How do you solve #2(3x-1)+2(4x+5)=8#?

1 Answer
Jan 15, 2017

Answer:

See the entire solution process below:

Explanation:

First, expand the terms in parenthesis by multiplying the by the term outside their respective parenthesis:

#color(red)(2)(3x - 1) + color(blue)(2)(4x + 5) = 8#

#(color(red)(2) xx 3x) - (color(red)(2) xx 1) + (color(blue)(2) xx 4x) + (color(blue)(2) xx 5) = 8#

#6x - 2 + 8x + 10 = 8#

Next, group and combine like terms:

#6x + 8x - 2+ 10 = 8#

#14x + 8 = 8#

Next, subtract #color(red)(8)# from each side of the equation to isolate the #x# term while keeping the equation balanced:

#14x + 8 - color(red)(8) = 8 - color(red)(8)#

#14x + 0 = 0#

#14x = 0#

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

#(14x)/color(red)(14) = 0/color(red)(14)#

#(color(red)(cancel(color(black)(14)))x)/cancel(color(red)(14)) = 0#

#x = 0#