How do you solve for n in #2n - 4m = 8#?

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Answer 1 · 2018-03-10T17:15:09

See a solution process below:

First, add #color(red)(4m)# to each side of the equation to isolate the #n# term while keeping the equation balanced:

#2n - 4m + color(red)(4m) = 8 + color(red)(4m)#

#2n - 0 = 8 + 4m#

#2n = 8 + 4m#

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

#(2n)/color(red)(2) = (8 + 4m)/color(red)(2)#

#(color(red)(cancel(color(black)(2)))n)/cancel(color(red)(2)) = 8/color(red)(2) + (4m)/color(red)(2)#

#n = 4 + 2m#

Answer 2 · 2018-03-10T17:18:25

#n=2(m+2)#

Add #4m# to both sides

#2n=8+4m#

Divide both sides by #2#

#n=4+2m#

Take a factor of #2# out of #4m+2#

#n=2(2+m)#