How do you solve for x: #6(x+1)=12(x-3)#?

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Answer 1 · 2017-02-04T12:25:47

See the entire solution process below:

First, expand all the terms in exponents:

#(6 xx x) + (6 xx 1) = (12 xx x) - (12 xx 3)#

#6x + 6 = 12x - 36#

Next, subtract #color(red)(6x)# and add #color(blue)(36)# to each side of the equation to isolate the #x# term while keeping the equation balanced:

#6x + 6 - color(red)(6x) + color(blue)(36) = 12x - 36 - color(red)(6x) + color(blue)(36)#

#6x - color(red)(6x) + 6 + color(blue)(36) = 12x - color(red)(6x) - 36 + color(blue)(36)#

#0 + 42 = 6x - 0#

#42 = 6x#

Now, divide each side of the equation by #color(red)(6)# to solve for #x# while keeping the equation balanced:

#42/color(red)(6) = (6x)/color(red)(6)#

#7 = (color(red)(cancel(color(black)(6)))x)/cancel(color(red)(6))#

#7 = x#

#x = 7#

Answer 2 · 2017-02-04T12:55:25

#x=7#

#6(x+1)=12(x-3)#

#:.6x+6=12x-36#

#:.6x-12x=-36-6#

#:.-6x=-42#

multiply both sides by# -1#

#:.6x=42#

#:.x=cancel42^7/cancel6^1#

#:.x=7#

substitute # x=7#

#:.6(7+1)=12(7-3)#

#:.6 xx 8=12 xx 4#

#:.48=48#