"},{"latex":"$$x = \text{\\htmlData{anchor=sub-val}{\\color{@sky}{y - 1}}}$"},{"latex":"$$(\\text{\\htmlData{anchor=sub-target}{\\color{@sky}{y - 1}}})^2 + y^2 = 13$"}],"highlights":[{"color":"sky","style":"box","anchor":"sub-val"},{"color":"sky","style":"box","anchor":"sub-target"}],"connections":[{"to":"sub-target","from":"sub-val","color":"sky","label":"Substitute","style":"solid"}]},"order":1,"title":"Substitute into circle equation","description":"We substitute our expression for $x$ into the quadratic circle equation."},{"type":"text","order":2,"title":"Expand and simplify","content":"Now, we expand the squared term and simplify the quadratic equation:\n\n$$\\begin{aligned} (y - 1)^2 + y^2 &= 13 \\\\ (y^2 - 2y + 1) + y^2 &= 13 \\\\ 2y^2 - 2y + 1 &= 13 \\end{aligned}$$\n\nSubtract $13$ from both sides to set the quadratic to $0$:\n$$2y^2 - 2y - 12 = 0$$\n\nDivide every term by $2$ to make the quadratic easier to solve:\n$$y^2 - y - 6 = 0$$"},{"type":"text","order":3,"title":"Solve the quadratic equation","content":"We can solve this quadratic by factorising. We need two numbers that multiply to $-6$ and add up to $-1$. These numbers are $-3$ and $+2$:\n\n$$(y - 3)(y + 2) = 0$$\n\nThis gives two possible values for $y$:\n$$y = 3 \\quad \\text{or} \\quad y = -2$$","calculator":[{"gdc":{"expression":"x^2 - x - 6 = 0","instructions":"Go to the Equation mode, select 'Polynomial', set the degree to 2, enter the coefficients $a=1, b=-1, c=-6$, and solve."},"ti84":{"expression":"x^2 - x - 6 = 0","instructions":"Press [apps], select 'PlySmlt2', then 'Polynomial Root Finder'. Set Order to 2, click 'Next', enter $a=1$, $b=-1$, $c=-6$, and press 'Solve'."},"desmos":{"expression":"y = x^2 - x - 6","instructions":"Graph the equation $y = x^2 - x - 6$ and locate the x-intercepts."},"description":"You can also use a calculator to solve the quadratic equation $y^2 - y - 6 = 0$ directly."}]},{"type":"text","order":4,"title":"Find the corresponding x values","content":"Now substitute both $y$ values back into our rearranged linear equation, $x = y - 1$, to find the matching $x$ values.\n\n- For $\\mathbf{y = 3}$:\n $$x = 3 - 1 = 2$$\n\n- For $\\mathbf{y = -2}$:\n $$x = -2 - 1 = -3$$"},{"type":"text","order":5,"title":"State final solution pairs","content":"The simultaneous equations have two pairs of solutions:\n\n$$x = 2, \\; y = 3$$\n$$\\text{and}$$\n$$x = -3, \\; y = -2$$"}]},{"id":"7c6a363c-a64d-4c9a-a369-dda3fdb28814:graphical","label":"Graphical Approach","steps":[{"type":"text","order":0,"title":"Understand the graphical method","content":"To solve simultaneous equations graphically, we find the points of intersection between the graphs of the two equations. In this case, we have a circle and a straight line:\n\n1. **Circle**: $$x^2 + y^2 = 13$$\n2. **Straight line**: $$y = x + 1$$\n\nAny point $(x, y)$ where the two graphs intersect is a solution to the system."},{"type":"text","order":1,"title":"Analyze the circle","content":"The equation of a circle with centre at the origin $(0, 0)$ and radius $r$ is given by $$x^2 + y^2 = r^2$$.\n\nComparing this to $$x^2 + y^2 = 13$$, we find:\n- **Centre**: $(0, 0)$\n- **Radius**: $r = \\sqrt{13} \\approx 3.61$\n\nTo sketch this circle, we can plot key points on the axes at approximately $3.6$ units away from the origin: $(3.6, 0)$, $(-3.6, 0)$, $(0, 3.6)$, and $(0, -3.6)$, and draw a smooth circle through them."},{"type":"text","order":2,"title":"Analyze the straight line","content":"The equation of the line is $$y = x + 1$$. \n\nThis is a straight line in the form $$y = mx + c$$:\n- **Gradient ($m$)**: $1$\n- **$y$-intercept ($c$)**: $1$, giving the point $(0, 1)$\n- **$x$-intercept**: Setting $y = 0$ gives $0 = x + 1 \\implies x = -1$, giving the point $(-1, 0)$\n\nWe can draw a straight line passing through $(-1, 0)$ and $(0, 1)$."},{"type":"text","order":3,"title":"Use a graphing tool","content":"We can plot both equations on the same set of coordinate axes to see exactly where they intersect.","calculator":[{"gdc":{"expression":"y_1 = \\sqrt{13 - x^2}, y_2 = -\\sqrt{13 - x^2}, y_3 = x + 1","instructions":"1. Open the Graph menu on your GDC.\n2. Since the circle equation is implicit, rearrange it to $y = \\pm\\sqrt{13 - x^2}$ or use relation mode if available.\n3. Input $Y1 = \\sqrt{13 - x^2}$, $Y2 = -\\sqrt{13 - x^2}$, and $Y3 = x + 1$.\n4. Draw the graphs and use the Intersect tool (G-Solve -> INTSECT) to find the intersection coordinates."},"ti84":{"expression":"Y1 = \\sqrt{13 - X^2}, Y2 = -\\sqrt{13 - X^2}, Y3 = X + 1","instructions":"1. Press [WINDOW] and set Xmin=-5, Xmax=5, Ymin=-5, Ymax=5.\n2. In [Y=], input the upper and lower halves of the circle by rearranging $x^2 + y^2 = 13$ to $y = \\pm\\sqrt{13 - x^2}$:\n - \\Y1 = \\sqrt{13 - X^2}\n - \\Y2 = -\\sqrt{13 - X^2}\n3. Input the line in \\Y3:\n - \\Y3 = X + 1\n4. Press [GRAPH].\n5. Press [2nd] [CALC] and select 5: intersect.\n6. Select the first curve, second curve, and provide a guess near each intersection point to find:\n - Point 1: $x = -3, y = -2$\n - Point 2: $x = 2, y = 3$"},"desmos":{"expression":"x^2 + y^2 = 13\\text{ and }y = x + 1","instructions":"1. In the first input bar, type: x^2 + y^2 = 13\n2. In the second input bar, type: y = x + 1\n3. Click on the gray intersection dots shown on the graph to reveal their coordinates."},"description":"Plot the circle and the line to find the intersection points."}]},{"type":"text","order":4,"title":"Identify the intersection coordinates","content":"By observing our sketch or using a graphing utility, we can clearly identify the two points where the circle and the line cross:\n\n- One intersection lies in Quadrant I at $$(2, 3)$$\n- The other intersection lies in Quadrant III at $$(-3, -2)$$\n\nWe can verify these points algebraically by substituting them back into our original equations:\n\nFor $(2, 3)$:\n- Circle: $2^2 + 3^2 = 4 + 9 = 13$ (True)\n- Line: $3 = 2 + 1$ (True)\n\nFor $(-3, -2)$:\n- Circle: $(-3)^2 + (-2)^2 = 9 + 4 = 13$ (True)\n- Line: $-2 = -3 + 1$ (True)\n\nThus, the solutions are:\n$$x = -3, y = -2 \\quad \\text{and} \\quad x = 2, y = 3$$"}]}],"generatedAt":"2026-06-12T21:35:20.090Z","modelVersion":"gemini-3.5-flash","explanationVersion":"v2"},"firstAttemptFullMarks":{"percent":66},"walkthroughVideo":{"url":"https://assets.mathsgenie.co.uk/walkthroughs/videos/7a6b6985-38bc-41be-818e-2673e2f2d760.mp4","durationSeconds":169,"posterUrl":"https://assets.mathsgenie.co.uk/walkthroughs/videos/7a6b6985-38bc-41be-818e-2673e2f2d760.jpg"}},{"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 = 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Quadratic Simultaneous Equations

Question 2

Quadratic Simultaneous Equations Questions

Quadratic Simultaneous Equations

Question 2

Quadratic Simultaneous Equations Questions