How do you solve #(n+0.3)/(n-0.3)=9/2#?

1 Answer
May 18, 2017

Answer:

See a solution process below:

Explanation:

First, multiply each side of the equation by #color(red)(2)(color(blue)(n - 0.3))# to eliminate the fractions while keeping the equation balanced:

#color(red)(2)(color(blue)(n - 0.3)) xx (n + 0.3)/(n - 0.3) = color(red)(2)(color(blue)(n - 0.3)) xx 9/2#

#color(red)(2)cancel((color(blue)(n - 0.3))) xx (n + 0.3)/color(blue)(cancel(color(black)(n - 0.3))) = cancel(color(red)(2))(color(blue)(n - 0.3)) xx 9/color(red)(cancel(color(black)(2)))#

#color(red)(2)(n + 0.3) = 9(color(blue)(n - 0.3))#

Next, expand the terms in parenthesis on both sides of the equation by multiplying each term within the parenthesis by the term outside the parenthesis:

#(color(red)(2) xx n) + (color(red)(2) xx 0.3) = (9 xx color(blue)(n)) - (9 xx color(blue)(0.3))#

#2n + 0.6 = 9n - 2.7#

Then, subtract #color(red)(2n)# and add #color(blue)(2.7)# to each side of the equation to isolate the #n# term while keeping the equation balanced:

#-color(red)(2n) + 2n + 0.6 + color(blue)(2.7) = -color(red)(2n) + 9n - 2.7 + color(blue)(2.7)#

#0 + 3.3 = (-color(red)(2) + 9)n - 0#

#3.3 = 7n#

Now, divide each side of the equation by #color(red)(7)# to solve for #n# while keeping the equation balanced:

#3.3/color(red)(7) = (7n)/color(red)(7)#

#0.47 = (color(red)(cancel(color(black)(7)))n)/cancel(color(red)(7))#

#0.47 = n#

#n = 0.47# rounded to the nearest hundredth.

Or

#(10/10 xx 3.3/color(red)(7)) = (color(red)(cancel(color(black)(7)))n)/cancel(color(red)(7))#

#33/70 = n#

#n = 33/70#