How do you solve #(x+2)/3+(x-3)/4=1#?

1 Answer
Feb 14, 2017

Answer:

See the entire solution process below:

Explanation:

First, we need to eliminate the fractions by multiplying each side of the equation by a common denominator for the two fractions.

#3 x 4 = color(red)(12)# is a common denominator.

#color(red)(12)((x + 2)/3 + (x - 3)/4) = color(red)(12) xx 1#

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

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

#4(x + 2) + 3(x - 3) = 12#

We can now expand the terms in parenthesis and group and combine like terms on the left side of the equation:

#(4 xx x) + (4 xx 2) + (3 xx x) - (3 xx 3) = 12#

#4x + 8 + 3x - 9 = 12#

#4x + 3x + 8 - 9 = 12#

#7x - 1 = 12#

Next, we can add #color(red)(1)# to each side of the equation to isolate the #x# term while keeping the equation balanced:

#7x - 1 + color(red)(1) = 12 + color(red)(1)#

#7x - 0 = 13#

#7x = 13#

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

#(7x)/color(red)(7) = 13/color(red)(7)#

#(color(red)(cancel(color(black)(7)))x)/cancel(color(red)(7)) = 13/7#

#x = 13/7#