How do you solve #3/(b+2)=4/b#?

2 Answers
Mar 18, 2018

Answer:

See a solution process below:

Explanation:

First, because both sides of the equation are pure fractions we can flip each equation to rewrite the equation as:

#(b + 2)/3 = b/4#

Next, multiply each side of the equation by #color(red)(12)# to eliminate the fractions while keeping the equation balanced. #color(red)(12)# is used because it is the lowest common denominator of the two fractions:

#color(red)(12) xx (b + 2)/3 = color(red)(12) xx b/4#

#cancel(color(red)(12))color(red)(4) xx (b + 2)/color(red)(cancel(color(black)(3))) = cancel(color(red)(12))color(red)(3) xx b/color(red)(cancel(color(black)(4)))#

#color(red)(4)(b + 2) = 3b#

#(color(red)(4) xx b) + (color(red)(4) xx 2) = 3b#

#4b + 8 = 3b#

Now, subtract #color(red)(8)# and #color(blue)(3b)# from each side of the equation to solve for #b# while keeping the equation balanced:

#4b - color(blue)(3b) + 8 - color(red)(8) = 3b - color(blue)(3b) - color(red)(8)#

#(4 - color(blue)(3))b + 0 = 0 - color(red)(8)#

#1b = -8#

#b = -8#

Mar 18, 2018

Answer:

I get #b=-8#

Explanation:

To solve equations with fractions use the rule of cross products. Each numerator is multiplied by the opposite denominator and the two products are equal.

#\frac{color(blue)(3)}{b+2}=\frac{4}{color(blue)(b)}#

#color(blue)(3b)=(4)(b+2)#

Now it's just an ordinary linear equation:

#3b=4b+8#

#-b=8#

#b=-8#

Check:

#\frac{3}{-8+2}=\frac{3}{-6}=-\frac{1}{2}#

#\frac{4}{-8}=-\frac{1}{2}#