How do you solve #0=4(n-4)-6(n-4)# using the distributive property?

1 Answer
Mar 25, 2017

Answer:

See the entire solution process below:

Explanation:

First, multiply each term with each parenthesis by the term outside the parenthesis:

#0 = color(red)(4)(n - 4) - color(blue)(6)(n - 4)#

#0 = (color(red)(4) xx n) - (color(red)(4) xx 4) - (color(blue)(6) xx n) - (color(blue)(6) xx -4)#

#0 = 4n - 16 - 6n - (-24)#

#0 = 4n - 16 - 6n + 24#

Next, on the right side of the equation, group and combine like terms:

#0 = 4n - 6n - 16 + 24#

#0 = (4 - 6)n + 8#

#0 = -2n + 8#

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

#0 - color(red)(8) = -2n + 8 - color(red)(8)#

#-8 = -2n + 0#

#-8 = -2n#

Now, divide each side of the equation by #color(red)(-2)# to solve for #n# while keeping the equation balanced:

#(-8)/color(red)(-2) = (-2n)/color(red)(-2)#

#4 = (color(red)(cancel(color(black)(-2)))n)/cancel(color(red)(-2))#

#4 = n#

#n = 4#