How do you solve #15/(7b+5)=5/(2b+1)#?

1 Answer
Jun 28, 2018

Answer:

See a solution process below:

Explanation:

Because both sides of the equation are pure fractions we can reverse the fractions and rewrite the equation as:

#(7b + 5)/15 = (2b + 1)/5#

Next, we can multiply each side of the equation by #color(red)(15)# to eliminate the fractions while keeping the equation balanced:

#color(red)(15) xx (7b + 5)/15 = color(red)(15) xx (2b + 1)/5#

#cancel(color(red)(15)) xx (7b + 5)/color(red)(cancel(color(black)(15))) = cancel(color(red)(15))color(red)(3) xx (2b + 1)/color(red)(cancel(color(black)(5)))#

#7b + 5 = color(red)(3)(2b + 1)#

#7b + 5 = (color(red)(3) xx 2b) + (color(red)(3) xx 1)#

#7b + 5 = 6b + 3#

Now, subtract #color(red)(5)# and #color(blue)(6b)# from each side of the equation to solve for #b# while keeping the equation balanced:

#7b - color(blue)(6b) + 5 - color(red)(5) = 6b - color(blue)(6b) + 3 - color(red)(5)#

#(7 - color(blue)(6))b + 0 = 0 - 2#

#1b = -2#

#b = -2#