How do you solve for x in #(5a)/x = (5b)/(x-1)#?

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Answer 1 · 2016-02-26T15:33:50

First, simplify by the property #a/b = m/n -> a xx n = b xx m#

#5bx = (x - 1)5a#

#5bx = 5ax - 5a#

#5bx - 5ax = -5a#

#5x(b - a) = -5a#

#5x = (-5a)/(b - a)#

#x = -((5a)/(b - a)) / 5#

#x = - (5a) / (5(b - a))#

#x = - (5a)/(5b - 5a)#

Practice exercises:

a) #(1 + 5a)/(2x)= (3z - 4x)/(5a)#

b). #1/(2x) + 1/(yz) = 3/(xzy)#

Good luck!