How do you solve #5/(b-5)=3/b#?

2 Answers
Jul 10, 2016

Answer:

#b=-15/2#

Explanation:

First we cross multiply and get:
#3(b-5)=5b#

Then we distribute the 3 on the left side
#3b-15=5b#

Now we collect terms by subtracting #3b# from both sides.
#-15=2b#

Divide both sides by 2.
#-15/2=b# which is our answer.

Jul 10, 2016

Answer:

#b=-15/2.#

Explanation:

#5/(b-5)=3/b.#
Multiplying by #b(b-5)# on both sides, we get,
#b(b-5)*5/(b-5)=b(b-5)*3/b#
#:. (5b(b-5))/(b-5)=(3b(b-5))/b#
#:. 5bcancel(b-5)/cancel(b-5)=3cancelb(b-5)/cancel(b)#
#:. 5b=3(b-5)#
#:. 5b=3b-15#
Adding, (-3b) on both sides, # 5b-3b=3b-15-3b#
#:. 2b=cancel3b-15-cancel3b#
#:. 2b=-15#
Multiplying by #1/2#,
#1/2(2b)=1/2(-15)#
#1/cancel2cancel2b=-15/2#
#:. b=-15/2.#
This is the reqd. soln. Enjoy, Maths.!