How do you solve #161= 4x + 3( 3x + 19)#?

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Answer 1 · 2017-08-26T01:57:57

See a solution process below: #x = 134/13#

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

#161 = 4x + color(red)(3)(3x + 19)#

#161 = 4x + (color(red)(3) xx 3x) + (color(red)(3) xx 19)#

#161 = 4x + 9x + 57#

#161 = (4 + 9)x + 57#

#161 = 13x + 57#

Next, subtract #color(red)(57)# from each side of the equation to isolate the #x# term while keeping the equation balanced:

#161 - color(red)(57) = 13x + 57 - color(red)(57)#

#104 = 13x + 0#

#104 = 13x#

Now, divide each side of the equation by #color(red)(13)# to solve for #x# while keeping the equation balanced:

#104/color(red)(13) = (13x)/color(red)(13)#

#8 = (color(red)(cancel(color(black)(13)))x)/cancel(color(red)(13))#

#8 = x#

#x = 8#