How do you solve #7x + 29= 4( x + 8)#?

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Answer 1 · 2017-06-22T21:01:17

See a solution process below:

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

#7x + 29 = color(red)(4)(x + 8)#

#7x + 29 = (color(red)(4) xx x) + (color(red)(4) xx 8)#

#7x + 29 = 4x + 32#

Next, subtract #color(red)(29)# and #color(blue)(4x)# from each side of the equation to isolate the #x# term while keeping the equation balanced:

#-color(blue)(4x) + 7x + 29- color(red)(29) = -color(blue)(4x) + 4x + 32 - color(red)(29)#

#(-color(blue)(4) + 7)x + 0 = 0 + 3#

#3x = 3#

Now, divide each side of the equation by #color(red)(3)# to solve for #x# while keeping the equation balanced:

#(3x)/color(red)(3) = 3/color(red)(3)#

#(color(red)(cancel(color(black)(3)))x)/cancel(color(red)(3)) = 1#

#x = 1#