// Numbas version: exam_results_page_options {"name": "Jinhua's copy of Simultaneous equations by elimination 2", "extensions": [], "custom_part_types": [], "resources": [], "navigation": {"allowregen": true, "showfrontpage": false, "preventleave": false}, "question_groups": [{"pickingStrategy": "all-ordered", "questions": [{"variable_groups": [], "advice": "

Solve the pair of equations

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\$\\begin{eqnarray} \\simplify{{n1}*x + {n2}*y} & = & \\var{ans1} &&&&&&&(1)\\\\ \\simplify{{n3}*x + {n4}*y} & = & \\var{ans2}&&&&&&&(2)\\end{eqnarray}\$

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We are going to solve for $y$ first. To do this, we need to eliminate $x$ from the equations. The quickest way to do this is to multiply the first equation by the co-efficient of $x$ in the second equation (here $\\var{n3}$), and multiply the second equation by the co-efficient of $x$ in the first equation (here $\\var{n1}$). We then get the equations:

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$\\simplify{{n3}*{n1}*x + {n3}*{n2}*y = {ans1}*{n3}}$

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$\\simplify{{n1}*{n3}*x + {n1}*{n4}*y = {ans2}*{n1}}$

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We then subtract one new equation from the other to get:

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$\\simplify{{yCoef}y = {ans3}}$

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Now we can work out $y$

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$y = \\var{b}$

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and substitute this value back in to any of the previous equations to get the value for $x$.

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$\\simplify{{n1}*x + {n2}*{b} = {ans1}}$

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which then solves to give $x = \\var{a}$.

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Solve the pair of equations

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\$\\begin{eqnarray} \\simplify{{n1}*x + {n2}*y} & = & \\var{ans1} &&&&&&&(1)\\\\ \\simplify{{n3}*x + {n4}*y} & = & \\var{ans2}&&&&&&&(2)\\end{eqnarray}\$

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We are going to solve for $y$ first. To do this, we need to eliminate $x$ from the equations. The quickest way to do this is to multiply the first equation by the co-efficient of $x$ in the second equation (here $\\var{n3}$), and multiply the second equation by the co-efficient of $x$ in the first equation (here $\\var{n1}$). We then get the equations:

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[[0]]$x +$[[1]]$y = \\simplify{{ans1}*{n3}}$

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[[2]]$x +$[[3]]$y = \\simplify{{ans2}*{n1}}$

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We then subtract one new equation from the other to get:

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[[4]]$\\simplify{y = {ans3}}$

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Now we can work out $y$

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$y =$[[5]]

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and substitute this value back in to any of the previous equations to get the value for $x$.

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$\\simplify{{n1}*x}$ + [[6]] = $\\var{ans1}$

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which then solves to give $x =$[[7]].

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