How do you solve #-28- 4n = 6( 4- 5n )#?

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Answer 1 · 2017-06-06T13:51:36

See a solution process below:

First, expand the terms on the right side of the equation by multiplying each term within the parenthesis by the term outside the parenthesis:

#-28 - 4n = color(red)(6)(4 - 5n)#

#-28 - 4n = (color(red)(6) xx 4) - (color(red)(6) xx 5n)#

#-28 - 4n = 24 - 30n#

Next, add #color(red)(30n)# and #color(blue)(28)# to each side of the equation to isolate the #n# term while keeping the equation balanced:

#color(blue)(28) - 28 - 4n + color(red)(30n) = color(blue)(28) + 24 - 30n + color(red)(30n)#

#0 + (-4 + color(red)(30))n = 52 - 0#

#26n = 52#

Now, divide each side of the equation by #color(red)(26)# to solve for #n# while keeping the equation balanced:

#(26n)/color(red)(26) = 52/color(red)(26)#

#(color(red)(cancel(color(black)(26)))n)/cancel(color(red)(26)) = 2#

#n = 2#