// Numbas version: exam_results_page_options
{"name": "Simultaneous Equations : Quadratics 3", "extensions": [], "custom_part_types": [], "resources": [], "navigation": {"allowregen": true, "showfrontpage": false, "preventleave": false, "typeendtoleave": false}, "question_groups": [{"pickingStrategy": "all-ordered", "questions": [{"name": "Simultaneous Equations : Quadratics 3", "tags": [], "metadata": {"description": "<p>Solving a pair of simultaneous equations of the form $a_1xy=c_1$ and $a_2x+b_2y=c_2$ 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+{b2}y} &amp;\\,= \\var{c2} \\end{split} \\]</p>\n<p><em>Give your answers correct to 2 decimal places.</em></p>", "advice": "<p>To solve a pair of simultaneous equations of this type we want to rearrange the&nbsp;first equation such that $y$ is the subject, which we can then substitute into the second equation. This can then be rearranged, resulting in a quadratic equation in terms of $x$ 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+{b2}y} &amp;\\,= \\var{c2} \\qquad\\qquad &amp;(2) \\end{split} \\]</p>\n<p>we want to rearrange equation (1)&nbsp;to make $y$&nbsp; the subject:</p>\n<p>\\[ y = \\simplify{{c1}/({a1}x)}. \\qquad\\qquad (3)\\]</p>\n<p>Substituting this into equation (2):</p>\n<p>\\[ \\simplify{{a2}x+{b2}({c1}/({a1}x))} = \\var{c2}. \\]</p>\n<p>Now multiplying this equation by $\\simplify{{mult1}x}$&nbsp;and&nbsp;rearranging:</p>\n<p>\\[ \\begin{split} \\simplify{{a2*mult1}x^2+{b2*mult1}({c1}/({a1}))} &amp;\\,= \\var{c2*mult1}x \\\\ \\Leftrightarrow \\qquad \\simplify{{a2*mult1}x^2 - {c2*mult1}x+ {b2*mult1}({c1}/({a1}))} &amp;\\,= 0. \\end{split} \\]</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(1..5)", "description": "", "templateType": "anything", "can_override": false}, "c1": {"name": "c1", "group": "Ungrouped variables", "definition": "random(1..5)", "description": "", "templateType": "anything", "can_override": false}, "a2": {"name": "a2", "group": "Ungrouped 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