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I got it wrong because when I did the calculation I swapped y on the right hand side to the left hand side to get \((3x-5)/2=0\) and then obviously I got the answer B rather than the OA C. I understand the OA however I would like to ask my silly question, which is why can I not do the way I did it. Is it wrong? It must be but WHY?

I got it wrong because when I did the calculation I swapped y on the right hand side to the left hand side to get \((3x-5)/2=0\) and then obviously I got the answer B rather than the OA C. I understand the OA however I would like to ask my silly question, which is why can I not do the way I did it. Is it wrong? It must be but WHY?

Thank you for enlightening me. I searched for the topic in the forum and I could not find it. Apologise if it has been created in the past.

\(y\frac{3x-5}{2}=y\)

\(y (\frac{3x-5}{2} -1) = 0\), and hence \(\frac{3x-5}{2} = 1\) and hence \(x = \frac{7}{3}\)

The mistake you made was equating it to zero. If you cancel y on both sides (which you can, since its not zero), the right hand side will be 1, not 0. Hope its clear.

Re: If y*(3x - 5)/2 = y and y is different from 0, then x = [#permalink]

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28 Sep 2015, 01:35

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