How do you solve #\frac { 2x } { 4} + \frac { 1} { 2} = \frac { 1- x } { 3} + 6#?

1 Answer
Apr 25, 2017

See the entire solution process below:

Explanation:

First, multiply each side of the equation by the Lowest Common Denominator of the three fractions which is #color(red)(12)# to eliminate the fractions while keeping the equation balanced:

#color(red)(12)((2x)/4 + 1/2) = color(red)(12)((1 - x)/3 + 6)#

#(color(red)(12) * (2x)/4) + (color(red)(12) * 1/2) = (color(red)(12) * (1 - x)/3) + (color(red)(12) * 6)#

#(cancel(color(red)(12)) 3 * (2x)/color(red)(cancel(color(black)(4)))) + (cancel(color(red)(12)) 6 * 1/color(red)(cancel(color(black)(2)))) = (cancel(color(red)(12)) 4 * ((1 - x))/color(red)(cancel(color(black)(3)))) + 72#

#6x + 6 = 4(1 - x) + 72#

#6x + 6 = (4 * 1) - (4 * x) + 72#

#6x + 6 = 4 - 4x + 72#

#6x + 6 = -4x + 76#

Next, subtract #color(red)(6)# and add #color(blue)(4x)# to each side of the equation to isolate the #x# term while keeping the equation balanced:

#color(blue)(4x) + 6x + 6 - color(red)(6) = color(blue)(4x) - 4x + 76 - color(red)(6)#

#(color(blue)(4) + 6)x + 0 = 0 + 70#

#10x = 70#

Now, divide each side of the equation by #color(red)(10)# to solve for #x# while keeping the equation balanced:

#(10x)/color(red)(10) = 70/color(red)(10)#

#(color(red)(cancel(color(black)(10)))x)/cancel(color(red)(10)) = 7#

#x = 7#