How do you solve #2/8=(n+4)/(n-4)#?

1 Answer
Jun 15, 2017

Answer:

See a solution process below:

Explanation:

First, multiply each side of the equation by #color(red)(8)color(blue)((n - 4))# to eliminate the fractions while keeping the equation balanced:

#color(red)(8)color(blue)((n - 4)) xx 2/8 = color(red)(8)color(blue)((n - 4)) xx (n + 4)/(n - 4)#

#cancel(color(red)(8))color(blue)((n - 4)) xx 2/color(red)(cancel(color(black)(8))) = color(red)(8)cancel(color(blue)((n - 4))) xx (n + 4)/color(blue)(cancel(color(black)(n - 4)))#

#2(n - 4) = 8(n + 4)#

Next, we expand the term in parenthesis by multiplying each term within the parenthesis by the term outside the parenthesis:

#color(red)(2)(n - 4) = color(blue)(8)(n + 4)#

#(color(red)(2) * n) - (color(red)(2) * 4) = (color(blue)(8) * n) + (color(blue)(8) * 4)#

#2n - 8 = 8n + 32#

Then, subtract #color(red)(2n)# and #color(blue)(32)# from each side of the equation to isolate the #n# term while keeping the equation balanced:

#-color(red)(2n) + 2n - 8 - color(blue)(32) = -color(red)(2n) + 8n + 32 - color(blue)(32)#

#0 - 40 = (-color(red)(2) + 8)n + 0#

#-40 = 6n#

Now, divide each side of the equation by #color(red)(6)# to solve for #n# while keeping the equation balanced:

#-40/color(red)(6) = (6n)/color(red)(6)#

#-20/3 = (color(red)(cancel(color(black)(6)))n)/cancel(color(red)(6))#

#-20/3 = n#

#n = -20/3#