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It may be helpful to factor out y. For example: 

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\\[\\simplify{x^{{n1}}*y+{n2}*y*x - {n3}*y}=y(\\simplify{x^{{n1}} + {n2}*x - {n3}})\\]

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Consider the equation:

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\\[\\simplify{x^{{n1}}*y + {n2}*y*x} = \\simplify{{n3}*y} + \\var{n4}\\]

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Re-arrange this equation to make $y$ the subject:

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$y = $[[0]]

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Another transposition question, which requires (basic) factorisation.

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rebelmaths

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Here, we first have to collect all terms involving $y$ on the same side. Hence, we get:

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\\[\\simplify{x^{{n1}}*y+{n2}*y*x - {n3}*y} = \\var{n4}\\]

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We then spot that $y$ appears exactly once in each term on the left, so factorise:

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\\[y(\\simplify{x^{{n1}} + {n2}*x - {n3}}) = \\var{n4}\\]

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and simple division gives the answer.

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