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Straightforward solving linear equations question.

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Adapted from 'Simultaneous equations by elimination 1 with parts' by Joshua Boddy.

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Solve the pair of simultaneous equations by working through parts a) to e)

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\$\\begin{eqnarray} \\simplify{{n1}*x + {n2}*y -{ans1}} = 0 &&&&&&&(1)\\\\ \\simplify{{n3}*x + {n4}*y -{ans2}} = 0 &&&&&&&(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]]$\\simplify{y +{ans1}*(-1)*{n3}} = 0$

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[[2]]$x$ + [[3]]$\\simplify{y +{ans2}*(-1)*{n1}} = 0$

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

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[[0]]$\\simplify{y - {ans3} = 0}$

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Now we can work out $y$

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$y =$[[0]]

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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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$\\simplify{{n1}*x}$ + [[0]] $+\\simplify{(-1)*{ans1}}$ = 0

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which then solves to give $x =$[[0]].

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Solve the pair of equations

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\$\\begin{eqnarray} \\simplify{{n1}*x + {n2}*y -{ans1}} = 0 &&&&&&&(1)\\\\ \\simplify{{n3}*x + {n4}*y -{ans2}} = 0 &&&&&&&(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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$\\simplify{{n3}*{n1}*x + {n3}*{n2}*y +{ans1}*(-1)*{n3}} = 0$

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$\\simplify{{n1}*{n3}*x + {n1}*{n4}*y +{ans2}*(-1)*{n1}} = 0$

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

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$\\simplify{{yCoef}y - {ans3} = 0}$

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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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$\\simplify{{n1}*x + {n2}*{b} + (-1)*{ans1}}$ = 0

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

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