How do you solve #-6( 1+ 4x ) = 22+ 4x#?

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Answer 1 · 2017-11-14T11:03:39

#x=-1#

First we will distribute parentheses/brackets using #x(y+z)=xy+xz#.

Therefore,
#-6(1+4x)# is the same as #-6*1-6*4x#.

#-6*1=-6# and #4*6x=24x#.

Whole equation now:
#-6-24x=22+4x#

Add #6# to both sides:
#-6-24xcolor(red)+color(red)6=22+4xcolor(red)+color(red)6#
Simplify:
#-24x=4x+28#

Subtract #4x# from both sides:
#-24xcolor(red)-color(red)4color(red)x=4x+28color(red)-color(red)4color(red)x#

Simplify:
#-28x=28#

Now, we divide both side by #-28#.
#(-28x)/-28=28/-28#

Which is equal to #x=-1#.

Final answer:
#x=-1#.