How do you solve #6/(3b+2)=3/5#?

2 Answers
Feb 10, 2017

Answer:

See the entire solution process below:

Explanation:

First, multiply each side of the equation by #color(red)(5)(color(blue)(3b + 2))# to eliminate the fraction while keeping the equation balanced:

#color(red)(5)(color(blue)(3b + 2)) xx 6/(3b + 2) = color(red)(5)(color(blue)(3b + 2)) xx 3/5#

#color(red)(5)cancel((color(blue)(3b + 2))) xx 6/color(blue)(cancel(color(black)((3b + 2)))) = cancel(color(red)(5))(color(blue)(3b + 2)) xx 3/color(red)(cancel(color(black)(5)))#

#5 xx 6 = (3b + 2) x 3#

#30 = (3b xx 3) + (2 xx 3)#

#30 = 9b + 6#

Next, subtract #color(red)(6)# from each side of the equation to isolate the #b# term while keeping the equation balanced:

#30 - color(red)(6) = 9b + 6 - color(red)(6)#

#24 = 9b + 0#

#24 = 9b#

Now, divide each side of the equation by #color(red)(9)# to solve for #b# while keeping the equation balanced:

#24/color(red)(9) = (9b)/color(red)(9)#

#(3 xx 8)/(3 xx 3) = (color(red)(cancel(color(black)(9)))b)/cancel(color(red)(9))#

#(cancel(3) xx 8)/(cancel(3) xx 3) = b#

#8/3 = b#

#b = 8/3#

Feb 10, 2017

Answer:

#2 2/3#

Explanation:

#6/(3b+2)=3/5#

multiply both sides by #3b+2#

#:.cancel(3b+2)^1/1 xx 6/cancel(3b+2)^1=3/5 xx (3b+2)/1#

#:.1/1 xx6/1=(9b+6)/5#

#:.6=(9b+6)/5#

multiply both sides by # 5/1#

#:.5/1 xx 6=(9b+6)/cancel5^1 xx cancel5^1/1#

#:.30=9b+6#

#:.9b+6=30#

#:.9b=30-6#

#:.9b=24#

#:.b=cancel24^8/cancel9^3#

#:.b=8/3# or #2 2/3#

substitute #x=8/3#

#:.6/(3(8/3)+2)=3/5#

#:.6/(cancel24^8/cancel33^1+2)=3/5#

#:.cancel6^3/cancel10^5=3/5#

#:.3/5=3/5#