If \(y(\frac{3x-5}{2}) =y\) and \(y\neq{0}\), then x =

(A) 2/3
(B) 5/3
(C) 7/3
(D) 1
(E) 4

Since \(y\neq{0}\), then we can reduce the equation by it: \(\frac{3x-5}{2}=1\) --> \(3x-5=2\) --> \(x=\frac{7}{3}\).

Answer: C.
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y((3x-5)/2)=y

y[((3x-5)/2) - 1)] = 0

since y <> 0

(3x-5)/2 -1 = 0

x= 7/3

It's sub-600 level question, so below 600.
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y gets cancelled out in observation itself

\(\frac{3x-5}{2}=1\)

3x-5=2

3x=7

\(x=\frac{7}{3}\)

Answer = C
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Y is common factor of LHS & RHS, so cancels out in the first instance
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The answer is A.

(Note: We can divide both sides by y because it’s given that y does not equal 0. Otherwise, we couldn't divide both sides by y.)
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( 3x-5)/2 = y/y
3x - 5 = 2
3x = 7
x = 7/3
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Regards,
Naveen
email: nkmungila@gmail.com
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\(y\neq{0}\) means that y is not 0. It can be any other value. It's given so that we could reduce by y. If y were 0, then we could not reduce because we cannot divide by 0. y being 0, would also imply that \(y(\frac{3x-5}{2}) =y\) would hold true for any value of x, because \(y(\frac{3x-5}{2})=0 =y\) and we would not be able to solve for x.

Hope it's clear.
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Hi niks18, can you advice if my solution is correct, and where from did Bunuel get 1 ?

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

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

\(y3x-5y = 2y\)

\(3xy= 2y +5y\) now lets divide by 3y since we need to isolate x

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

\(x= 7y/3y\)

hence OA = C