How do you solve #(x+5)-2(4x-1)=0#?

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Answer 1 · 2016-06-29T18:13:49

#x=1#

#(x+5) - 2 ( 4x - 1) = 0#

#(x+5) - 2 * ( 4x - 1) = 0#

#x+5 + (-2) * (4x) + ( -2) * (- 1) = 0#

#x+5 -8x +2 = 0#

#x-8x +2 + 5= 0#

#-7x + 7= 0#

#-7x = - 7#

#x = (-7)/(-7)#

#x=1#