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bg-background","children":["$","$L39",null,{"backHref":"/gcse/maths/ccea/questions","basePath":"/gcse/maths/ccea/grade-8-9-8-1-quadratic-simultaneous-equations/questions","formulaUrl":"https://ccea.org.uk/downloads/docs/Qual-Assessment/Live/GCSE%20Controlled%20Assessment/Examination%20Materials/GCSE%20Mathematics%20(2017)/2023-Summer/GCSE%20Mathematics%20Additional%20Support%20Materials%20(Higher%20Tier)-Summer2023_0.pdf","initialIndex":10,"initialQuestionWindow":[{"id":"7212abe4-9d9f-42ce-9fdd-8579125a4d9d","revisionId":"7d905a21-3ec7-4e2d-8778-f3fe369c56be","questionNumber":"1","questionType":"LA","specification":"","difficultyBand":"medium","tier":null,"totalMarks":5,"sourceTitle":"Quadratic Simultaneous Equations","calculatorUse":"non_calc","isCalculation":true,"parts":[{"id":"d82b8177-a41d-4597-a9f8-0a3881b3f9fb","label":"","content":"Solve the simultaneous equations\n\n$$x^2 + y^2 = 13$$\n\n$$x = y - 5$$","markscheme":"- Substitutes $x = y - 5$ into the first equation: $(y - 5)^2 + y^2 = 13$ M1\n- Expands and simplifies to a quadratic equation: $2y^2 - 10y + 12 = 0$ or $y^2 - 5y + 6 = 0$ M1\n- Factorises or uses formula to find $y$: $(y - 3)(y - 2) = 0$ so $y = 3$ or $y = 2$ M1\n- Substitutes $y$ values back to find $x$: $x = 3 - 5 = -2$ and $x = 2 - 5 = -3$ 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":"Substitutes the linear equation into the quadratic equation.","code":"M"},{"id":"p2","marks":1,"description":"Expands brackets and simplifies to a three-term quadratic equal to zero.","code":"M"},{"id":"p3","marks":1,"description":"Solves the resulting quadratic equation to find two values for one variable.","code":"M"},{"id":"p4","marks":1,"description":"Substitutes found values back into the linear equation to find the second variable.","code":"M"},{"id":"p5","marks":1,"description":"Provides both pairs of correct solutions: $(-2, 3)$ and $(-3, 2)$. ","code":"A"}],"finalAnswers":["x = -2, y = 3","x = -3, y = 2"],"generatedAt":"2026-05-05T15:56:19.822Z","modelVersion":"gemini-3-flash-preview"},"tableImages":[],"explanation":{"methods":[{"id":"substitute_and_factorise","label":"Substitute and factorise","steps":[{"type":"hyper_latex","block":{"nodes":[{"latex":"$$x^2+y^2=13$"},{"latex":"$$x=\\color{@sky}{\\htmlData{anchor=xexpr}{y-5}}$"},{"latex":"$$\\color{@sky}{\\htmlData{anchor=subbed}{(y-5)^2}}+y^2=13$"}]},"order":0,"title":"Substitute for x","description":"The second equation already gives x in terms of y, so replace x in the first equation. Writing this substitution earns the first M1."},{"type":"hyper_latex","block":{"nodes":[{"latex":"$$\\color{@orange}{\\htmlData{anchor=bracket}{(y-5)^2}}$"},{"latex":"$$\\color{@orange}{\\htmlData{anchor=expanded}{y^2-10y+25}}$"},{"latex":"$$y^2-10y+25+y^2=13$"},{"latex":"$$\\color{@green}{\\htmlData{anchor=collected}{2y^2-10y+12=0}}$"},{"latex":"$$\\color{@green}{\\htmlData{anchor=divided}{y^2-5y+6=0}}$"}],"connections":[{"to":"expanded","from":"bracket","color":"orange","label":"expand"},{"to":"divided","from":"collected","color":"green","label":"divide by 2"}]},"order":1,"title":"Form the quadratic","description":"Expand the bracket, substitute the expansion, collect like terms, then divide every term by 2. Reaching either quadratic form shown earns the second M1."},{"type":"hyper_latex","block":{"nodes":[{"latex":"$$y^2-5y+6=0$"},{"latex":"$$\\bigl(\\color{@teal}{\\htmlData{anchor=factor3}{y-3}}\\bigr)\\bigl(\\color{@purple}{\\htmlData{anchor=factor2}{y-2}}\\bigr)=0$"},{"latex":"$$\\color{@teal}{\\htmlData{anchor=root3}{y=3}}$"},{"latex":"$$\\color{@purple}{\\htmlData{anchor=root2}{y=2}}$"}]},"order":2,"title":"Factorise for y","description":"Find two numbers that multiply to 6 and add to −5. Factorising and obtaining both y-values earns the third M1."},{"type":"hyper_latex","block":{"nodes":[{"latex":"$$x=y-5$"},{"latex":"$$\\color{@sky}{\\htmlData{anchor=line3}{x=3-5}}$"},{"latex":"$$\\color{@sky}{\\htmlData{anchor=result3}{x=-2}}$"},{"latex":"$$\\color{@orange}{\\htmlData{anchor=line2}{x=2-5}}$"},{"latex":"$$\\color{@orange}{\\htmlData{anchor=result2}{x=-3}}$"}]},"order":3,"title":"Find each x","description":"Substitute each y-value into x = y − 5. Finding both matching x-values earns the fourth M1."},{"type":"text","order":4,"title":"State both solutions","content":"Keep each value of $x$ paired with the $y$-value that produced it:\n\n$$\n(x,y)=(-2,3)\\quad\\text{or}\\quad(x,y)=(-3,2).\n$$\n\nGiving both complete pairs earns the final A1."}]},{"id":"substitute_and_formula","label":"Substitute and use the formula","steps":[{"type":"hyper_latex","block":{"nodes":[{"latex":"$$x^2+y^2=13$"},{"latex":"$$x=\\color{@sky}{\\htmlData{anchor=xexpr}{y-5}}$"},{"latex":"$$\\color{@sky}{\\htmlData{anchor=subbed}{(y-5)^2}}+y^2=13$"}]},"order":0,"title":"Substitute for x","description":"Replace x in the first equation using the expression from the second equation. This earns the first M1."},{"type":"hyper_latex","block":{"nodes":[{"latex":"$$\\color{@orange}{\\htmlData{anchor=bracket}{(y-5)^2}}$"},{"latex":"$$\\color{@orange}{\\htmlData{anchor=expanded}{y^2-10y+25}}$"},{"latex":"$$y^2-10y+25+y^2=13$"},{"latex":"$$\\color{@green}{\\htmlData{anchor=collected}{2y^2-10y+12=0}}$"},{"latex":"$$\\color{@green}{\\htmlData{anchor=divided}{y^2-5y+6=0}}$"}],"connections":[{"to":"expanded","from":"bracket","color":"orange","label":"expand"},{"to":"divided","from":"collected","color":"green","label":"divide by 2"}]},"order":1,"title":"Form the quadratic","description":"Expand the bracket, substitute the expansion, collect like terms, then divide every term by 2. Reaching either quadratic form shown earns the second M1."},{"type":"hyper_latex","block":{"nodes":[{"latex":"$$y=\\dfrac{-b\\pm\\sqrt{b^2-4ac}}{2a}$"},{"latex":"$$y=\\dfrac{-(-5)\\pm\\sqrt{(-5)^2-4(1)(6)}}{2(1)}$"},{"latex":"$$\\htmlData{anchor=both_signs}{\\color{@purple}{y=\\dfrac{5\\pm\\sqrt{1}}{2}}}$"},{"latex":"$$\\htmlData{anchor=plus_branch}{\\color{@teal}{y=\\dfrac{5+1}{2}=3}}$"},{"latex":"$$\\htmlData{anchor=minus_branch}{\\color{@orange}{y=\\dfrac{5-1}{2}=2}}$"},{"latex":"$$y=3\\text{ or }y=2$"}],"connections":[{"to":"plus_branch","from":"both_signs","color":"teal","label":"use +"},{"to":"minus_branch","from":"both_signs","color":"orange","label":"use −"}]},"order":2,"title":"Apply the formula","description":"For the quadratic y² − 5y + 6 = 0, the coefficients are a = 1, b = −5 and c = 6. Substitute them carefully: because b is negative, −b becomes −(−5). Keeping both signs earns the third M1."},{"type":"hyper_latex","block":{"nodes":[{"latex":"$$x=y-5$"},{"latex":"$$\\color{@sky}{x=3-5}$"},{"latex":"$$\\color{@sky}{x=-2}$"},{"latex":"$$\\color{@orange}{x=2-5}$"},{"latex":"$$\\color{@orange}{x=-3}$"}]},"order":3,"title":"Find each x","description":"Return to the linear equation for each root. Finding both corresponding x-values earns the fourth M1."},{"type":"text","order":4,"title":"State both solutions","content":"Keep each value of $x$ paired with the $y$-value that produced it:\n\n$$\n(x,y)=(-2,3)\\quad\\text{or}\\quad(x,y)=(-3,2).\n$$\n\nGiving both complete pairs earns the final A1."}]}],"generatedAt":"2026-07-19T19:16:21.115Z","modelVersion":"gpt-5.6-sol","explanationVersion":"v3"},"options":[]}],"options":[],"explanation":{"methods":[{"id":"d82b8177-a41d-4597-a9f8-0a3881b3f9fb:substitute_x","label":"Substitution for x","steps":[{"type":"hyper_latex","block":{"nodes":[{"latex":"$$x = \\htmlData{anchor=sub-val}{\\color{@sky}{y - 5}}$"},{"latex":"$$\\htmlData{anchor=x-sq}{\\color{@sky}{x}}^2 + y^2 = 13$"},{"latex":"$$\\htmlData{anchor=sub-target}{\\color{@sky}{(y - 5)}}^2 + y^2 = 13$"}],"highlights":[{"color":"sky","style":"text-color","anchor":"sub-val"},{"color":"sky","style":"text-color","anchor":"sub-target"}],"connections":[{"to":"sub-target","from":"sub-val","color":"sky","label":"Substitute","style":"solid"}]},"order":0,"title":"Substitute the linear expression","description":"We substitute the expression for $x$ from our linear equation directly into the quadratic equation. This earns you the first mark **(M1)**."},{"type":"text","order":1,"title":"Expand and simplify","content":"Now, let's expand the brackets and simplify the equation:\\n\\n$$(y - 5)^2 + y^2 = 13$$\\n\\nExpanding the term $(y - 5)^2$ gives:\\n\\n$$(y^2 - 10y + 25) + y^2 = 13$$\\n\\nNow we combine the like terms $y^2 + y^2 = 2y^2$:\\n\\n$$2y^2 - 10y + 25 = 13$$\\n\\nSubtract $13$ from both sides to form a quadratic equation set to $0$:\\n\\n$$2y^2 - 10y + 12 = 0$$\\n\\nTo make factorising easier, we can divide the entire equation by $2$:\\n\\n$$y^2 - 5y + 6 = 0$$\\n\\nThis simplified equation earns a mark **(M1)**."},{"type":"text","order":2,"title":"Factorise and solve for y","content":"Next, we solve the quadratic equation $y^2 - 5y + 6 = 0$ by factorising.\\n\\nWe need to find two numbers that multiply to $+6$ and add up to $-5$. These numbers are $-2$ and $-3$.\\n\\nThus, we factorise the quadratic equation as:\\n\\n$$(y - 2)(y - 3) = 0$$\\n\\nSetting each bracket to zero gives us two solutions for $y$ **(M1)**:\\n\\n$$y = 2 \\quad \\text{or} \\quad y = 3$$"},{"type":"text","order":3,"title":"Substitute back to find x","content":"Now we substitute our $y$ values back into the linear equation $x = y - 5$ to find the corresponding values of $x$ **(M1)**:\\n\\n- For $y = 2$:\\n $$x = 2 - 5 = -3$$\\n\\n- For $y = 3$:\\n $$x = 3 - 5 = -2$$"},{"type":"text","order":4,"title":"State the final solutions","content":"Pair up your values clearly to present the final coordinates **(A1)**:\\n\\n- $x = -3, y = 2$\\n- $x = -2, y = 3$"}]},{"id":"d82b8177-a41d-4597-a9f8-0a3881b3f9fb:substitute_y","label":"Substitution for y","steps":[{"type":"text","order":0,"title":"Rearrange the linear equation","content":"To use the **Substitution for $y$** approach, we begin by rearranging the linear equation to express $y$ in terms of $x$.\n\nThe given linear equation is:\n$$x = y - 5$$\n\nAdding $5$ to both sides gives:\n$$y = x + 5$$","highlights":[{"text":"x = y - 5","location":"specification"}]},{"type":"hyper_latex","block":{"nodes":[{"latex":"$$$x^2 + \\htmlData{anchor=y-var}{\\color{@sky}{y}}^2 = 13$$"},{"latex":"$$$\\htmlData{anchor=y-expr}{\\color{@sky}{y}} = \\htmlData{anchor=x-expr}{\\color{@amber}{x + 5}}$$"},{"latex":"$$$x^2 + (\\htmlData{anchor=sub-target}{\\color{@amber}{x + 5}})^2 = 13$$"}],"highlights":[{"color":"sky","style":"box","anchor":"y-var"},{"color":"sky","style":"text-color","anchor":"y-expr"},{"color":"amber","style":"text-color","anchor":"x-expr"},{"color":"amber","style":"box","anchor":"sub-target"}],"connections":[{"to":"y-var","from":"y-expr","color":"sky","label":"substitute","style":"solid"},{"to":"sub-target","from":"x-expr","color":"amber","label":"substituted value","style":"solid"}]},"order":1,"title":"Substitute into the quadratic","description":"Now, substitute the expression for $y$ into the quadratic equation $x^2 + y^2 = 13$ so we obtain an equation with only one variable, $x$."},{"type":"text","order":2,"title":"Expand and simplify","content":"We expand the bracket and simplify the equation to form a standard quadratic equation of the form $ax^2 + bx + c = 0$:\n\n$$\\begin{aligned} x^2 + (x + 5)^2 &= 13 \\\\ x^2 + (x^2 + 10x + 25) &= 13 \\\\ 2x^2 + 10x + 25 &= 13 \\end{aligned}$$\n\nSubtract $13$ from both sides to set the equation to $0$:\n\n$$2x^2 + 10x + 12 = 0$$\n\nTo make factorising simpler, we can divide the entire equation by $2$:\n\n$$x^2 + 5x + 6 = 0$$"},{"type":"text","order":3,"title":"Solve for x","content":"We solve the quadratic equation $x^2 + 5x + 6 = 0$ by factorising.\n\nWe need two numbers that multiply to $6$ and add to $5$. These numbers are $2$ and $3$:\n\n$$(x + 2)(x + 3) = 0$$\n\nThis gives two possible values for $x$:\n\n$$x = -2 \\quad \\text{or} \\quad x = -3$$"},{"type":"text","order":4,"title":"Find the corresponding y values","content":"Substitute the $x$ values back into our rearranged linear equation $y = x + 5$ to find the corresponding values of $y$.\n\n- **When $x = -2$:**\n $$y = -2 + 5 = 3$$\n\n- **When $x = -3$:**\n $$y = -3 + 5 = 2$$\n\nTherefore, the two pairs of solutions are:\n$$x = -2, y = 3 \\quad \\text{and} \\quad x = -3, y = 2$$"}]}],"generatedAt":"2026-06-12T20:24:12.765Z","modelVersion":"gemini-3.5-flash","explanationVersion":"v2"},"firstAttemptFullMarks":{"percent":25},"walkthroughVideo":{"url":"https://assets.mathsgenie.co.uk/walkthroughs/review/0ecf0e0d-12e7-4ad9-881c-0f7e1f01180e.mp4","durationSeconds":197,"posterUrl":"https://assets.mathsgenie.co.uk/walkthroughs/review/0ecf0e0d-12e7-4ad9-881c-0f7e1f01180e.jpg"}},{"id":"5b64e26f-c984-4b4f-a998-70eee47a327f","revisionId":"9fa2d1bb-5cdb-43c4-92b2-3e694f74cdb7","questionNumber":null,"questionType":"LA","specification":"","difficultyBand":"medium","tier":null,"totalMarks":5,"sourceTitle":null,"calculatorUse":"non_calc","isCalculation":true,"parts":[{"id":"5d191032-59ec-4448-8689-6de450763475","label":"a","content":"Solve the simultaneous equations\n\n$$x^2 + y^2 = 25$$\n\n$$y = x + 1$$","markscheme":"- Substitutes $y = x + 1$ into the first equation: $x^2 + (x + 1)^2 = 25$ M1\n- Expands and simplifies to a quadratic equation: $2x^2 + 2x - 24 = 0$ or $x^2 + x - 12 = 0$ M1\n- Factorises or uses formula to find $x$: $(x + 4)(x - 3) = 0$ so $x = -4$ or $x = 3$ M1\n- Substitutes $x$ values back to find $y$: $y = -4 + 1 = -3$ and $y = 3 + 1 = 4$ M1\n- Final answers: $x = -4, y = -3$ and $x = 3, y = 4$ A1","marks":5,"order":0,"markingRubric":{"totalMarks":5,"points":[{"id":"p1","marks":1,"description":"Evidence of substituting the linear equation into the quadratic equation.","code":"M","acceptable":["x^2 + (x + 1)^2 = 25"]},{"id":"p2","marks":1,"description":"Expands brackets and simplifies to a three-term quadratic equal to zero.","code":"M","acceptable":["2x^2 + 2x - 24 = 0","x^2 + x - 12 = 0"]},{"id":"p3","marks":1,"description":"Solving the quadratic equation to find two values for the first variable.","code":"M","acceptable":["x = -4 and x = 3"]},{"id":"p4","marks":1,"description":"Substituting found values into the linear equation to find the corresponding second variable values.","code":"M","acceptable":["y = -3 and y = 4"]},{"id":"p5","marks":1,"description":"Both pairs of solutions given correctly and paired properly.","code":"A"}],"finalAnswers":["(-4, -3), (3, 4)"],"generatedAt":"2026-05-06T14:07:11.467Z","modelVersion":"gemini-3-flash-preview"},"tableImages":[],"explanation":{"methods":[{"id":"substitute_and_factorise","label":"Substitute and factorise","steps":[{"type":"hyper_latex","block":{"nodes":[{"latex":"$$x^2+\\htmlData{anchor=yterm}{\\color{@sky}{y^2}}=25$"},{"latex":"$$x^2+\\htmlData{anchor=substitution}{\\color{@sky}{(x+1)^2}}=25$"}],"highlights":[{"color":"sky","style":"text-color","anchor":"yterm"},{"color":"sky","style":"text-color","anchor":"substitution"}],"connections":[{"to":"substitution","from":"yterm","color":"sky","label":"use y = x + 1"}]},"order":0,"title":"Substitute for y","description":"Replace y in the first equation with x + 1. Writing this substitution earns the first method mark (M1)."},{"type":"hyper_latex","block":{"nodes":[{"latex":"$$x^2+\\htmlData{anchor=bracket}{\\color{@orange}{(x+1)^2}}=25$"},{"latex":"$$x^2+\\htmlData{anchor=expanded}{\\color{@orange}{x^2+2x+1}}=25$"},{"latex":"$$\\htmlData{anchor=collected}{\\color{@teal}{2x^2+2x-24}}=0$"},{"latex":"$$\\htmlData{anchor=divided}{\\color{@teal}{x^2+x-12}}=0$"}],"highlights":[{"color":"orange","style":"text-color","anchor":"bracket"},{"color":"orange","style":"text-color","anchor":"expanded"},{"color":"teal","style":"text-color","anchor":"collected"},{"color":"teal","style":"text-color","anchor":"divided"}],"connections":[{"to":"expanded","from":"bracket","color":"orange","label":"expand"},{"to":"collected","from":"expanded","color":"teal","label":"collect terms"},{"to":"divided","from":"collected","color":"teal","label":"divide by 2"}]},"order":1,"title":"Form the quadratic","description":"Expand the bracket, collect like terms, then divide every term by 2. Reaching either quadratic form earns the second method mark (M1)."},{"type":"hyper_latex","block":{"nodes":[{"latex":"$$\\htmlData{anchor=quadratic}{\\color{@purple}{x^2+x-12}}=0$"},{"latex":"$$\\htmlData{anchor=factorised}{\\color{@purple}{(x+4)(x-3)=0}}$"},{"latex":"$$\\htmlData{anchor=leftfactor}{\\color{@orange}{x+4=0}}$"},{"latex":"$$\\htmlData{anchor=rightfactor}{\\color{@sky}{x-3=0}}$"}],"highlights":[{"color":"purple","style":"text-color","anchor":"quadratic"},{"color":"purple","style":"text-color-with-background","anchor":"factorised"},{"color":"orange","style":"text-color","anchor":"leftfactor"},{"color":"sky","style":"text-color","anchor":"rightfactor"}],"connections":[{"to":"factorised","from":"quadratic","color":"purple","label":"factorise"},{"to":"leftfactor","from":"factorised","color":"orange","label":"first factor is zero"},{"to":"rightfactor","from":"factorised","color":"sky","label":"second factor is zero"}]},"order":2,"title":"Factorise and split","description":"Factorise the quadratic, then use the zero-product rule: if a product is zero, at least one factor must be zero. This sets up both possible roots."},{"type":"hyper_latex","block":{"nodes":[{"latex":"$$\\htmlData{anchor=lefteq}{\\color{@orange}{x+4=0}}$"},{"latex":"$$\\htmlData{anchor=leftroot}{\\color{@orange}{x=-4}}$"},{"latex":"$$\\htmlData{anchor=righteq}{\\color{@sky}{x-3=0}}$"},{"latex":"$$\\htmlData{anchor=rightroot}{\\color{@sky}{x=3}}$"}],"highlights":[{"color":"orange","style":"text-color","anchor":"lefteq"},{"color":"orange","style":"text-color-with-background","anchor":"leftroot"},{"color":"sky","style":"text-color","anchor":"righteq"},{"color":"sky","style":"text-color-with-background","anchor":"rightroot"}],"connections":[{"to":"leftroot","from":"lefteq","color":"orange","label":"subtract 4"},{"to":"rightroot","from":"righteq","color":"sky","label":"add 3"}]},"order":3,"title":"Solve both branches","description":"Solve each one-step equation. Obtaining both values of x earns the third method mark (M1)."},{"type":"hyper_latex","block":{"nodes":[{"latex":"$$y=\\htmlData{anchor=xminusfour}{\\color{@orange}{-4}}+1$"},{"latex":"$$\\htmlData{anchor=yminusthree}{\\color{@orange}{y=-3}}$"},{"latex":"$$y=\\htmlData{anchor=xthree}{\\color{@sky}{3}}+1$"},{"latex":"$$\\htmlData{anchor=yfour}{\\color{@sky}{y=4}}$"}],"highlights":[{"color":"orange","style":"text-color","anchor":"xminusfour"},{"color":"orange","style":"text-color-with-background","anchor":"yminusthree"},{"color":"sky","style":"text-color","anchor":"xthree"},{"color":"sky","style":"text-color-with-background","anchor":"yfour"}],"connections":[{"to":"yminusthree","from":"xminusfour","color":"orange","label":"substitute"},{"to":"yfour","from":"xthree","color":"sky","label":"substitute"}]},"order":4,"title":"Find both y-values","description":"Put each value of x into y = x + 1. Finding both corresponding values of y earns the fourth method mark (M1)."},{"type":"text","order":5,"title":"State both solutions","content":"Pair each $x$-value with the $y$-value found from it:\n\n$$\n(x,y)=(-4,-3) \\quad \\text{or} \\quad (x,y)=(3,4).\n$$\n\nStating both complete ordered pairs earns the final accuracy mark (A1)."}]},{"id":"substitute_and_formula","label":"Substitute and use formula","steps":[{"type":"hyper_latex","block":{"nodes":[{"latex":"$$x^2+\\htmlData{anchor=yterm}{\\color{@sky}{y^2}}=25$"},{"latex":"$$x^2+\\htmlData{anchor=substitution}{\\color{@sky}{(x+1)^2}}=25$"}],"highlights":[{"color":"sky","style":"text-color","anchor":"yterm"},{"color":"sky","style":"text-color","anchor":"substitution"}],"connections":[{"to":"substitution","from":"yterm","color":"sky","label":"use y = x + 1"}]},"order":0,"title":"Substitute for y","description":"Replace y in the first equation with x + 1. Writing this substitution earns the first method mark (M1)."},{"type":"hyper_latex","block":{"nodes":[{"latex":"$$x^2+\\htmlData{anchor=bracket}{\\color{@orange}{(x+1)^2}}=25$"},{"latex":"$$x^2+\\htmlData{anchor=expanded}{\\color{@orange}{x^2+2x+1}}=25$"},{"latex":"$$\\htmlData{anchor=collected}{\\color{@teal}{2x^2+2x-24}}=0$"},{"latex":"$$\\htmlData{anchor=divided}{\\color{@teal}{x^2+x-12}}=0$"}],"highlights":[{"color":"orange","style":"text-color","anchor":"bracket"},{"color":"orange","style":"text-color","anchor":"expanded"},{"color":"teal","style":"text-color","anchor":"collected"},{"color":"teal","style":"text-color","anchor":"divided"}],"connections":[{"to":"expanded","from":"bracket","color":"orange","label":"expand"},{"to":"collected","from":"expanded","color":"teal","label":"collect terms"},{"to":"divided","from":"collected","color":"teal","label":"divide by 2"}]},"order":1,"title":"Form the quadratic","description":"Expand the bracket, collect like terms, then divide every term by 2. Reaching either quadratic form earns the second method mark (M1)."},{"type":"hyper_latex","block":{"nodes":[{"latex":"$$x=\\dfrac{-b\\pm\\sqrt{b^2-4ac}}{2a}$"},{"latex":"$$\\htmlData{anchor=substituted}{x=\\dfrac{-1\\pm\\sqrt{1^2-4(1)(-12)}}{2(1)}}$"},{"latex":"$$\\htmlData{anchor=simplified}{x=\\dfrac{-1\\pm7}{2}}$"},{"latex":"$$\\htmlData{anchor=plusroot}{\\color{@sky}{x=3}}$"},{"latex":"$$\\htmlData{anchor=minusroot}{\\color{@orange}{x=-4}}$"}],"highlights":[{"color":"sky","style":"text-color-with-background","anchor":"plusroot"},{"color":"orange","style":"text-color-with-background","anchor":"minusroot"}],"connections":[{"to":"simplified","from":"substituted","color":"purple","label":"square root is 7"},{"to":"plusroot","from":"simplified","color":"sky","label":"use plus"},{"to":"minusroot","from":"simplified","color":"orange","label":"use minus"}]},"order":2,"title":"Apply the formula","description":"Use a = 1, b = 1 and c = -12 in the quadratic formula. Evaluate the plus and minus cases separately; obtaining both roots earns the third method mark (M1)."},{"type":"hyper_latex","block":{"nodes":[{"latex":"$$y=\\htmlData{anchor=xthree}{\\color{@sky}{3}}+1$"},{"latex":"$$\\htmlData{anchor=yfour}{\\color{@sky}{y=4}}$"},{"latex":"$$y=\\htmlData{anchor=xminusfour}{\\color{@orange}{-4}}+1$"},{"latex":"$$\\htmlData{anchor=yminusthree}{\\color{@orange}{y=-3}}$"}],"highlights":[{"color":"sky","style":"text-color","anchor":"xthree"},{"color":"sky","style":"text-color-with-background","anchor":"yfour"},{"color":"orange","style":"text-color","anchor":"xminusfour"},{"color":"orange","style":"text-color-with-background","anchor":"yminusthree"}],"connections":[{"to":"yfour","from":"xthree","color":"sky","label":"substitute"},{"to":"yminusthree","from":"xminusfour","color":"orange","label":"substitute"}]},"order":3,"title":"Find both y-values","description":"Substitute each root into y = x + 1. Finding both corresponding y-values earns the fourth method mark (M1)."},{"type":"text","order":4,"title":"State both solutions","content":"Match each root with its corresponding $y$-value:\n\n$$\n(x,y)=(-4,-3) \\quad \\text{or} \\quad (x,y)=(3,4).\n$$\n\nGiving both complete ordered pairs earns the final accuracy mark (A1)."}]}],"generatedAt":"2026-07-19T16:26:08.436Z","modelVersion":"gpt-5.6-sol","explanationVersion":"v3"},"options":[]}],"options":[],"explanation":{"methods":[{"id":"5d191032-59ec-4448-8689-6de450763475:substitution_y","label":"Substitution for y","steps":[{"type":"hyper_latex","block":{"nodes":[{"latex":"$$y = \frac{}{}\begin{aligned} &x^2 + \big(\\htmlData{anchor=y-var}{\\color{@sky}{y}}\\big)^2 = 25 \\\\ &\\htmlData{anchor=y-expr}{\\color{@sky}{x + 1}} \\end{aligned}$"},{"latex":"$$x^2 + (\\htmlData{anchor=subbed-expr}{\\color{@sky}{x + 1}})^2 = 25$"}],"highlights":[{"color":"sky","style":"box","anchor":"y-var"},{"color":"sky","style":"underline","anchor":"y-expr"}],"connections":[{"to":"subbed-expr","from":"y-expr","color":"sky","label":"substitute","style":"solid"}]},"order":0,"title":"Substitute y into quadratic","description":"We are given two equations:\n\n1. $x^2 + y^2 = 25$\n2. $y = x + 1$\n\nTo solve them using the substitution method for $y$, we take the expression for $y$ from the linear equation and substitute it directly into the quadratic equation."},{"type":"text","order":1,"title":"Expand and simplify","content":"Now, we expand the squared term $(x + 1)^2$ and simplify the equation:\n\n$$(x + 1)^2 = x^2 + 2x + 1$$\n\nSubstitute this back into our equation:\n\n$$x^2 + (x^2 + 2x + 1) = 25$$\n\nCombine the like terms:\n\n$$2x^2 + 2x + 1 = 25$$"},{"type":"text","order":2,"title":"Form a quadratic equation","content":"To solve the quadratic equation, we must set it to $0$ by subtracting $25$ from both sides:\n\n$$2x^2 + 2x - 24 = 0$$\n\nWe can make the equation simpler by dividing every term by $2$:\n\n$$x^2 + x - 12 = 0$$"},{"type":"text","order":3,"title":"Solve for x","content":"Next, we solve $x^2 + x - 12 = 0$ by factorising. We need two numbers that multiply to $-12$ and add up to $1$. These numbers are $+4$ and $-3$:\n\n$$(x + 4)(x - 3) = 0$$\n\nThis gives us two possible values for $x$:\n\n$$x = -4 \\quad \\text{or} \\quad x = 3$$"},{"type":"text","order":4,"title":"Find the corresponding y values","content":"Now substitute both $x$ values back into the linear equation $y = x + 1$ to find their corresponding $y$ values:\n\n- **If $x = -4$:**\n $$y = -4 + 1 = -3$$\n\n- **If $x = 3$:**\n $$y = 3 + 1 = 4$$\n\nThis gives us our two pairs of solutions:\n\n$$(x = -4, y = -3) \\quad \\text{and} \\quad (x = 3, y = 4)$$"}]},{"id":"5d191032-59ec-4448-8689-6de450763475: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 start with the linear equation:\n\n$$y = x + 1$$\n\nWe rearrange this equation to express $x$ in terms of $y$ by subtracting $1$ from both sides:\n\n$$x = y - 1$$"},{"type":"hyper_latex","block":{"nodes":[{"latex":"$$x^2 + y^2 = 25$"},{"latex":"$$\text{Substitute } \begin{aligned} \bullet \text{ linear: } \rightarrow &\boxed{\bar{x} = \text{\\htmlData{anchor=sub-val}{\\color{@sky}{y - 1}}}} \\\\ \bullet \text{ quadratic: } \rightarrow &\boxed{(\\text{\\htmlData{anchor=sub-x}{\\color{@sky}{x}}})^2 + y^2 = 25} \\end{aligned}$"},{"latex":"$$(\\color{@sky}{y - 1})^2 + y^2 = 25$"}],"highlights":[{"color":"sky","style":"text-color","anchor":"sub-val"},{"color":"sky","style":"text-color","anchor":"sub-x"}],"connections":[{"to":"sub-x","from":"sub-val","color":"sky","label":"substitute","style":"solid"}]},"order":1,"title":"Substitute into quadratic","description":"Substitute $x = y - 1$ into the quadratic equation $x^2 + y^2 = 25$."},{"type":"text","order":2,"title":"Expand and simplify","content":"Now we expand the brackets and simplify the equation to form a standard quadratic equation in terms of $y$:\n\n$$(y - 1)^2 + y^2 = 25$$\n\n$$(y^2 - 2y + 1) + y^2 = 25$$\n\n$$2y^2 - 2y + 1 = 25$$\n\nSubtract $25$ from both sides to set the quadratic to $0$:\n\n$$2y^2 - 2y - 24 = 0$$\n\nTo make it simpler to solve, divide the entire equation by $2$:\n\n$$y^2 - y - 12 = 0$$"},{"type":"text","order":3,"title":"Solve for y","content":"Next, we factorise the quadratic equation $y^2 - y - 12 = 0$. We need two numbers that multiply to $-12$ and add to $-1$:\n\nThese numbers are $-4$ and $+3$:\n\n$$(y - 4)(y + 3) = 0$$\n\nThis gives two solutions for $y$:\n\n$$y = 4 \\quad \\text{or} \\quad y = -3$$"},{"type":"text","order":4,"title":"Find the x values","content":"Now we substitute our $y$ values back into our rearranged linear equation, $x = y - 1$, to find the corresponding values of $x$:\n\n* **If $y = 4$:**\n $$x = 4 - 1 = 3$$\n\n* **If $y = -3$:**\n $$x = -3 - 1 = -4$$\n\nThus, our two pairs of solutions are:\n\n$$x = 3, y = 4 \\quad \\text{and} \\quad x = -4, y = -3$$"}]},{"id":"5d191032-59ec-4448-8689-6de450763475:graphical","label":"Graphical Method","steps":[{"type":"text","order":0,"title":"Understand the graphical approach","content":"To solve simultaneous equations graphically, we plot both equations on the same coordinate grid. The points where the two graphs intersect represent the solutions to the simultaneous equations."},{"type":"text","order":1,"title":"Identify the circle shape","content":"The first equation is $$x^2 + y^2 = 25$$\n\nThis is a circle centered at the origin $(0, 0)$ with a radius $r$ of:\n$$r = \\sqrt{25} = 5$$\n\nTo sketch this circle, we can plot its key intercepts on the axes:\n- $(5, 0)$ and $(-5, 0)$ on the $x$-axis\n- $(0, 5)$ and $(0, -5)$ on the $y$-axis"},{"type":"text","order":2,"title":"Identify the line shape","content":"The second equation is $$y = x + 1$$\n\nThis is a straight line in the form $y = mx + c$ with gradient $m = 1$ and $y$-intercept $(0, 1)$.\n\nWe can find a few points to plot the line:\n- When $x = 0$, $y = 1 \\implies (0, 1)$\n- When $x = 3$, $y = 3 + 1 = 4 \\implies (3, 4)$\n- When $x = -4$, $y = -4 + 1 = -3 \\implies (-4, -3)$"},{"type":"text","order":3,"title":"Plot the equations","content":"Draw both shapes on a grid:\n1. Draw a perfect circle centered at $(0,0)$ passing through $5$ and $-5$ on both axes.\n2. Draw a straight line passing through $(0, 1)$, $(3, 4)$, and $(-4, -3)$.\n\nObserve where the line crosses the circle. They intersect at exactly two points.","calculator":[{"gdc":{"instructions":"1. Enter the relations or equations into your graphing menu. If your GDC supports relation graphing, enter \$x^2 + y^2 = 25\$ and \$y = x + 1\$.\n2. If not, rewrite the circle as \$y = \\pm\\sqrt{25 - x^2}\$ and graph both halves alongside the line.\n3. Use the 'Intersection' tool (usually under G-Solv or Analyze Graph) to find the intersection points."},"ti84":{"instructions":"To graph a circle on a TI-84, solve for $y$: $y = \\pm \\sqrt{25 - x^2}$.\n1. Press [Y=] and enter:\n \$Y_1 = \\sqrt{25 - X^2}\$\n \$Y_2 = -\\sqrt{25 - X^2}\$\n \$Y_3 = X + 1\$\n2. Press [ZOOM] [5] (ZSquare) so the circle doesn't look like an ellipse.\n3. Press [2nd] [TRACE] [5] (intersect) to find the intersections by selecting the curves and pressing enter near the intersection points."},"desmos":{"expression":"x^2 + y^2 = 25 \\\\ y = x + 1","instructions":"1. In the first expression box, type: `x^2 + y^2 = 25`\n2. In the second expression box, type: `y = x + 1`\n3. Click on the gray intersection dots shown on the graph to reveal their coordinates."},"description":"Plotting a circle and a line to find their intersections using a graphing utility."}]},{"type":"text","order":4,"title":"Read the intersection points","content":"From your graph, read the coordinates of the two points where the line cuts through the circle:\n- One intersection is in the first quadrant at $$(3, 4)$$\n- The other intersection is in the third quadrant at $$(-4, -3)$$\n\nThese coordinate pairs are the solutions to our simultaneous equations."},{"type":"text","order":5,"title":"State the final solutions","content":"The final simultaneous solutions are:\n\n$$x = 3, y = 4$$\n$$\\text{and}$$\n$$x = -4, y = -3$$"}]}],"generatedAt":"2026-06-12T17:20:57.894Z","modelVersion":"gemini-3.5-flash","explanationVersion":"v2"},"firstAttemptFullMarks":{"percent":67},"walkthroughVideo":{"url":"https://assets.mathsgenie.co.uk/walkthroughs/review/0a87ac81-18eb-4ec2-beab-f6456c554d27.mp4","durationSeconds":185,"posterUrl":"https://assets.mathsgenie.co.uk/walkthroughs/review/0a87ac81-18eb-4ec2-beab-f6456c554d27.jpg"}},{"id":"6a051e5d-1130-4612-9b39-cd388c35e097","revisionId":"eec0c428-9ed7-4df8-8a7a-ca41df6e993a","questionNumber":null,"questionType":"LA","specification":"","difficultyBand":"medium","tier":null,"totalMarks":5,"sourceTitle":null,"calculatorUse":"calc","isCalculation":true,"parts":[{"id":"ac0b0b5c-cc8e-4601-94b1-223610d57c74","label":"a","content":"Solve the simultaneous equations\nGive your answers to 3 significant figures\n\n$$x^2 + y^2 = 13$$\n\n$$2x + y = 3$$","markscheme":"- Rearranges linear equation to make $y$ the subject: $y = 3 - 2x$ M1\n- Substitutes linear expression into quadratic: $x^2 + (3 - 2x)^2 = 13$ M1\n- Simplifies to a standard quadratic form: $5x^2 - 12x - 4 = 0$ M1\n- Uses the quadratic formula or completing the square to solve for $x$: $x = \\frac{12 \\pm \\sqrt{(-12)^2 - 4(5)(-4)}}{10}$ M1\n- Final answers rounded to 3sf: $x = 2.70, y = -2.39$ and $x = -0.297, y = 3.59$ A1","marks":5,"order":0,"markingRubric":{"totalMarks":5,"points":[{"id":"p1","marks":1,"description":"Rearranges the linear equation to isolate one variable.","code":"M"},{"id":"p2","marks":1,"description":"Substitutes the linear expression into the quadratic equation.","code":"M"},{"id":"p3","marks":1,"description":"Simplifies the resulting equation into a standard quadratic form $ax^2 + bx + c = 0$.","code":"M"},{"id":"p4","marks":1,"description":"Correct application of the quadratic formula or completing the square for their quadratic.","code":"M"},{"id":"p5","marks":1,"description":"Provides both sets of $x$ and $y$ values correctly rounded to 3 significant figures.","code":"A"}],"finalAnswers":[],"generatedAt":"2026-05-06T14:07:08.600Z","modelVersion":"gemini-3-flash-preview"},"tableImages":[],"explanation":{"methods":[{"id":"quadratic_formula","label":"Substitution and quadratic formula","steps":[{"type":"hyper_latex","block":{"nodes":[{"latex":"$$2x+y=3$"},{"latex":"$$\\htmlData{anchor=y_subject}{\\color{@sky}{y}}=3-2x$"}],"highlights":[{"color":"sky","style":"text-color","anchor":"y_subject"}]},"order":0,"title":"Make y the subject","description":"Start with the linear equation because it is straightforward to rearrange. This earns the first M1 mark."},{"type":"hyper_latex","block":{"nodes":[{"latex":"$$13=x^2+\\htmlData{anchor=old_y}{\\color{@orange}{y^2}}$"},{"latex":"$$13=x^2+\\htmlData{anchor=new_y}{\\color{@orange}{(3-2x)^2}}$"}],"highlights":[{"color":"orange","style":"text-color","anchor":"old_y"},{"color":"orange","style":"text-color","anchor":"new_y"}],"connections":[{"to":"new_y","from":"old_y","color":"orange","label":"substitute"}]},"order":1,"title":"Substitute for y","description":"Replace y in the circle equation with 3 - 2x. Writing this substitution earns the second M1 mark."},{"type":"hyper_latex","block":{"nodes":[{"latex":"$$13=x^2+\\htmlData{anchor=squared_bracket}{\\color{@purple}{(3-2x)^2}}$"},{"latex":"$$13=x^2+\\htmlData{anchor=expanded_bracket}{\\color{@purple}{9-12x+4x^2}}$"},{"latex":"$$5x^2-12x+9=13$"},{"latex":"$$\\htmlData{anchor=standard_form}{\\color{@green}{5x^2-12x-4=0}}$"}],"highlights":[{"color":"green","style":"text-color","anchor":"standard_form"}],"connections":[{"to":"expanded_bracket","from":"squared_bracket","color":"purple","label":"expand the square"}]},"order":2,"title":"Form the quadratic","description":"Expand the squared bracket carefully, then collect like terms. Reaching the standard quadratic form earns the third M1 mark."},{"type":"hyper_latex","block":{"nodes":[{"latex":"$$x=\\frac{-(-12)\\pm\\sqrt{(-12)^2-4(5)(-4)}}{2(5)}$"},{"latex":"$$x=\\frac{12\\pm\\sqrt{224}}{10}$"},{"latex":"$$\\htmlData{anchor=positive_root}{\\color{@sky}{x=2.696662\\ldots}}$"},{"latex":"$$\\htmlData{anchor=negative_root}{\\color{@sky}{x=-0.296662\\ldots}}$"}],"highlights":[{"color":"sky","style":"text-color","anchor":"positive_root"},{"color":"sky","style":"text-color","anchor":"negative_root"}]},"order":3,"title":"Apply the formula","description":"Here a = 5, b = -12 and c = -4; keep both negative signs when substituting. Use the quadratic formula shown below. Correct use earns the fourth M1 mark."},{"type":"hyper_latex","block":{"nodes":[{"latex":"$$y=3-2x$"},{"latex":"$$y_1=3-2\\left(2.696662\\ldots\\right)$"},{"latex":"$$\\htmlData{anchor=y_one}{\\color{@sky}{y_1=-2.393325\\ldots}}$"},{"latex":"$$y_2=3-2\\left(-0.296662\\ldots\\right)$"},{"latex":"$$\\htmlData{anchor=y_two}{\\color{@orange}{y_2=3.593325\\ldots}}$"}],"highlights":[{"color":"sky","style":"text-color","anchor":"y_one"},{"color":"orange","style":"text-color","anchor":"y_two"}]},"order":4,"title":"Find each y-value","description":"Substitute each unrounded x-value into y = 3 - 2x. Keeping the calculator values unrounded prevents premature-rounding errors."},{"type":"text","order":5,"title":"Round and pair answers","content":"Round only at the end, and keep each $x$-value with the $y$-value calculated from it:\n\n$$\n(x,y)=(2.70,-2.39)\\quad\\text{or}\\quad(x,y)=(-0.297,3.59)\n$$\n\nBoth correctly paired solutions to $3$ significant figures earn the final A1 mark.","highlights":[{"text":"Give your answers to 3 significant figures","location":"specification"}]}]},{"id":"completing_square","label":"Substitution and completing the square","steps":[{"type":"hyper_latex","block":{"nodes":[{"latex":"$$2x+y=3$"},{"latex":"$$\\color{@sky}{\\htmlData{anchor=y_subject}{y}}=3-2x$"}],"highlights":[{"color":"sky","style":"text-color","anchor":"y_subject"}]},"order":0,"title":"Make y the subject","description":"Rearrange the linear equation first. This earns the first M1 mark."},{"type":"hyper_latex","block":{"nodes":[{"latex":"$$13=x^2+\\color{@orange}{\\htmlData{anchor=y_squared}{y^2}}$"},{"latex":"$$13=x^2+\\color{@orange}{\\htmlData{anchor=substitution}{(3-2x)^2}}$"}],"highlights":[{"color":"orange","style":"text-color","anchor":"y_squared"},{"color":"orange","style":"text-color","anchor":"substitution"}],"connections":[{"to":"substitution","from":"y_squared","color":"orange","label":"substitute"}]},"order":1,"title":"Substitute for y","description":"Replace y in the circle equation with 3 - 2x. This correct substitution earns the second M1 mark."},{"type":"hyper_latex","block":{"nodes":[{"latex":"$$13=x^2+\\color{@purple}{\\htmlData{anchor=square}{(3-2x)^2}}$"},{"latex":"$$13=x^2+\\color{@purple}{\\htmlData{anchor=expanded}{9-12x+4x^2}}$"},{"latex":"$$5x^2-12x+9=13$"},{"latex":"$$\\color{@green}{\\htmlData{anchor=standard}{5x^2-12x-4=0}}$"}],"highlights":[{"color":"purple","style":"text-color","anchor":"square"},{"color":"purple","style":"text-color","anchor":"expanded"},{"color":"green","style":"text-color","anchor":"standard"}],"connections":[{"to":"expanded","from":"square","color":"purple","label":"expand the square"}]},"order":2,"title":"Form the quadratic","description":"Expand the squared bracket carefully, then collect like terms. Reaching the standard quadratic form earns the third M1 mark."},{"type":"hyper_latex","block":{"nodes":[{"latex":"$$5x^2-12x-4=0$"},{"latex":"$$x^2-\\frac{12}{5}x=\\frac45$"},{"latex":"$$\\left(x-\\frac65\\right)^2=\\frac45+\\frac{36}{25}$"},{"latex":"$$\\left(x-\\frac65\\right)^2=\\frac{56}{25}$"}]},"order":3,"title":"Complete the square","description":"Divide through by 5, then halve the coefficient of x to build the square."},{"type":"hyper_latex","block":{"nodes":[{"latex":"$$\\left(x-\\frac65\\right)^2=\\frac{56}{25}$"},{"latex":"$$x-\\frac65=\\pm\\frac{\\sqrt{56}}5$"},{"latex":"$$x=\\frac{6\\pm\\sqrt{56}}5$"},{"latex":"$$\\color{@sky}{\\htmlData{anchor=root_one}{x=2.696662\\ldots}}$"},{"latex":"$$\\color{@orange}{\\htmlData{anchor=root_two}{x=-0.296662\\ldots}}$"}],"highlights":[{"color":"sky","style":"text-color","anchor":"root_one"},{"color":"orange","style":"text-color","anchor":"root_two"}]},"order":4,"title":"Solve for x","description":"Square-root both sides, remembering the positive and negative roots, then isolate x. Completing the square and solving for x earns the fourth M1 mark."},{"type":"hyper_latex","block":{"nodes":[{"latex":"$$y=3-2x$"},{"latex":"$$y_1=3-2(2.696662\\ldots)$"},{"latex":"$$\\color{@sky}{\\htmlData{anchor=y_one}{y_1=-2.393325\\ldots}}$"},{"latex":"$$y_2=3-2(-0.296662\\ldots)$"},{"latex":"$$\\color{@orange}{\\htmlData{anchor=y_two}{y_2=3.593325\\ldots}}$"}],"highlights":[{"color":"sky","style":"text-color","anchor":"y_one"},{"color":"orange","style":"text-color","anchor":"y_two"}]},"order":5,"title":"Calculate both y-values","description":"Use the unrounded roots in the rearranged linear equation so that the final rounding remains accurate."},{"type":"text","order":6,"title":"Round both pairs","content":"Give each coordinate to three significant figures, keeping each $x$-value paired with the $y$-value calculated from it:\n\n$$ (x,y)=(2.70,-2.39) \\quad \\text{or} \\quad (x,y)=(-0.297,3.59) $$\n\nBoth complete pairs, correctly rounded, earn the final A1 mark.","highlights":[{"text":"Give your answers to 3 significant figures","location":"specification"}]}]}],"generatedAt":"2026-07-19T19:03:03.569Z","modelVersion":"gpt-5.6-sol","explanationVersion":"v3"},"options":[]}],"options":[],"explanation":{"methods":[{"id":"ac0b0b5c-cc8e-4601-94b1-223610d57c74:substitution_y","label":"Substitution (making y the subject)","steps":[{"type":"text","order":0,"title":"Rearrange the linear equation","content":"To solve these simultaneous equations using substitution, we start by rearranging the simpler linear equation to isolate one variable. \n\nLet's make $y$ the subject of the linear equation:\n\n$$2x + y = 3$$\n\nSubtract $2x$ from both sides:\n\n$$y = 3 - 2x$$","highlights":[{"text":"2x + y = 3","location":"specification"}]},{"type":"hyper_latex","block":{"nodes":[{"latex":"$$\text{Substitute } \begin{aligned} \\htmlData{anchor=y-def}{\\color{@sky}{y}} &= \\htmlData{anchor=y-expr}{\\color{@sky}{3 - 2x}} \\end{aligned} \text{ into the quadratic equation:}$"},{"latex":"$$x^2 + (\\htmlData{anchor=y-sub}{\\color{@sky}{y}})^2 = 13$"},{"latex":"$$x^2 + (\\htmlData{anchor=sub-target}{\\color{@sky}{3 - 2x}})^2 = 13$"}],"highlights":[{"color":"sky","style":"text-color","anchor":"y-def"},{"color":"sky","style":"text-color","anchor":"y-sub"}],"connections":[{"to":"sub-target","from":"y-expr","color":"sky","label":"substitute","style":"solid"}]},"order":1,"title":"Substitute into quadratic","description":"We replace $y$ in the second equation with the expression we just found."},{"type":"text","order":2,"title":"Expand and simplify","content":"Now expand the bracket $(3 - 2x)^2$ and simplify the equation to the standard quadratic form $ax^2 + bx + c = 0$:\n\n$$(3 - 2x)^2 = (3 - 2x)(3 - 2x) = 9 - 6x - 6x + 4x^2 = 4x^2 - 12x + 9$$\\ \nSubstitute this back into our equation:\n\n$$x^2 + (4x^2 - 12x + 9) = 13$$\n\nCombine like terms:\n\n$$5x^2 - 12x + 9 = 13$$\n\nSubtract $13$ from both sides to set the equation to $0$:\n\n$$5x^2 - 12x - 4 = 0$$"},{"type":"text","order":3,"title":"Apply the quadratic formula","content":"Since this quadratic does not factorise easily, we use the quadratic formula where $a = 5$, $b = -12$, and $c = -4$:\n\n$$x = \\frac{-b \\pm \\sqrt{b^2 - 4ac}}{2a}$$\n\nSubstituting our values in:\n\n$$x = \\frac{-(-12) \\pm \\sqrt{(-12)^2 - 4(5)(-4)}}{2(5)}$$\n\n$$x = \\frac{12 \\pm \\sqrt{144 + 80}}{10}$$\n\n$$x = \\frac{12 \\pm \\sqrt{224}}{10}$$"},{"type":"text","order":4,"title":"Solve for x","content":"Now, we compute the two values for $x$:\n\n$$\\begin{aligned} x_1 &= \\frac{12 + \\sqrt{224}}{10} \\approx \\frac{12 + 14.967}{10} = 2.697 \\approx 2.70 \\\\ x_2 &= \\frac{12 - \\sqrt{224}}{10} \\approx \\frac{12 - 14.967}{10} = -0.297 \\end{aligned}$$\n\nTo $3$ significant figures, our $x$ values are:\n- $x = 2.70$\n- $x = -0.297$"},{"type":"text","order":5,"title":"Solve for y","content":"Finally, substitute each value of $x$ back into our linear expression $y = 3 - 2x$ to find the corresponding values of $y$:\n\n**Case 1: $x = 2.697...$**\n\n$$y = 3 - 2(2.697...) \\approx -2.39$$\n\n**Case 2: $x = -0.2966...$**\n\n$$y = 3 - 2(-0.2966...) \\approx 3.59$$\n\nThus, our two pairs of solutions to $3$ significant figures are:\n- $x = 2.70, y = -2.39$\n- $x = -0.297, y = 3.59$"}]},{"id":"ac0b0b5c-cc8e-4601-94b1-223610d57c74:substitution_x","label":"Substitution (making x the subject)","steps":[{"type":"text","order":0,"title":"Rearrange the linear equation","content":"To solve these simultaneous equations using the substitution method with $x$ as the subject, we first rearrange the linear equation to express $x$ in terms of $y$.\n\nStart with the linear equation:\n$$2x + y = 3$$\n\nSubtract $y$ from both sides:\n$$2x = 3 - y$$\n\nDivide by 2 to make $x$ the subject:\n$$x = \\frac{3-y}{2}$$\n\nThis rearranges the linear equation to get our expression for substitution.","highlights":[{"text":"$$$2x + y = 3$$","location":"specification"}]},{"type":"hyper_latex","block":{"nodes":[{"latex":"$$\\% \\text{We have the quadratic equation:}$"},{"latex":"$$\\htmlData{anchor=quad}{\\color{@sky}{x}}^2 + y^2 = 13$"},{"latex":"$$\\% \\text{Substitute } \\htmlData{anchor=xsub}{\\color{@sky}{x = \\frac{3-y}{2}}} \\text{ into it:}$"},{"latex":"$$\\color{@sky}{\\left(\\frac{3-y}{2}\\right)}^2 + y^2 = 13$"}],"highlights":[{"color":"sky","style":"text-color","anchor":"quad"},{"color":"sky","style":"text-color","anchor":"xsub"}],"connections":[{"to":"xsub","from":"quad","color":"sky","label":"substitute x","style":"solid"}]},"order":1,"title":"Substitute into the quadratic","description":"Now we substitute our expression for $x$ into the quadratic equation, $x^2 + y^2 = 13$."},{"type":"text","order":2,"title":"Simplify and expand","content":"Now we expand the terms to form a quadratic equation in terms of $y$ only:\n\n$$\\left(\\frac{3-y}{2}\\right)^2 + y^2 = 13$$\n\n$$\\frac{(3-y)^2}{4} + y^2 = 13$$\n\nExpand the numerator $(3-y)^2 = 9 - 6y + y^2$:\n\n$$\\frac{9 - 6y + y^2}{4} + y^2 = 13$$\n\nTo eliminate the fraction, multiply every term in the entire equation by 4:\n\n$$(9 - 6y + y^2) + 4y^2 = 52$$\n\nCombine like terms:\n\n$$5y^2 - 6y + 9 = 52$$\n\nSubtract 52 from both sides to set the quadratic equation equal to 0:\n\n$$5y^2 - 6y - 43 = 0$$"},{"type":"text","order":3,"title":"Solve for y","content":"Now we solve the quadratic equation $5y^2 - 6y - 43 = 0$ using the quadratic formula:\n\n$$y = \\frac{-b \\pm \\sqrt{b^2 - 4ac}}{2a}$$\n\nHere, $a = 5$, $b = -6$, and $c = -43$:\n\n$$y = \\frac{-(-6) \\pm \\sqrt{(-6)^2 - 4(5)(-43)}}{2(5)}$$\n\n$$y = \\frac{6 \\pm \\sqrt{36 + 860}}{10}$$\n\n$$y = \\frac{6 \\pm \\sqrt{896}}{10}$$\n\nCalculating the two possible values for $y$ to 3 significant figures:\n\n$$y_1 = \\frac{6 + \\sqrt{896}}{10} \\approx 3.5933 \\approx 3.59$$\n\n$$y_2 = \\frac{6 - \\sqrt{896}}{10} \\approx -2.3933 \\approx -2.39$$"},{"type":"text","order":4,"title":"Calculate corresponding x-values","content":"Substitute our $y$-values back into our substitution equation $x = \\frac{3-y}{2}$ to find the corresponding values of $x$.\n\n**Case 1: When $y \\approx 3.5933$**\n\n$$x = \\frac{3 - 3.5933}{2} = \\frac{-0.5933}{2} \\approx -0.297\\text{ (to 3 sf)}$$\n\n**Case 2: When $y \\approx -2.3933$**\n\n$$x = \\frac{3 - (-2.3933)}{2} = \\frac{5.3933}{2} \\approx 2.70\\text{ (to 3 sf)}$$\n\nThis gives us the two pairs of solutions:\n* $x = -0.297$, $y = 3.59$\n* $x = 2.70$, $y = 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Quadratic Simultaneous Equations

Question 2

Quadratic Simultaneous Equations Questions

Quadratic Simultaneous Equations

Question 2

Quadratic Simultaneous Equations Questions