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Solving a pair of simultaneous equations of the form $a_1x+y=c_1$ and $a_2x^2+b_2xy=c_2$ by forming a quadratic equation.
", "licence": "Creative Commons Attribution-NonCommercial-ShareAlike 4.0 International"}, "statement": "Solve the following simultaneous equations:
\n\\[ \\begin{split} \\simplify{{a1}x+y} &\\,= \\var{c1} \\\\ \\simplify{{a2}x^2+{b2}x*y} &\\,= \\var{c2} \\end{split} \\]
\n\nGive your answers to 2 decimal places where necessary.
", "advice": "To solve a pair of simultaneous equations of this type we want to rearrange the linear equation such that $y$ is the subject, which we can then substitute into the equation with the quadratic $x$-term. This will result in a quadratic equation in terms of $x$ only.
\nFor the equations
\n\\[ \\begin{split} \\simplify{{a1}x+y} &\\,= \\var{c1} \\qquad \\qquad &(1) \\\\\\simplify{{a2}x^2+{b2}x*y} &\\,= \\var{c2} \\qquad \\qquad &(2) \\end{split} \\]
\nwe can rearrange equation (1) to make $y$ the subject:
\n\\[ y = \\simplify{{c1}-{a1}x}. \\qquad\\qquad (3)\\]
\nSubstituting this into equation (2):
\n\\[ \\begin{split}\\simplify{{a2}x^2+{b2}x({c1}-{a1}x)} &\\,=\\var{c2} \\\\ \\simplify[!cancelTerms,unitFactor]{{a2}x^2+{b2*c1}x-{b2*a1}x^2} &\\,=\\var{c2}. \\end{split} \\]
\nCollecting similar terms:
\n\\[ \\simplify{({a2}-{b2*a1})x^2+{b2*c1}x-{c2}} =0. \\qquad\\qquad (4) \\]
\nUsing the quadratic formula, we find two solutions for $x$:
\n{check}
\nTherefore, the 2 pairs of solutions for these simultaneous equations are
\n\\[ (x_1,y_1) = (\\var{x1dp},\\var{y1dp}) \\] and \\[ (x_2,y_2) = (\\var{x2dp},\\var{y2dp}). \\]
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\\nTo find the corresponding $y$-values, we can plug these solutions for $x$ back into equation (3), which gives:
\\n\\\\[ y_1 = \\\\var{y1dp} \\\\, \\\\quad \\\\text{and} \\\\quad y_2=\\\\var{y2dp} \\\\, \\\\text{(2 d.p.)} \\\\]
\"", "description": "", "templateType": "long string", "can_override": false}, "text2": {"name": "text2", "group": "Ungrouped variables", "definition": "\"\\\\[ x_1 = \\\\var{x1dp} \\\\, \\\\text{ (2 d.p.)}\\\\quad \\\\text{and} \\\\quad x_2=\\\\var{x2dp} \\\\]
\\nTo find the corresponding $y$-values, we can plug these solutions for $x$ back into equation (3), which gives:
\\n\\\\[ y_1 = \\\\var{y1dp} \\\\, \\\\text{(2 d.p.)} \\\\quad \\\\text{and} \\\\quad y_2=\\\\var{y2dp} \\\\]
\"", "description": "", "templateType": "long string", "can_override": false}, "text3": {"name": "text3", "group": "Ungrouped variables", "definition": "\"\\\\[ x_1 = \\\\var{x1dp} \\\\, \\\\text{ (2 d.p.)}\\\\quad \\\\text{and} \\\\quad x_2=\\\\var{x2dp} \\\\, \\\\text{(2 d.p.)} \\\\]
\\nTo find the corresponding $y$-values, we can plug these solutions for $x$ back into equation (3), which gives:
\\n\\\\[ y_1 = \\\\var{y1dp} \\\\, \\\\text{(2 d.p.)} \\\\quad \\\\text{and} \\\\quad y_2=\\\\var{y2dp} \\\\, \\\\text{(2 d.p.)} \\\\]
\"", "description": "", "templateType": "long string", "can_override": false}, "text": {"name": "text", "group": "Ungrouped variables", "definition": "\"\\\\[ x_1 = \\\\var{x1dp} \\\\, \\\\quad \\\\text{and} \\\\quad x_2=\\\\var{x2dp}\\\\]
\\nTo find the corresponding $y$-values, we can plug these solutions for $x$ back into equation (3), which gives:
\\n\\\\[ y_1 = \\\\var{y1dp} \\\\, \\\\quad \\\\text{and} \\\\quad y_2=\\\\var{y2dp} \\\\]
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