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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":95,"initialQuestionWindow":[{"id":"914d10a2-a8a9-4c2d-8999-98c94c6d067d","revisionId":"b387e7f9-0500-4112-9c81-655a1d0daed7","questionNumber":null,"questionType":"LA","specification":"","difficulty":5,"difficultyBand":"medium","totalMarks":5,"sourceTitle":null,"calculatorUse":"non_calc","isCalculation":true,"parts":[{"id":"a8d01198-2e41-4f2c-93a2-d952bcbf518a","label":"a","content":"Solve algebraically the simultaneous equations\n\n$$3x^2 - y^2 = 11$$\n\n$$y - x = 1$$","markscheme":"- Rearranges linear: $y = x + 1$ M1\n- Substitutes into quadratic: $3x^2 - (x + 1)^2 = 11$ M1\n- Simplifies to a quadratic in $x$: $2x^2 - 2x - 12 = 0$ (or $x^2 - x - 6 = 0$) M1\n- Solves for $x$: $(x - 3)(x + 2) = 0$ so $x = 3$ or $x = -2$ M1\n- Finds $y$: $y = 4$ or $y = -1$ 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","acceptable":["y = x + 1","x = y - 1"]},{"id":"p2","marks":1,"description":"Correct substitution into the quadratic equation.","code":"M"},{"id":"p3","marks":1,"description":"Simplifies to a quadratic equation.","code":"M"},{"id":"p4","marks":1,"description":"Finds two values for the first variable.","code":"M"},{"id":"p5","marks":1,"description":"Finds the corresponding values for the second variable.","code":"A"}],"finalAnswers":["x = 3, y = 4; x = -2, y = -1"],"generatedAt":"2026-05-06T14:07:21.855Z","modelVersion":"gemini-3-flash-preview"},"tableImages":[],"explanation":null,"options":[]}],"options":[],"explanation":{"methods":[{"id":"a8d01198-2e41-4f2c-93a2-d952bcbf518a:substitution_y","label":"Substitution for y","steps":[{"type":"hyper_latex","block":{"nodes":[{"latex":"$$$\\text{Our linear equation is: } \\htmlData{anchor=linear-eq}{y - x = 1}$$"},{"latex":"$$$\\text{Add } x \\text{ to both sides to solve for } y:$$"},{"latex":"$$$\\htmlData{anchor=y-expressed}{\\color{@sky}{y}} = \\htmlData{anchor=x-expressed}{\\color{@amber}{x + 1}}$$"}],"highlights":[{"color":"sky","style":"text-color","anchor":"y-expressed"},{"color":"amber","style":"text-color","anchor":"x-expressed"}],"connections":[{"to":"y-expressed","from":"linear-eq","color":"sky","label":"rearrange","style":"solid"}]},"order":0,"title":"Rearrange the linear equation","description":"To use the substitution method, we first rearrange our linear equation to make $y$ the subject."},{"type":"hyper_latex","block":{"nodes":[{"latex":"$$$\\text{Substitute } \\htmlData{anchor=sub-val}{\\color{@amber}{(x + 1)}} \\text{ in place of } \\htmlData{anchor=y-target}{\\color{@sky}{y}} \\text{ in the quadratic equation:}$$"},{"latex":"$$$3x^2 - \\htmlData{anchor=y-sq}{\\color{@sky}{y}}^2 = 11$$"},{"latex":"$$$3x^2 - \\htmlData{anchor=subbed-eq}{\\color{@amber}{(x + 1)}}^2 = 11$$"}],"highlights":[{"color":"sky","style":"text-color","anchor":"y-sq"},{"color":"amber","style":"text-color","anchor":"subbed-eq"}],"connections":[{"to":"subbed-eq","from":"sub-val","color":"amber","label":"substitute","style":"solid"}]},"order":1,"title":"Substitute into the quadratic","description":"Now we substitute this expression for $y$ into the quadratic equation."},{"type":"text","order":2,"title":"Expand and simplify","content":"Expand the bracket $(x + 1)^2$ and simplify the equation:\n\n$$\\begin{aligned} 3x^2 - (x^2 + 2x + 1) &= 11 \\\\ 3x^2 - x^2 - 2x - 1 &= 11 \\\\ 2x^2 - 2x - 1 &= 11 \\end{aligned}$$\n\nSubtract $11$ from both sides to form a quadratic equation set to $0$:\n\n$$2x^2 - 2x - 12 = 0$$"},{"type":"text","order":3,"title":"Solve the quadratic equation","content":"To make solving easier, we can divide the entire equation by $2$:\n\n$$x^2 - x - 6 = 0$$\n\nNow, factorise the quadratic by finding two numbers that multiply to $-6$ and add to $-1$. These numbers are $-3$ and $2$:\n\n$$(x - 3)(x + 2) = 0$$\n\nThis gives two possible values for $x$:\n\n$$x = 3 \\quad \\text{or} \\quad x = -2$$"},{"type":"text","order":4,"title":"Find the corresponding y-values","content":"Substitute each $x$-value back into our linear expression $y = x + 1$ to find the corresponding values of $y$.\n\n- For $x = 3$:\n $$y = 3 + 1 = 4$$\n\n- For $x = -2$:\n $$y = -2 + 1 = -1$$\n\nOur two pairs of solutions are:\n\n$$x = 3, y = 4 \\quad \\text{and} \\quad x = -2, y = -1$$"}]},{"id":"a8d01198-2e41-4f2c-93a2-d952bcbf518a:substitution_x","label":"Substitution for x","steps":[{"type":"text","order":0,"title":"Rearrange the linear equation","content":"To use the substitution method for $x$, we first rearrange the linear equation to express $x$ in terms of $y$.\n\nWe start with the linear equation:\n$$y - x = 1$$\n\nRearranging to make $x$ the subject:\n$$x = y - 1$$","highlights":[{"text":"y - x = 1","location":"specification"}]},{"type":"hyper_latex","block":{"nodes":[{"latex":"$$3 \\htmlData{anchor=x-var}{\\color{@sky}{x}}^2 - y^2 = 11$"},{"latex":"$$\\htmlData{anchor=x-sub}{\\color{@sky}{x = y - 1}}$"},{"latex":"$$3\\htmlData{anchor=subbed-expr}{\\color{@sky}{(y - 1)}}^2 - y^2 = 11$"}],"highlights":[{"color":"sky","style":"text-color","anchor":"x-var"},{"color":"sky","style":"text-color","anchor":"x-sub"},{"color":"sky","style":"text-color","anchor":"subbed-expr"}],"connections":[{"to":"subbed-expr","from":"x-sub","color":"sky","label":"Substitute x","style":"solid"}]},"order":1,"title":"Substitute into quadratic equation","description":"Now, we substitute this expression for $x$ into our quadratic equation $3x^2 - y^2 = 11$."},{"type":"text","order":2,"title":"Expand and simplify","content":"Next, we expand the squared term and simplify the equation to form a standard quadratic equation in terms of $y$:\n\n$$\\begin{aligned} 3(y - 1)^2 - y^2 &= 11 \\\\ 3(y^2 - 2y + 1) - y^2 &= 11 \\\\ 3y^2 - 6y + 3 - y^2 &= 11 \\\\ 2y^2 - 6y + 3 &= 11 \\end{aligned}$$\n\nSubtract $11$ from both sides to set the quadratic to $0$:\n\n$$2y^2 - 6y - 8 = 0$$"},{"type":"text","order":3,"title":"Solve for y","content":"To make factorising easier, we can divide the entire equation by $2$:\n\n$$y^2 - 3y - 4 = 0$$\n\nNow, we find two numbers that multiply to $-4$ and add to $-3$. These numbers are $-4$ and $1$:\n\n$$(y - 4)(y + 1) = 0$$\n\nThis gives two possible values for $y$:\n\n$$y = 4 \\quad \\text{or} \\quad y = -1$$"},{"type":"text","order":4,"title":"Find corresponding x values","content":"Finally, we substitute our $y$-values back into our rearranged linear equation, $x = y - 1$, to find the corresponding values of $x$.\n\n**Case 1: When $y = 4$**\n$$x = 4 - 1 = 3$$\n\n**Case 2: When $y = -1$**\n$$x = -1 - 1 = -2$$\n\nSo, the solutions to the simultaneous equations are:\n$$x = 3, y = 4 \\quad \\text{and} \\quad x = -2, y = -1$$"}]}],"generatedAt":"2026-06-12T23:15:55.694Z","modelVersion":"gemini-3.5-flash","explanationVersion":"v2"},"firstAttemptFullMarks":null,"walkthroughVideo":{"url":"https://assets.mathsgenie.co.uk/walkthroughs/videos/914d10a2-a8a9-4c2d-8999-98c94c6d067d.mp4","durationSeconds":167,"posterUrl":"https://assets.mathsgenie.co.uk/walkthroughs/videos/914d10a2-a8a9-4c2d-8999-98c94c6d067d.jpg"}},{"id":"9f7d08f2-aecf-44a1-9da2-5da49492166d","revisionId":"1c0d198d-30dd-4c8f-bb62-ee766ccf7c73","questionNumber":null,"questionType":"LA","specification":"","difficulty":3,"difficultyBand":"medium","totalMarks":5,"sourceTitle":null,"calculatorUse":"non_calc","isCalculation":true,"parts":[{"id":"69a6d920-c31a-4861-a16a-d77d91beef7e","label":"a","content":"Find the possible values of $x$ and $y$ that satisfy the following equations:\n\n$$x^2 + y^2 = 34$$\n\n$$x + y = 8$$","markscheme":"- Rearranges linear equation to $y = 8 - x$ (or $x = 8 - y$) and substitutes: $x^2 + (8 - x)^2 = 34$ M1\n- Expands and simplifies: $2x^2 - 16x + 30 = 0$ or $x^2 - 8x + 15 = 0$ M1\n- Solves the quadratic: $(x - 3)(x - 5) = 0$ so $x = 3$ or $x = 5$ M1\n- Finds corresponding $y$ values: $y = 8 - 3 = 5$ and $y = 8 - 5 = 3$ M1\n- Final answers: $x = 3, y = 5$ and $x = 5, y = 3$ A1","marks":5,"order":0,"markingRubric":{"totalMarks":5,"points":[{"id":"p1","marks":1,"description":"Rearranging the linear equation and substituting into the quadratic equation.","code":"M"},{"id":"p2","marks":1,"description":"Expansion of the squared term and simplification to a standard quadratic equation.","code":"M"},{"id":"p3","marks":1,"description":"Correct method for solving the simplified quadratic equation.","code":"M"},{"id":"p4","marks":1,"description":"Obtaining both values for the second variable correctly.","code":"M"},{"id":"p5","marks":1,"description":"Correct identification of both pairs of solutions.","code":"A"}],"finalAnswers":["x = 3, y = 5 and x = 5, y = 3","(3, 5) and (5, 3)"],"generatedAt":"2026-05-06T14:07:21.854Z","modelVersion":"gemini-3-flash-preview"},"tableImages":[],"explanation":{"methods":[{"id":"substitution","label":"Substitution and factorising","steps":[{"type":"hyper_latex","block":{"nodes":[{"latex":"$$x+y=8$"},{"latex":"$$\\htmlData{anchor=y_expression}{\\color{@sky}{y=8-x}}$"},{"latex":"$$x^2+y^2=34$"},{"latex":"$$\\htmlData{anchor=substituted_expression}{x^2+\\color{@sky}{(8-x)^2}=34}$"}],"connections":[{"to":"substituted_expression","from":"y_expression","color":"sky","label":"substitute for y"}]},"order":0,"title":"Make one variable","description":"Rearrange the linear equation, then replace y in the other equation. This complete substitution earns the first M1 mark."},{"type":"hyper_latex","block":{"nodes":[{"latex":"$$\\htmlData{anchor=square}{x^2+\\color{@orange}{(8-x)^2}=34}$"},{"latex":"$$\\htmlData{anchor=expanded}{x^2+\\color{@orange}{64-16x+x^2}=34}$"},{"latex":"$$\\htmlData{anchor=collected}{\\color{@teal}{2x^2-16x+30=0}}$"},{"latex":"$$\\htmlData{anchor=halved}{\\color{@teal}{x^2-8x+15=0}}$"}],"connections":[{"to":"expanded","from":"square","color":"orange","label":"expand the square"},{"to":"collected","from":"expanded","color":"teal","label":"collect terms"},{"to":"halved","from":"collected","color":"teal","label":"divide by 2"}]},"order":1,"title":"Expand the quadratic","description":"Expand the bracket carefully, collect like terms, then divide every term by 2. Reaching either of the last two equations earns the second M1 mark."},{"type":"hyper_latex","block":{"nodes":[{"latex":"$$x^2-8x+15=0$"},{"latex":"$$\\htmlData{anchor=factorised}{\\color{@purple}{(x-3)(x-5)}=0}$"},{"latex":"$$\\htmlData{anchor=roots}{\\color{@green}{x=3\\text{ or }x=5}}$"}],"connections":[{"to":"roots","from":"factorised","color":"green","label":"zero-product rule"}]},"order":2,"title":"Factorise and solve","description":"Find two numbers whose product is 15 and whose sum is −8. Solving the factorised quadratic earns the third M1 mark."},{"type":"hyper_latex","block":{"nodes":[{"latex":"$$\\color{@purple}{y=8-x}$"},{"latex":"$$\\htmlData{anchor=x_three}{x=3}$"},{"latex":"$$\\htmlData{anchor=y_five}{\\color{@sky}{y=8-3=5}}$"},{"latex":"$$\\htmlData{anchor=x_five}{x=5}$"},{"latex":"$$\\htmlData{anchor=y_three}{\\color{@orange}{y=8-5=3}}$"}],"connections":[{"to":"y_five","from":"x_three","color":"sky"},{"to":"y_three","from":"x_five","color":"orange"}]},"order":3,"title":"Find each partner","description":"Use y = 8 − x separately for each value of x. Finding both corresponding values of y earns the fourth M1 mark."},{"type":"text","order":4,"title":"State both pairs","content":"Pair each $x$-value with the $y$-value found from it:\n\n$$\\boxed{(x,y)=(3,5)\\text{ or }(5,3)}$$\n\nBoth ordered pairs are needed for the final A1 mark. Do not list the possible $x$- and $y$-values separately, because that could incorrectly suggest other pairings."}]}],"generatedAt":"2026-07-19T18:21:00.737Z","modelVersion":"gpt-5.6-sol","explanationVersion":"v3"},"options":[]}],"options":[],"explanation":{"methods":[{"id":"69a6d920-c31a-4861-a16a-d77d91beef7e:substitution","label":"Substitution Method","steps":[{"type":"text","order":0,"title":"Rearrange the linear equation","content":"To solve this system of equations using the **Substitution Method**, we start with the simpler, linear equation:\n\n$$x + y = 8$$\n\nWe rearrange this to express one variable in terms of the other. Let's make $y$ the subject by subtracting $x$ from both sides:\n\n$$y = 8 - x$$","highlights":[{"text":"x + y = 8","location":"specification"}]},{"type":"hyper_latex","block":{"nodes":[{"latex":"$$y = \\htmlData{anchor=sub-expr}{\\color{@sky}{8 - x}}$"},{"latex":"$$x^2 + \\htmlData{anchor=quad-y}{\\color{@sky}{y}}^2 = 34$"},{"latex":"$$x^2 + (\\htmlData{anchor=quad-sub}{\\color{@sky}{8 - x}})^2 = 34$"}],"highlights":[{"color":"sky","style":"text-color","anchor":"sub-expr"},{"color":"sky","style":"text-color","anchor":"quad-y"},{"color":"sky","style":"text-color","anchor":"quad-sub"}],"connections":[{"to":"quad-sub","from":"sub-expr","color":"sky","label":"substitute","style":"solid"}]},"order":1,"title":"Substitute into the quadratic","description":"Now, we substitute this expression for $y$ into the quadratic equation $x^2 + y^2 = 34$."},{"type":"text","order":2,"title":"Expand and simplify","content":"Next, we expand the squared term and simplify the resulting equation into standard quadratic form ($ax^2 + bx + c = 0$):\n\n$$\\begin{aligned} x^2 + (8 - x)^2 &= 34 \\\\ x^2 + (64 - 16x + x^2) &= 34 \\\\ 2x^2 - 16x + 64 &= 34 \\end{aligned}$$\n\nSubtract $34$ from both sides to set the quadratic to $0$:\n\n$$2x^2 - 16x + 30 = 0$$\n\nTo make it simpler to factorise, we divide every term by $2$:\n\n$$x^2 - 8x + 15 = 0$$"},{"type":"text","order":3,"title":"Solve the quadratic equation","content":"We need to find two numbers that multiply to $+15$ and add up to $-8$. These numbers are $-3$ and $-5$. \n\nWe can write the quadratic in factorised form:\n\n$$(x - 3)(x - 5) = 0$$\n\nSetting each factor to $0$ gives our possible $x$ values:\n\n$$x = 3 \\quad \\text{or} \\quad x = 5$$","calculator":[{"gdc":{"expression":"x^2 - 8x + 15 = 0","instructions":"Go to Equation mode (or Polynomial solver). Select Degree 2. Input a=1, b=-8, c=15, and solve."},"ti84":{"expression":"x^2 - 8x + 15 = 0","instructions":"Press [apps], select 'PLYSMLT2', choose 'POLY ROOT FINDER'. Set Order to 2, click 'NEXT'. Enter a=1, b=-8, c=15, and press [SOLVE]."},"desmos":{"expression":"y = x^2 - 8x + 15","instructions":"Type the equation in the input bar. The x-intercepts of the graph correspond to the solutions."},"description":"You can also find the roots of the quadratic equation using your graphing calculator."}]},{"type":"text","order":4,"title":"Find corresponding y values","content":"Finally, we substitute our $x$ values back into our rearranged linear equation, $y = 8 - x$, to find the corresponding $y$ values.\n\n- If $\\mathbf{x = 3}$:\n $$y = 8 - 3 = 5$$\n\n- If $\\mathbf{x = 5}$:\n $$y = 8 - 5 = 3$$\n\nSo the solutions to the simultaneous equations are:\n\n$$\\begin{aligned} x = 3, & \\quad y = 5 \\\\ \\text{and} \\quad x = 5, & \\quad y = 3 \\end{aligned}$$"}]},{"id":"69a6d920-c31a-4861-a16a-d77d91beef7e:symmetric_identities","label":"Using Algebraic Identities","steps":[{"type":"text","order":0,"title":"Identify the identity","content":"We are given the simultaneous equations:\n1) $x^2 + y^2 = 34$\n2) $x + y = 8$\n\nTo find the values of $x$ and $y$, we can use the algebraic identity:\n$$(x+y)^2 = x^2 + y^2 + 2xy$$"},{"type":"hyper_latex","block":{"nodes":[{"latex":"$$(\\htmlData{anchor=sum-val}{\\color{@sky}{8}})^2 = \\htmlData{anchor=sq-sum}{\\color{@amber}{34}} + 2xy$"},{"latex":"$$64 = 34 + 2xy$"},{"latex":"$$2xy = 30 \\implies \\htmlData{anchor=prod-val}{\\color{@purple}{xy = 15}}$"}],"highlights":[{"color":"sky","style":"text-color","anchor":"sum-val"},{"color":"amber","style":"text-color","anchor":"sq-sum"},{"color":"purple","style":"box","anchor":"prod-val"}],"connections":[{"to":"prod-val","from":"sum-val","color":"purple","label":"solve","style":"solid"}]},"order":1,"title":"Substitute and solve for xy","description":"Substitute the known values $x + y = 8$ and $x^2 + y^2 = 34$ into the identity to find the product $xy$."},{"type":"text","order":2,"title":"Form a quadratic equation","content":"Now we know both the sum and the product of the two numbers $x$ and $y$:\n- Sum: $x + y = 8$\n- Product: $xy = 15$\n\nAny two numbers with a known sum $S$ and product $P$ are the roots of the quadratic equation:\n$$t^2 - St + P = 0$$\n\nSubstituting our values $S = 8$ and $P = 15$:\n$$t^2 - 8t + 15 = 0$$"},{"type":"text","order":3,"title":"Solve the quadratic equation","content":"To find the solutions, we factorise the quadratic equation by looking for two numbers that multiply to $+15$ and add up to $-8$. These numbers are $-3$ and $-5$:\n$$(t - 3)(t - 5) = 0$$\n\nThis gives two possible values for $t$:\n$$t = 3 \\quad \\text{or} \\quad t = 5$$"},{"type":"text","order":4,"title":"State the final pairs","content":"Since $x$ and $y$ represent the roots of our equation, the possible pairs of solutions are:\n- If $x = 3$, then $y = 5$\n- If $x = 5$, then $y = 3$\n\nThus, the solutions are:\n$$(x, y) = (3, 5) \\quad \\text{and} \\quad (x, y) = (5, 3)$$"}]},{"id":"69a6d920-c31a-4861-a16a-d77d91beef7e:graphical","label":"Graphical Approach","steps":[{"type":"text","order":0,"title":"Understand the graphical method","content":"To solve the simultaneous equations graphically, we need to plot both equations on the same set of coordinate axes:\n\n1. The circle equation: \n$$x^2 + y^2 = 34$$\n\n2. The linear equation: \n$$x + y = 8 \\implies y = 8 - x$$\n\nThe solutions to this system of equations are the coordinates $(x, y)$ where the circle and the straight line intersect.","highlights":[{"text":"x^2 + y^2 = 34","location":"specification"},{"text":"x + y = 8","location":"specification"}]},{"type":"text","order":1,"title":"Analyze the circle","content":"The equation $$x^2 + y^2 = 34$$ represents a circle centered at the origin $(0, 0)$.\n\nThe radius of the circle is:\n$$r = \\sqrt{34} \\approx 5.83$$\n\nTo help us sketch or graph it accurately, we can identify some key integer coordinate points on this circle by finding pairs of squared integers that sum to $34$:\n$$\\begin{aligned}\n3^2 + 5^2 &= 9 + 25 = 34 \\\\\n5^2 + 3^2 &= 25 + 9 = 34\n\\end{aligned}$$\n\nThis means the circle passes exactly through points like $(3, 5)$, $(5, 3)$, $(-3, 5)$, $(5, -3)$, etc."},{"type":"text","order":2,"title":"Analyze the line","content":"The equation $$x + y = 8$$ can be rearranged to the slope-intercept form:\n$$y = -x + 8$$\n\nThis is a straight line with a $y$-intercept at $(0, 8)$ and a gradient of $-1$.\n\nLet's find a few points on this line to plot it:\n- If $x = 0$, $y = 8$ → $(0, 8)$\n- If $x = 3$, $y = 8 - 3 = 5$ → $(3, 5)$\n- If $x = 5$, $y = 8 - 5 = 3$ → $(5, 3)$\n- If $x = 8$, $y = 0$ → $(8, 0)$"},{"type":"text","order":3,"title":"Plot and find intersections","content":"Now, let's use a graphing calculator or graph paper to plot both equations.\n\nLooking at where the line and circle cross on our graph, they intersect at exactly two points:\n$$\\text{Intersection 1: } (3, 5)$$\n$$\\text{Intersection 2: } (5, 3)$$\n\nThese represent our two solution pairs.","calculator":[{"gdc":{"instructions":"1. Go to the Graph menu.\n2. Enter the circle relation \$x^2 + y^2 = 34\$ (if your GDC supports relation graphing) or enter the upper and lower semi-circles: \$y_1 = \\sqrt{34 - x^2}\$ and \$y_2 = -\\sqrt{34 - x^2}\$.\n3. Enter the line \$y_3 = 8 - x\$.\n4. Press [DRAW]/[GRAPH].\n5. Use the G-Solve tool ([F5]) and select Intersect ([F5]) to find the coordinates of the two intersection points."},"ti84":{"instructions":"1. Press [Y=].\n2. Since the TI-84 requires equations in terms of y, rewrite them:\n - For the circle, enter: \n \$Y_1 = \\sqrt{34 - X^2}\$\n \$Y_2 = -\\sqrt{34 - X^2}\$\n - For the line, enter:\n \$Y_3 = 8 - X\$\n3. Press [WINDOW] and set Xmin = -8, Xmax = 8, Ymin = -8, Ymax = 8.\n4. Press [GRAPH].\n5. To find the intersections, press [2nd] [TRACE] (CALC) and select '5: intersect'.\n6. Select the circle's upper curve (Y1) as the first curve, the line (Y3) as the second curve, and move the cursor near the first intersection to guess. Repeat for the other intersection."},"desmos":{"expression":"x^2 + y^2 = 34, x + y = 8","instructions":"1. Type the circle equation in the first row: `x^2 + y^2 = 34`\n2. Type the line equation in the second row: `x + y = 8`\n3. Click on the gray intersection points on the graph to display their coordinates."},"description":"Graph both equations to find their points of intersection."}]},{"type":"text","order":4,"title":"State the final solutions","content":"The coordinates of the points of intersection give us the values of $x$ and $y$ that satisfy both equations simultaneously:\n\n$$\\begin{aligned}\nx = 3, & \\quad y = 5 \\\\\n\\text{and} \\\\\nx = 5, & \\quad y = 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