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

1 Answer
Feb 26, 2016

Answer:

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

Explanation:

#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:

  1. Solve for x. Hint for b): place on an equivalent denominator.

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

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

Good luck!