How do you solve #\frac { 3x + 15+ ( 11x - 5) } { 2} = 47#?

1 Answer
Mar 14, 2017

See the entire solution process below:

Explanation:

First, multiply each side of the equation by #color(red)(2)# to eliminate the fraction and keep the equation balanced:

#color(red)(2) xx (3x + 15 + (11x - 5))/2 = color(red)(2) xx 47#

#cancel(color(red)(2)) xx (3x + 15 + (11x - 5))/color(red)(cancel(color(black)(2))) = 94#

#3x + 15 + (11x - 5) = 94#

Next, remove the terms on the left hand side of the equation from parenthesis, group and combine like terms:

#3x + 15 + 11x - 5 = 94#

#3x + 11x + 15 - 5 = 94#

#(3 + 11)x + (15 - 5) = 94#

#14x + 10 = 94#

Then, subtract #color(red)(10)# from each side of the equation to isolate the #x# term while keeping the equation balanced:

#14x + 10 - color(red)(10) = 94 - color(red)(10)#

#14x + 0 = 84#

#14x = 84#

Now, divide each side of the equation by #color(red)(14)# to solve for #x# while keeping the equation balanced:

#(14x)/color(red)(14) = 84/color(red)(14)#

#(color(red)(cancel(color(black)(14)))x)/cancel(color(red)(14)) = 6#

#x = 6#