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Above eq can written as:
11^2x = (5/10)^3y = 22^2z
or 11^2x = (2)^(-3y) = 2^2z . 11^2z
multiply first two and equate with square of 3rd as all are equal
11^2x . 2^(-3y) = (2^2z . 11^2z ) ^2
now equate LHS & RHS powers which having same base:
for power of 11 => 2x = 4z => z= x/2 -------------(a)
for power of 2 => -3y = 4z => z= -3y/4 --------------(b)
above two x/2 = -3y/4 => 2x = -3y -----------------------------------(c)
now check option (hit and trial)
try option 2: z = 3xy/(3y-2x)
put value of 2x in denominator from eq c ===> z = 3xy / (3y +3y) ==> z= 3xy/6y => z=x/2
which is same as eq a
HENCE option B:
z = 3xy / (3y-2x)
+91-85588-96644
Above eq can written as: 11^2x = (5/10)^3y = 22^2z or 11^2x = (2)^(-3y) = 2^2z . 11^2z multiply first two and equate with square of 3rd as all are equal 11^2x . 2^(-3y) = (2^2z . 11^2z ) ^2 now equate LHS & RHS powers which having same base: for power of 11 => 2x = 4z => z= x/2 -------------(a) for power of 2 => -3y = 4z => z= -3y/4 --------------(b) above two x/2 = -3y/4 => 2x = -3y -----------------------------------(c) now check option (hit and trial) try option 2: z = 3xy/(3y-2x) put value of 2x in denominator from eq c ===> z = 3xy / (3y +3y) ==> z= 3xy/6y => z=x/2 which is same as eq a HENCE option B: z = 3xy / (3y-2x)