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Re: x+y=1/6
[#permalink]
18 May 2021, 08:09
Solving the equations we get
\(x+y=\frac{1}{6}\)
\(3x+2y=\frac{1}{6}\)
Lets multiple the first equation by 3 we get
\(3x+3y=\frac{3}{6}\) ....(1)
\(3x+2y=\frac{1}{6}\) ....(2)
(1) - (2) we get
3x + 3y - 3x - 2y = \(\frac{3}{6} - \frac{1}{6}\) = \(\frac{2}{6}\)
=> y = \(\frac{2}{6}\)
Using \(x+y=\frac{1}{6}\), we get
x + \(\frac{2}{6}\) = \(\frac{1}{6}\)
=> x = \(\frac{1}{6}\) - \(\frac{2}{6}\) = -\(\frac{1}{6}\)
Clearly y > x => Quantity B > Quantity A
So, Answer will be B
Hope it helps!