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Bunuel
If \(2yz = 1\) and \(\frac{3}{3 + \frac{1}{z}} = 2\), then y =

A. -3/4

B. -2/3

C. -1/3

D. 1/3

E. 2/3


\(\frac{3}{3 + \frac{1}{z}} = 2\)

\(\frac{3z}{3z+1} = 2\)

\(3z = 6z + 2\)

\(3z = -2\)

\(z = \frac{-2}{3}\)


\(2yz = 1\)

\(2 * y *\frac{ -2}{3} = 1\)

\(-4y = 3 \)

\(y = - \frac{3}{4}\)


Answer A
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Bunuel
If \(2yz = 1\) and \(\frac{3}{3 + \frac{1}{z}} = 2\), then y =

A. -3/4

B. -2/3

C. -1/3

D. 1/3

E. 2/3
\(2yz = 1\) and \(\frac{3}{3 + \frac{2yz}{z}} = 2\)

Or,\(\frac{3}{3 + 2y} = 2\)

Or, \(6 + 4y = 3\)

Or, \(4y = -3\)

Or, \(y = -\frac{3}{4}\), Answer must be (A)
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First, use equation 2 (the fraction). Get a common denominator in the denominator, and flip so that you only have one fraction. That fraction is equal to 2.
Then, solve z.

Next, use equation 1. Plug in z and solve y.

See picture below, y = -3/4. Answer is A.

Attachment:
Best method.png
Best method.png [ 18 KiB | Viewed 1599 times ]
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