How do you solve #4(4 - w) = 3(2w + 2) #?

1 Answer
Mar 27, 2017

Answer:

See the entire solution process below:

Explanation:

First, expand the terms in parenthesis by multiplying each term within the parenthesis by the term outside the parethesis:

#color(red)(4)(4 - w) = color(blue)(3)(2w + 2)#

#(color(red)(4) xx 4) - (color(red)(4) xx w) = (color(blue)(3) xx 2w) + (color(blue)(3) xx 2)#

#16 - 4w = 6w + 6#

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

#16 - 4w + color(red)(4w) - color(blue)(6) = 6w + 6 + color(red)(4w) - color(blue)(6)#

#16 - color(blue)(6) - 4w + color(red)(4w) = 6w + color(red)(4w) + 6 - color(blue)(6)#

#10 - 0 = (6 + color(red)(4))w + 0#

#10 = 10w#

#10/color(red)(10) = (10w)/color(red)(10)#

#1 = (color(red)(cancel(color(black)(10)))w)/cancel(color(red)(10))#

#1 = w#

#w = 1#