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We are going to solve for $x$ first. To do this, we need to eliminate $y$ from the equations.

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We start by rearranging equation (1) like so:

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$\\simplify{y = 1/{n2}*({ans1}-{n1}*x)}$

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$\\simplify{{n3}*x + {n4}/{n2}({ans1}-{n1}*x) - {ans2} = 0}$

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$\\simplify{({n3}-{n1}*{n4}/{n2})*x + {n4}*{ans1}/{n2} - {ans2} = 0}$

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$\\simplify{x=({n4}*{ans1}-{n2}*{ans2})/({n1}*{n4}-{n3}*{n2})}$

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Substitute the value of x back into equation (1) to find

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$\\simplify{y=({ans1}*{n3}-{n1}*{ans2})/({n2}*{n3}-{n1}*{n4})}$

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", "rulesets": {}, "parts": [{"prompt": "

We are going to solve for $x$ first. To do this, we need to eliminate $y$ from the equations.

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By equating equation (1) and equation (2), we find that:

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[[0]]$\\simplify{x = {ans2} - {n4}*{ans1}/{n2}}$

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Solve this linear equation to give  $x = $[[0]]

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Substitute the value of x back into equation (1) or equation (2) to find

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$y = $[[0]]

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Find the point of intersection of the following pair of straight lines by working through parts a) to d)

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\\[\\begin{eqnarray} \\simplify{{n2}*y = {ans1} - {n1}*x} &&&&&&&(1)\\\\ \\simplify{{n4}*y = {ans2} - {n3}*x} &&&&&&&(2)\\end{eqnarray}\\]

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Straightforward solving linear equations question.

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Adapted from 'Simultaneous equations by substitution 3 with parts' by Joshua Boddy.

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