How do you solve #(8-3x)/4=32#?

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Answer 1 · 2016-12-07T19:14:11

#x = -40#

In #color(blue)((8-3x))/4 = 32# , you need to isolate the part that contains #x#

To get rid of the 4 in the denominator, multiply both sides by 4:

#(cancel4xxcolor(blue)((8-3x)))/cancel4 = 4xx32#

#8color(red)(-3x) = 128" "larr# now isolate the x-term

#-3x = 128-8#

#-3x = 120" "# divide both sides by -3.

#(-3x)/-3 = 120/-3#

#x =-40#