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{"name": "Simultaneous Equations : Quadratics 1", "extensions": [], "custom_part_types": [], "resources": [], "navigation": {"allowregen": true, "showfrontpage": false, "preventleave": false, "typeendtoleave": false}, "question_groups": [{"pickingStrategy": "all-ordered", "questions": [{"name": "Simultaneous Equations : Quadratics 1", "tags": [], "metadata": {"description": "<p>Solving a pair of simultaneous equations of the form $a_1x+b_1y=c_1$ and $a_2 x^2+b_2y^2=c_2$ by forming a quadratic equation.</p>", "licence": "Creative Commons Attribution-NonCommercial-ShareAlike 4.0 International"}, "statement": "<p>Solve the following simultaneous equations:</p>\n<p>\\[ \\begin{split} \\simplify{{a1}x+y} &amp;\\,= \\var{c1} \\\\ \\simplify{{a2}x^2+{b2}y^2} &amp;\\,= \\var{c2^2} \\end{split} \\]</p>\n<p>Give your answers to 2 decimal places where necessary.</p>", "advice": "<p>To solve a pair of simultaneous equations of this type we want to rearrange the linear equation such that it is in terms of $x$ or $y$, which we can then substitute into the equation with the quadratic terms. This will result in a quadratic equation in terms of one variable only.</p>\n<p>For the equations&nbsp;</p>\n<p>\\[ \\begin{split} \\simplify{{a1}x+y} &amp;\\,= \\var{c1} \\qquad \\qquad &amp;(1) \\\\ \\simplify{{a2}x^2+{b2}y^2} &amp;\\,= \\var{c2^2}&nbsp;\\qquad \\qquad &amp;(2) \\end{split} \\]</p>\n<p>we can rearrange equation (1)&nbsp;to make $y$ the subject:</p>\n<p>\\[ y = \\simplify{{c1}-{a1}x}. \\qquad\\qquad (3)\\]</p>\n<p>Substituting this into equation (2):</p>\n<p>\\[ \\begin{split}\\simplify{{a2}x^2+{b2}({c1}-{a1}x)^2} &amp;\\,=\\var{c2^2} \\\\ \\simplify{{a2}x^2+{b2}({c1}^2-2{c1}{a1}x+{a1}^2x^2)} &amp;\\,=\\var{c2^2} \\\\ \\simplify[!cancelTerms,unitFactor]{{a2}x^2+{b2*c1^2}-{2*b2*c1*a1}x+{b2*a1^2}x^2} &amp;\\,=\\var{c2^2}. \\end{split} \\]</p>\n<p>Collecting similar terms:</p>\n<p>\\[ \\simplify{({a2}+{b2*a1^2})x^2-{2*b2*c1*a1}x+({b2*c1^2}-{c2^2})} =0. \\qquad\\qquad (4) \\]</p>\n<p>Using the quadratic formula, we find two solutions for $x$:</p>\n<p>\\[ x_1 = \\var{x1dp} \\, \\text{(2 d.p.)} \\quad \\text{and} \\quad x_2=\\var{x2dp} \\,&nbsp;\\text{(2 d.p.)} \\]</p>\n<p>To find the corresponding $y$-values, we can plug these solutions for $x$ back into equation (3), which gives:</p>\n<p>\\[ y_1 = \\var{y1dp} \\, \\text{(2 d.p.)} \\quad \\text{and} \\quad y_2=\\var{y2dp} \\,&nbsp;\\text{(2 d.p.)} \\]</p>\n<p></p>\n<p>Therefore, the 2 pairs of solutions for these simultaneous equations are</p>\n<p>\\[ (x_1,y_1) = (\\var{x1dp},\\var{y1dp}) \\] and&nbsp;\\[ (x_2,y_2) = (\\var{x2dp},\\var{y2dp}). \\]</p>", "rulesets": {}, "extensions": [], "builtin_constants": {"e": true, "pi,\u03c0": true, "i": true}, "constants": [], "variables": {"a1": {"name": "a1", "group": "Ungrouped variables", "definition": "random(-5..5 except 0)", "description": "", "templateType": "anything", "can_override": false}, "c1": {"name": "c1", "group": "Ungrouped variables", "definition": "random(1..10)", "description": "", "templateType": "anything", "can_override": false}, "a2": 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