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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":50,"initialQuestionWindow":[{"id":"63469825-9d6d-4832-8c4b-31a38f6d3780","revisionId":"3c77f927-3b90-4d00-bdd7-6fb8d2499cd6","questionNumber":null,"questionType":"LA","specification":"","difficulty":4,"difficultyBand":"medium","totalMarks":5,"sourceTitle":null,"calculatorUse":"non_calc","isCalculation":true,"parts":[{"id":"648a52ab-6c06-4415-893c-f2753f90e21b","label":"a","content":"Solve the simultaneous equations\n\n$$x^2 + y^2 = 13$$\n\n$$x = y + 1$$","markscheme":"- Substitutes $x = y + 1$: $(y + 1)^2 + y^2 = 13$ M1\n- Expands and simplifies: $2y^2 + 2y - 12 = 0$ or $y^2 + y - 6 = 0$ M1\n- Solves for $y$: $(y + 3)(y - 2) = 0$ so $y = -3$ or $y = 2$ M1\n- Finds $x$: $x = -3 + 1 = -2$ and $x = 2 + 1 = 3$ M1\n- Final answers: $x = -2, y = -3$ and $x = 3, y = 2$ A1","marks":5,"order":0,"markingRubric":{"totalMarks":5,"points":[{"id":"p1","marks":1,"description":"Substitute the linear expression for $x$ into the circle equation.","code":"M"},{"id":"p2","marks":1,"description":"Expand brackets and simplify to a quadratic in terms of $y$.","code":"M"},{"id":"p3","marks":1,"description":"Solve the resulting quadratic for $y$.","code":"M"},{"id":"p4","marks":1,"description":"Substitute $y$ values back into the linear equation to find $x$.","code":"M"},{"id":"p5","marks":1,"description":"State both correct coordinate pairs.","code":"A"}],"finalAnswers":["x = -2, y = -3 and x = 3, y = 2","(-2, -3) and (3, 2)"],"generatedAt":"2026-05-06T14:07:21.329Z","modelVersion":"gemini-3-flash-preview"},"tableImages":[],"explanation":null,"options":[]}],"options":[],"explanation":{"methods":[{"id":"648a52ab-6c06-4415-893c-f2753f90e21b:substitution_x","label":"Substitution for x","steps":[{"type":"hyper_latex","block":{"nodes":[{"latex":"$$x = \text{\\htmlData{anchor=sub-x}{\\color{@sky}{y + 1}}}$"},{"latex":"$$\\left(\\text{\\htmlData{anchor=sub-target}{\\color{@sky}{y + 1}}}\\right)^2 + y^2 = 13$"}],"highlights":[{"color":"sky","style":"box","anchor":"sub-x"},{"color":"sky","style":"box","anchor":"sub-target"}],"connections":[{"to":"sub-target","from":"sub-x","color":"sky","label":"Substitute x","style":"solid"}]},"order":0,"title":"Substitute the linear expression","description":"We start by substituting the expression for $x$ from the linear equation, $x = y + 1$, into the quadratic equation $x^2 + y^2 = 13$. This gives us a single equation in terms of $y$ only."},{"type":"text","order":1,"title":"Expand and simplify","content":"Now, we expand the squared term and collect like terms to form a quadratic equation:\n\n$$(y + 1)^2 + y^2 = 13$$\n\nFirst, expand $(y + 1)^2$:\n\n$$(y^2 + 2y + 1) + y^2 = 13$$\n\nCombine the $y^2$ terms:\n\n$$2y^2 + 2y + 1 = 13$$"},{"type":"text","order":2,"title":"Rearrange to equal zero","content":"To solve a quadratic equation, we must make one side equal to zero. Subtract $13$ from both sides:\n\n$$2y^2 + 2y - 12 = 0$$\n\nTo make the equation easier to solve, we can divide every term by $2$:\n\n$$y^2 + y - 6 = 0$$"},{"type":"text","order":3,"title":"Solve the quadratic equation","content":"We can solve the quadratic equation $y^2 + y - 6 = 0$ by factorising. We need two numbers that multiply to $-6$ and add up to $1$:\n- The numbers are $3$ and $-2$.\n\nFactorise the quadratic:\n\n$$(y + 3)(y - 2) = 0$$\n\nThis gives two possible solutions for $y$:\n\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 (F1), choose degree 2, and enter the coefficients: 1, 1, and -6. Press Solve."},"ti84":{"expression":"x^2 + x - 6 = 0","instructions":"Press [APPS], select 'PlySmlt2', then 'Polynomial Root Finder'. Set degree to 2 and enter the coefficients: a = 1, b = 1, c = -6. Press [SOLVE] to see the solutions."},"desmos":{"expression":"y = x^2 + x - 6","instructions":"Type y = x^2 + x - 6 in the input bar and look at the x-intercepts of the parabola to find the roots."},"description":"Verify the roots of the quadratic equation $y^2 + y - 6 = 0$ using a solver."}]},{"type":"hyper_latex","block":{"nodes":[{"latex":"$$y = \\text{\\htmlData{anchor=y-neg}{\\color{@red}{-3}}} \\quad \\Rightarrow \\quad x = \\text{\\htmlData{anchor=y-sub-neg}{\\color{@red}{-3}}} + 1 = \\text{\\htmlData{anchor=x-neg}{\\color{@pink}{-2}}}$"},{"latex":"$$y = \\text{\\htmlData{anchor=y-pos}{\\color{@teal}{2}}} \\quad \\Rightarrow \\quad x = \\text{\\htmlData{anchor=y-sub-pos}{\\color{@teal}{2}}} + 1 = \\text{\\htmlData{anchor=x-pos}{\\color{@green}{3}}}$"}],"highlights":[{"color":"red","style":"text-color","anchor":"y-neg"},{"color":"pink","style":"text-color","anchor":"x-neg"},{"color":"teal","style":"text-color","anchor":"y-pos"},{"color":"green","style":"text-color","anchor":"x-pos"}],"connections":[{"to":"y-sub-neg","from":"y-neg","color":"red","label":"substitute y","style":"solid"},{"to":"y-sub-pos","from":"y-pos","color":"teal","label":"substitute y","style":"solid"}]},"order":4,"title":"Find the corresponding x-values","description":"We substitute our $y$-values back into the linear equation $x = y + 1$ to find each corresponding $x$-value."},{"type":"text","order":5,"title":"State final solution pairs","content":"The final solutions are paired together as coordinates:\n\n$$x = -2, \\; y = -3$$\n\n$$x = 3, \\; y = 2$$"}]},{"id":"648a52ab-6c06-4415-893c-f2753f90e21b:substitution_y","label":"Substitution for y","steps":[{"type":"text","order":0,"title":"Rearrange the linear equation","content":"To solve these simultaneous equations using substitution, we first rearrange the linear equation to make $y$ the subject.\n\nWe start with the linear equation:\n$$x = y + 1$$\n\nSubtract $1$ from both sides to isolate $y$:\n$$y = x - 1$$","highlights":[{"text":"x = y + 1","location":"specification"}]},{"type":"hyper_latex","block":{"nodes":[{"latex":"$$x^2 + \\htmlData{anchor=y-sub}{\\color{@sky}{y}}^2 = 13$"},{"latex":"$$x^2 + (\\htmlData{anchor=y-expr}{\\color{@sky}{x - 1}})^2 = 13$"}],"highlights":[{"color":"sky","style":"text-color","anchor":"y-expr"},{"color":"sky","style":"text-color","anchor":"y-sub"}],"connections":[{"to":"y-sub","from":"y-expr","color":"sky","label":"substitute","style":"solid"}]},"order":1,"title":"Substitute into the quadratic","description":"Now, we substitute our expression for $y$ into the quadratic equation."},{"type":"text","order":2,"title":"Expand and simplify","content":"Next, we expand the bracket $(x - 1)^2$ and simplify the equation to form a standard quadratic equation:\n\n$$\\begin{aligned} x^2 + (x^2 - 2x + 1) &= 13 \\\\ 2x^2 - 2x + 1 &= 13 \\end{aligned}$$\n\nSubtract $13$ from both sides to set the quadratic to $0$:\n$$2x^2 - 2x - 12 = 0$$\n\nTo make it simpler to factorise, divide every term by $2$:\n$$x^2 - x - 6 = 0$$"},{"type":"text","order":3,"title":"Solve for x","content":"We need to solve the quadratic equation $x^2 - x - 6 = 0$ by factorising.\n\nWe look for two numbers that multiply to $-6$ and add up to $-1$. These numbers are $-3$ and $+2$:\n$$(x - 3)(x + 2) = 0$$\n\nSetting each factor to zero gives us the $x$-coordinates:\n$$x - 3 = 0 \\implies x = 3$$\n$$x + 2 = 0 \\implies x = -2$$"},{"type":"text","order":4,"title":"Find the corresponding y values","content":"Substitute our $x$ values back into the rearranged linear equation, $y = x - 1$, to find the corresponding $y$ values.\n\n- For $\\mathbf{x = 3}$:\n $$y = 3 - 1 = 2$$\n\n- For $\\mathbf{x = -2}$:\n $$y = -2 - 1 = -3$$\n\nThus, the solutions to the simultaneous equations are:\n$$x = 3, y = 2 \\quad \\text{and} \\quad x = -2, y = -3$$"}]},{"id":"648a52ab-6c06-4415-893c-f2753f90e21b:graphical","label":"Graphical Method","steps":[{"type":"text","order":0,"title":"Understand the graphical approach","content":"To solve these simultaneous equations graphically, we need to plot both equations on the same coordinate grid and find where they intersect:\n\n1. The circle: $$x^2 + y^2 = 13$$\n2. The straight line: $$x = y + 1$$"},{"type":"text","order":1,"title":"Analyze the circle","content":"The equation $$x^2 + y^2 = 13$$ represents a circle centered at the origin $(0, 0)$ with a radius of $r = √13 ≈ 3.61$.\n\nTo help us sketch this circle accurately, we can identify key integer coordinates that lie exactly on the circle. Since $13 = 4 + 9 = 2^2 + 3^2$, the circle passes through points where the coordinate values are combinations of $\\pm 2$ and $\\pm 3$:\n- $(3, 2)$, $(3, -2)$, $(-3, 2)$, $(-3, -2)$\n- $(2, 3)$, $(2, -3)$, $(-2, 3)$, $(-2, -3)$"},{"type":"text","order":2,"title":"Analyze the line","content":"The equation for the line is $x = y + 1$. Rearranging this into the slope-intercept form ($y = mx + c$) gives:\n\n$$y = x - 1$$\n\nThis is a straight line with a gradient ($m$) of $1$ and a $y$-intercept ($c$) at $(0, -1)$. \n\nWe can find a few points on this line to plot it:\n- When $x = 0$, $y = -1 \\Rightarrow (0, -1)$\n- When $x = 1$, $y = 0 \\Rightarrow (1, 0)$\n- When $x = 3$, $y = 2 \\Rightarrow (3, 2)$\n- When $x = -2$, $y = -3 \\Rightarrow (-2, -3)$"},{"type":"text","order":3,"title":"Plot and find intersections","content":"By plotting both curves on a coordinate grid, we look for the points where the line crosses the circle.\n\nFrom our plotted points, we can see that the line and circle intersect at exactly two points:\n- The first point of intersection is at $$(3, 2)$$\n- The second point of intersection is at $$(-2, -3)$$ \n\nThese coordinate pairs are the solutions to our simultaneous equations.","calculator":[{"gdc":{"instructions":"1. Open the graphing application.\n2. Enter the functions: f1(x) = x - 1, f2(x) = √(13 - x²), and f3(x) = -√(13 - x²).\n3. Graph the curves and use the 'Analyze Graph' -> 'Intersection' tool to select the intersection points."},"ti84":{"instructions":"1. Press [Y=].\n2. For the line, enter Y1 = X - 1.\n3. For the circle, rearrange to get Y = ±√(13 - X²). Enter Y2 = √(13 - X²) and Y3 = -√(13 - X²).\n4. Press [GRAPH]. Adjust the window if needed using [ZOOM] [Square] to keep the circle proportioned.\n5. Press [2nd] [CALC] and select '5: intersect' to find the intersection points."},"desmos":{"instructions":"1. In the first expression bar, type: x^2 + y^2 = 13\n2. In the second expression bar, type: x = y + 1\n3. Click on the gray intersection points on the graph to display their coordinates: (3, 2) and (-2, -3)."},"description":"Plot the equations and find their intersection points."}]},{"type":"text","order":4,"title":"State final answers","content":"The final solutions to the simultaneous equations are:\n\n$$x = 3, y = 2$$\n$$\\text{and}$$\n$$x = -2, y = -3$$"}]}],"generatedAt":"2026-06-12T18:15:31.889Z","modelVersion":"gemini-3.5-flash","explanationVersion":"v2"},"firstAttemptFullMarks":null,"walkthroughVideo":{"url":"https://assets.mathsgenie.co.uk/walkthroughs/videos/63469825-9d6d-4832-8c4b-31a38f6d3780.mp4","durationSeconds":181,"posterUrl":"https://assets.mathsgenie.co.uk/walkthroughs/videos/63469825-9d6d-4832-8c4b-31a38f6d3780.jpg"}},{"id":"684f9fa4-33a0-4a05-9bc8-7d7fc063dd54","revisionId":"94b1d939-9bf7-4e0b-8b49-1f5c634c2b04","questionNumber":null,"questionType":"LA","specification":"","difficulty":3,"difficultyBand":"medium","totalMarks":5,"sourceTitle":null,"calculatorUse":"non_calc","isCalculation":true,"parts":[{"id":"e158a59d-e1c1-4d2a-ad52-608e6566046b","label":"a","content":"Solve the simultaneous equations\n\n$$x^2 + y^2 = 34$$\n\n$$y = x + 2$$","markscheme":"- Substitutes $y = x + 2$ into the first equation: $x^2 + (x + 2)^2 = 34$ M1\n- Expands and simplifies to a quadratic equation: $2x^2 + 4x - 30 = 0$ or $x^2 + 2x - 15 = 0$ M1\n- Factorises or uses formula to find $x$: $(x + 5)(x - 3) = 0$ so $x = -5$ or $x = 3$ M1\n- Substitutes $x$ values back to find $y$: $y = -5 + 2 = -3$ and $y = 3 + 2 = 5$ M1\n- Final answers: $x = -5, y = -3$ and $x = 3, y = 5$ 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 bracket and simplifies to form a three-term quadratic equation equal to zero.","code":"M"},{"id":"p3","marks":1,"description":"Solves the quadratic equation to find two values for one variable.","code":"M"},{"id":"p4","marks":1,"description":"Substitutes both found values back into the linear equation to find the corresponding values of the second variable.","code":"M"},{"id":"p5","marks":1,"description":"Provides both correct pairs of solutions.","code":"A"}],"finalAnswers":[],"generatedAt":"2026-05-06T11:36:31.024Z","modelVersion":"gemini-3-flash-preview"},"tableImages":[],"explanation":{"methods":[{"id":"substitution_factorising","label":"Substitution and factorising","steps":[{"type":"hyper_latex","block":{"nodes":[{"latex":"$$x^2+\\htmlData{anchor=old_y}{\\color{@sky}{y^2}}=34$"},{"latex":"$$x^2+\\htmlData{anchor=new_y}{\\color{@sky}{(x+2)^2}}=34$"}],"highlights":[{"color":"sky","style":"text-color","anchor":"old_y"},{"color":"sky","style":"text-color","anchor":"new_y"}]},"order":0,"title":"Substitute for y","description":"Replace y in the first equation with x + 2. Writing this substitution earns the first M1 mark."},{"type":"hyper_latex","block":{"nodes":[{"latex":"$$x^2+\\htmlData{anchor=bracket}{\\color{@orange}{(x+2)^2}}=34$"},{"latex":"$$x^2+x^2+4x+4=34$"},{"latex":"$$\\htmlData{anchor=quadratic}{\\color{@orange}{2x^2+4x-30=0}}$"}],"highlights":[{"color":"orange","style":"text-color","anchor":"bracket"},{"color":"orange","style":"text-color","anchor":"quadratic"}]},"order":1,"title":"Form the quadratic","description":"Expand the bracket, then collect terms. This earns the second M1 mark."},{"type":"hyper_latex","block":{"nodes":[{"latex":"$$\\htmlData{anchor=original}{\\color{@sky}{2x^2+4x-30=0}}$"},{"latex":"$$\\htmlData{anchor=simplified}{\\color{@sky}{x^2+2x-15=0}}$"},{"latex":"$$\\htmlData{anchor=factored}{\\color{@purple}{(x+5)(x-3)=0}}$"},{"latex":"$$\\htmlData{anchor=roots}{\\color{@green}{x=-5\\text{ or }x=3}}$"}],"highlights":[{"color":"sky","style":"text-color","anchor":"original"},{"color":"sky","style":"text-color","anchor":"simplified"},{"color":"purple","style":"text-color","anchor":"factored"},{"color":"green","style":"text-color","anchor":"roots"}],"connections":[{"to":"simplified","from":"original","color":"sky","label":"divide by 2"},{"to":"factored","from":"simplified","color":"purple","label":"factorise"},{"to":"roots","from":"factored","color":"green","label":"zero-product rule"}]},"order":2,"title":"Factorise the quadratic","description":"Simplify, factorise, then use the zero-product rule: if two factors multiply to zero, at least one factor is zero. This earns the third M1 mark."},{"type":"hyper_latex","block":{"nodes":[{"latex":"$$\\htmlData{anchor=xneg}{\\color{@sky}{x=-5}}$"},{"latex":"$$y=\\htmlData{anchor=subneg}{\\color{@sky}{-5}}+2=-3$"},{"latex":"$$\\htmlData{anchor=xpos}{\\color{@orange}{x=3}}$"},{"latex":"$$y=\\htmlData{anchor=subpos}{\\color{@orange}{3}}+2=5$"}],"highlights":[{"color":"sky","style":"text-color","anchor":"xneg"},{"color":"sky","style":"text-color","anchor":"subneg"},{"color":"orange","style":"text-color","anchor":"xpos"},{"color":"orange","style":"text-color","anchor":"subpos"}]},"order":3,"title":"Find both y-values","description":"Use y = x + 2 for each value of x. Showing both substitutions earns the fourth M1 mark."},{"type":"text","order":4,"title":"State both solutions","content":"The two solutions are\n\n$$\n(x,y)=(-5,-3) \\quad \\text{and} \\quad (x,y)=(3,5).\n$$\n\nGiving both complete pairs earns the final A1 mark."}]}],"generatedAt":"2026-07-20T10:40:32.252Z","modelVersion":"gpt-5.6-sol","explanationVersion":"v3"},"options":[]}],"options":[],"explanation":{"methods":[{"id":"e158a59d-e1c1-4d2a-ad52-608e6566046b:substitution_y","label":"Substitution for y","steps":[{"type":"hyper_latex","block":{"nodes":[{"latex":"$$$y = \\htmlData{anchor=lin-eq}{\\color{@sky}{x + 2}}$$"},{"latex":"$$$x^2 + (\\htmlData{anchor=quad-sub}{\\color{@sky}{y}})^2 = 34$$"},{"latex":"$$$x^2 + (\\htmlData{anchor=expanded-term}{\\color{@sky}{x + 2}})^2 = 34$$"}],"highlights":[{"color":"sky","style":"box","anchor":"lin-eq"},{"color":"sky","style":"box","anchor":"expanded-term"}],"connections":[{"to":"quad-sub","from":"lin-eq","color":"sky","label":"substitute","style":"solid"}]},"order":0,"title":"Substitute the expression","description":"We are given a system of simultaneous equations where one equation is quadratic and the other is linear. To find $x$ first, we substitute the expression for $y$ from the linear equation, $y = x + 2$, into the quadratic equation."},{"type":"text","order":1,"title":"Expand and simplify","content":"Now we expand the bracket $(x + 2)^2$ and simplify the equation:\n\n$$\\begin{aligned}\nx^2 + (x + 2)(x + 2) &= 34 \\\\\nx^2 + (x^2 + 4x + 4) &= 34 \\\\\n2x^2 + 4x + 4 &= 34\n\\end{aligned}$$\n\nSubtract $34$ from both sides to set the quadratic to $0$:\n\n$$2x^2 + 4x - 30 = 0$$"},{"type":"text","order":2,"title":"Divide by common factor","content":"To make factorising easier, we can divide every term by $2$:\n\n$$x^2 + 2x - 15 = 0$$"},{"type":"text","order":3,"title":"Factorise the quadratic","content":"We look for two numbers that multiply to $-15$ and add up to $+2$. These numbers are $+5$ and $-3$.\n\nSo, we factorise the equation:\n\n$$(x + 5)(x - 3) = 0$$\n\nSetting each bracket to zero gives our two possible values for $x$:\n\n$$x = -5 \\quad \\text{or} \\quad x = 3$$"},{"type":"hyper_latex","block":{"nodes":[{"latex":"$$$\\text{When } \\htmlData{anchor=x1}{\\color{@purple}{x = -5}} \\implies y = (\\htmlData{anchor=x1-sub}{\\color{@purple}{-5}}) + 2 = \\htmlData{anchor=y1}{\\color{@pink}{-3}}$$"},{"latex":"$$$\\text{When } \\htmlData{anchor=x2}{\\color{@teal}{x = 3}} \\implies y = (\\htmlData{anchor=x2-sub}{\\color{@teal}{3}}) + 2 = \\htmlData{anchor=y2}{\\color{@orange}{5}}$$"}],"highlights":[{"color":"pink","style":"text-color","anchor":"y1"},{"color":"orange","style":"text-color","anchor":"y2"}],"connections":[{"to":"x1-sub","from":"x1","color":"purple","label":"substitute x","style":"dashed"},{"to":"x2-sub","from":"x2","color":"teal","label":"substitute x","style":"dashed"}]},"order":4,"title":"Find corresponding y values","description":"Now substitute each $x$ value back into the simpler linear equation, $y = x + 2$, to find the corresponding values for $y$."},{"type":"text","order":5,"title":"State final pairs","content":"The solutions to the simultaneous equations are the coordinate pairs:\n\n$$x = -5, y = -3 \\quad \\text{and} \\quad x = 3, y = 5$$","calculator":[{"gdc":{"instructions":"Graph both the circle and the line, then use the intersection tool to read off the coordinate pairs."},"ti84":{"instructions":"1. Go to [Y=] and enter Y1 = X + 2.\n2. Rearrange the first equation to Y2 = √(34 - X²) and Y3 = -√(34 - X²).\n3. Press [GRAPH] to display the intersection of the line and the circle.\n4. Press [2nd] [CALC] and select 'intersect' (option 5) to find the coordinate intersections at (-5, -3) and (3, 5)."},"desmos":{"expression":"x^2 + y^2 = 34\ny = x + 2","instructions":"Type `x^2 + y^2 = 34` in the first row and `y = x + 2` in the second row. Click on the gray intersection points on the graph to reveal $(-5, -3)$ and $(3, 5)$."},"description":"You can verify these solutions graphically by plotting both equations and finding their intersection points."}]}]},{"id":"e158a59d-e1c1-4d2a-ad52-608e6566046b:substitution_x","label":"Substitution for x","steps":[{"type":"text","order":0,"title":"Rearrange the linear equation","content":"To use the **Substitution for x** approach, we first rearrange our linear equation to make $x$ the subject.\n\nOur linear equation is:\n$$y = x + 2$$\n\nSubtract $2$ from both sides to isolate $x$:\n$$x = y - 2$$"},{"type":"hyper_latex","block":{"nodes":[{"latex":"$$x = \blank{\\htmlData{anchor=x-expr}{\\color{@sky}{y - 2}}}$"},{"latex":"$$(\\blank{\\htmlData{anchor=x-sub}{\\color{@sky}{y - 2}}})^2 + y^2 = 34$"}],"highlights":[{"color":"sky","style":"text-color","anchor":"x-expr"},{"color":"sky","style":"text-color","anchor":"x-sub"}],"connections":[{"to":"x-sub","from":"x-expr","color":"sky","label":"substitute","style":"solid"}]},"order":1,"title":"Substitute into the quadratic","description":"Now we substitute our expression for $x$ into the quadratic equation:\n\n$$x^2 + y^2 = 34$$"},{"type":"text","order":2,"title":"Expand and simplify","content":"Now, we expand the squared term and simplify the equation to form a standard quadratic equation:\n\n$$\\begin{aligned} (y - 2)^2 + y^2 &= 34 \\\\ (y^2 - 4y + 4) + y^2 &= 34 \\\\ 2y^2 - 4y + 4 &= 34 \\end{aligned}$$\n\nSubtract $34$ from both sides to set the quadratic equation to $0$:\n\n$$2y^2 - 4y - 30 = 0$$\n\nTo make factorising easier, we can divide the entire equation by $2$:\n\n$$y^2 - 2y - 15 = 0$$"},{"type":"text","order":3,"title":"Solve for y","content":"We need to find two numbers that multiply to $-15$ and add to $-2$. These numbers are $-5$ and $3$.\n\nFactorising the quadratic expression gives:\n$$(y - 5)(y + 3) = 0$$\n\nSetting each factor to $0$, we get our two values for $y$:\n$$y = 5 \\quad \\text{or} \\quad y = -3$$"},{"type":"text","order":4,"title":"Find corresponding x values","content":"Finally, we substitute our $y$ values back into the rearranged linear formula, $x = y - 2$, to find the corresponding $x$ values.\n\n**Case 1:** When $y = 5$\n$$x = 5 - 2 = 3$$\n\n**Case 2:** When $y = -3$\n$$x = -3 - 2 = -5$$\n\nThis gives us our final solution pairs:\n$$x = 3, y = 5 \\quad \\text{and} \\quad x = -5, y = -3$$"}]},{"id":"e158a59d-e1c1-4d2a-ad52-608e6566046b:graphical","label":"Graphical Approach","steps":[{"type":"text","order":0,"title":"Understand the graphical method","content":"To solve these simultaneous equations graphically, we need to find where the graphs of the two equations intersect on a coordinate grid.\n\n1. The first equation, $$x^2 + y^2 = 34$$, represents a circle centered at the origin $(0, 0)$ with a radius of $r = √{34} ≈ 5.83$.\n2. The second equation, $$y = x + 2$$, represents a straight line with a $y$-intercept of $2$ and a gradient (slope) of $1$.","highlights":[{"text":"x^2 + y^2 = 34","location":"specification"},{"text":"y = x + 2","location":"specification"}]},{"type":"text","order":1,"title":"Plot both equations","content":"We can input both equations into a graphing calculator or tool to plot them visually. \n\nOn our graph:\n- The circle will be drawn centered at $(0, 0)$ passing through approximately $(5.83, 0)$, $(0, 5.83)$, $(-5.83, 0)$, and $(0, -5.83)$.\n- The straight line will pass through points like $(0, 2)$, $(1, 3)$, and $(-2, 0)$.","calculator":[{"gdc":{"instructions":"1. Open the graphing menu on your GDC.\n2. Input the equations. If your GDC supports relation graphing, you can type 'x^2 + y^2 = 34' directly. Otherwise, rearrange the circle to y = ±√(34 - x^2) and input both branches.\n3. Input the line 'y = x + 2'.\n4. Use the 'Intersect' tool (often found under G-Solv or Analyze Graph) to find the points where the line crosses the circle."},"ti84":{"instructions":"1. Press [WINDOW] and set Xmin = -10, Xmax = 10, Ymin = -10, Ymax = 10.\n2. To graph the circle, we must solve for y: y = ±√(34 - x²). Press [Y=] and enter:\n Y₁ = √(34 - X²)\n Y₂ = -√(34 - X²)\n3. Enter the line in Y₃:\n Y₃ = X + 2\n4. Press [GRAPH].\n5. To find the intersections, press [2nd] [CALC] and select '5: intersect'.\n6. Select the first curve (Y₁ or Y₂), the second curve (Y₃), and move the cursor near an intersection point to guess. Repeat for both intersection points."},"desmos":{"expression":"x^2 + y^2 = 34\ny = x + 2","instructions":"1. In the first input bar, type: x^2 + y^2 = 34\n2. In the second input bar, type: y = x + 2\n3. Click on the gray intersection points on the graph to display their coordinates."},"description":"Plot and find the intersection points using your graphing tool."}]},{"type":"text","order":2,"title":"Identify the intersection points","content":"By looking at the graph, we can see that the line crosses the circle at exactly two points:\n\n1. In the first quadrant, the line and circle intersect at the point $$(3, 5)$$.\n2. In the third quadrant, they intersect at the point $$(-5, -3)$$.\n\nThese coordinate pairs are the solutions to our simultaneous equations."},{"type":"text","order":3,"title":"Verify your solutions","content":"Let's algebraically double-check the coordinates we found from our graph:\n\n**For the point $(3, 5)$:**\n- Circle: $3^2 + 5^2 = 9 + 25 = 34$ (True)\n- Line: $5 = 3 + 2$ (True)\n\n**For the point $(-5, -3)$:**\n- Circle: $(-5)^2 + (-3)^2 = 25 + 9 = 34$ (True)\n- Line: $-3 = -5 + 2$ (True)\n\nBoth pairs perfectly satisfy both equations."},{"type":"text","order":4,"title":"State the final answer","content":"The solutions to the simultaneous equations are:\n\n$$x = 3, y = 5$$\nand\n$$x = -5, y = 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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 7

Quadratic Simultaneous Equations Questions

Quadratic Simultaneous Equations

Question 7

Quadratic Simultaneous Equations Questions