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bg-background","children":["$","$L39",null,{"backHref":"/gcse/maths/wjec/questions","basePath":"/gcse/maths/wjec/grade-8-9-8-1-quadratic-simultaneous-equations/questions","formulaUrl":null,"initialIndex":42,"initialQuestionWindow":[{"id":"cfa4a6c3-9c8b-42bd-b6bd-132ac0dc9f19","revisionId":"5b3e5908-776d-4c7a-8f1b-f3d84b1da0d1","questionNumber":null,"questionType":"LA","specification":"","difficultyBand":"medium","tier":null,"totalMarks":5,"sourceTitle":null,"calculatorUse":"non_calc","isCalculation":true,"parts":[{"id":"f9f5142d-836a-44dd-ac46-99bbcff781ab","label":"a","content":"Solve algebraically the simultaneous equations\n\n$$3x^2 + xy = 4$$\n\n$$y - 2x = -8$$","markscheme":"- Rearranges linear: $y = 2x - 8$ M1\n- Substitutes into quadratic: $3x^2 + x(2x - 8) = 4$ M1\n- Simplifies to a quadratic in $x$: $5x^2 - 8x - 4 = 0$ M1\n- Solves for $x$: $(5x + 2)(x - 2) = 0$ so $x = 2$ or $x = -0.4$ M1\n- Finds $y$: $y = -4$ or $y = -8.8$ A1","marks":5,"order":0,"markingRubric":{"totalMarks":5,"points":[{"id":"p1","marks":1,"description":"Rearranges the linear equation to make $y$ or $x$ the subject.","code":"M"},{"id":"p2","marks":1,"description":"Substitutes the linear expression into the quadratic equation.","code":"M"},{"id":"p3","marks":1,"description":"Simplifies to a standard 3-term quadratic equation equal to zero.","code":"M"},{"id":"p4","marks":1,"description":"Solves the quadratic equation to find two values for $x$ (or $y$).","code":"M"},{"id":"p5","marks":1,"description":"Finds the corresponding values for the second variable to give two pairs of solutions.","code":"A"}],"finalAnswers":["x = 2, y = -4","x = -0.4, y = -8.8"],"generatedAt":"2026-05-06T14:07:37.896Z","modelVersion":"gemini-3-flash-preview"},"tableImages":[],"explanation":{"methods":[{"id":"substitution","label":"Substitution","steps":[{"type":"hyper_latex","block":{"nodes":[{"latex":"$$\\htmlData{anchor=line_start}{y-2x=-8}$"},{"latex":"$$\\htmlData{anchor=line_rearranged}{y=2x-8}$"}],"connections":[{"to":"line_rearranged","from":"line_start","color":"sky","label":"add 2x to both sides"}]},"order":0,"title":"Rearrange the line","description":"Make y the subject of the linear equation. Adding 2x to both sides earns the first M1."},{"type":"hyper_latex","block":{"nodes":[{"latex":"$$\\htmlData{anchor=original_equation}{3x^2+xy=4}$"},{"latex":"$$\\htmlData{anchor=substituted_equation}{3x^2+x(2x-8)=4}$"}],"connections":[{"to":"substituted_equation","from":"original_equation","color":"orange","label":"use y=2x-8"}]},"order":1,"title":"Substitute for y","description":"Replace y in the quadratic equation with 2x - 8. Writing this substitution correctly earns the second M1."},{"type":"hyper_latex","block":{"nodes":[{"latex":"$$\\htmlData{anchor=before_expand}{4=3x^2+x(2x-8)}$"},{"latex":"$$\\htmlData{anchor=after_expand}{4=3x^2+2x^2-8x}$"},{"latex":"$$\\htmlData{anchor=before_subtract}{5x^2-8x=4}$"},{"latex":"$$\\htmlData{anchor=quadratic}{5x^2-8x-4=0}$"}],"connections":[{"to":"after_expand","from":"before_expand","color":"sky","label":"expand"},{"to":"before_subtract","from":"after_expand","color":"amber","label":"collect like terms"},{"to":"quadratic","from":"before_subtract","color":"green","label":"subtract 4"}]},"order":2,"title":"Form the quadratic","description":"Expand the bracket, collect like terms, then subtract 4 from both sides. Reaching the final quadratic earns the third M1."},{"type":"hyper_latex","block":{"nodes":[{"latex":"$$\\htmlData{anchor=quadratic_start}{5x^2-8x-4=0}$"},{"latex":"$$\\htmlData{anchor=split_middle}{5x^2+2x-10x-4=0}$"},{"latex":"$$\\htmlData{anchor=grouped}{x(5x+2)-2(5x+2)=0}$"},{"latex":"$$\\htmlData{anchor=factors}{(5x+2)(x-2)=0}$"}],"connections":[{"to":"split_middle","from":"quadratic_start","color":"purple","label":"split -8x"},{"to":"grouped","from":"split_middle","color":"sky","label":"factor each pair"},{"to":"factors","from":"grouped","color":"green","label":"take out 5x+2"}]},"order":3,"title":"Factorise the quadratic","description":"To split the middle term, choose 2 and -10: they multiply to -20, which is 5 times -4, and add to -8. Factor the two pairs, then take out the repeated bracket."},{"type":"hyper_latex","block":{"nodes":[{"latex":"$$\\htmlData{anchor=factors_again}{(5x+2)(x-2)=0}$"},{"latex":"$$\\htmlData{anchor=factor_one}{5x+2=0}$"},{"latex":"$$\\htmlData{anchor=root_one}{x=-0.4}$"},{"latex":"$$\\htmlData{anchor=factor_two}{x-2=0}$"},{"latex":"$$\\htmlData{anchor=root_two}{x=2}$"}],"connections":[{"to":"factor_one","from":"factors_again","color":"sky","label":"first factor is zero"},{"to":"root_one","from":"factor_one","color":"green","label":"solve"},{"to":"factor_two","from":"factors_again","color":"purple","label":"second factor is zero"},{"to":"root_two","from":"factor_two","color":"amber","label":"solve"}]},"order":4,"title":"Solve both factors","description":"Use the zero-product rule: if two factors multiply to zero, at least one factor must equal zero. Solve both resulting linear equations to earn the fourth M1."},{"type":"hyper_latex","block":{"nodes":[{"latex":"$$\\htmlData{anchor=sub_one}{y=2(2)-8}$"},{"latex":"$$\\htmlData{anchor=y_one}{y=-4}$"},{"latex":"$$\\htmlData{anchor=pair_one}{(x,y)=(2,-4)}$"},{"latex":"$$\\htmlData{anchor=sub_two}{y=2(-0.4)-8}$"},{"latex":"$$\\htmlData{anchor=y_two}{y=-8.8}$"},{"latex":"$$\\htmlData{anchor=pair_two}{(x,y)=(-0.4,-8.8)}$"}],"connections":[{"to":"y_one","from":"sub_one","color":"green","label":"calculate"},{"to":"pair_one","from":"y_one","color":"sky","label":"pair with x=2"},{"to":"y_two","from":"sub_two","color":"green","label":"calculate"},{"to":"pair_two","from":"y_two","color":"sky","label":"pair with x=-0.4"}]},"order":5,"title":"Find matching y-values","description":"Substitute each x-value into the rearranged linear equation. Keep each y-value paired with the x-value that produced it; both ordered pairs earn the final A1."}]}],"generatedAt":"2026-07-19T16:13:54.458Z","modelVersion":"gpt-5.6-sol","explanationVersion":"v3"},"options":[]}],"options":[],"explanation":null,"firstAttemptFullMarks":null,"walkthroughVideo":{"url":"https://assets.mathsgenie.co.uk/walkthroughs/review/06f432de-da7a-47dd-bedf-d51f61b75def.mp4","durationSeconds":161,"posterUrl":"https://assets.mathsgenie.co.uk/walkthroughs/review/06f432de-da7a-47dd-bedf-d51f61b75def.jpg"}},{"id":"eb3fe762-6dfd-445a-98fc-aa3ebf296a31","revisionId":"f3d8dc2f-085f-42e5-82ae-4fbf5bca4b50","questionNumber":null,"questionType":"LA","specification":"","difficultyBand":"medium","tier":null,"totalMarks":5,"sourceTitle":null,"calculatorUse":"non_calc","isCalculation":true,"parts":[{"id":"93ff7ea6-0cc2-4714-85c4-dc4c40ae60c8","label":"a","content":"Solve the simultaneous equations\n\n$$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- Solves the quadratic: $(x + 4)(x - 3) = 0$ so $x = -4$ or $x = 3$ M1\n- Finds the corresponding $y$ values: $y = -4 + 1 = -3$ 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 substitution of the linear equation into the quadratic equation.","code":"M"},{"id":"p2","marks":1,"description":"Correct expansion and simplification to a standard form quadratic equation (e.g., $x^2 + x - 12 = 0$).","code":"M"},{"id":"p3","marks":1,"description":"Method to solve their quadratic equation, either by factoring or using the formula, to find two $x$ values.","code":"M"},{"id":"p4","marks":1,"description":"Correct method to find both $y$ values by substituting $x$ values into one of the original 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Quadratic Simultaneous Equations

Question 7

Quadratic Simultaneous Equations Questions

Quadratic Simultaneous Equations

Question 7

Quadratic Simultaneous Equations Questions