How do you solve #4( a - 1) = 7a + 2#?

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Answer 1 · 2018-04-21T10:31:47

#a=-2#

To get to those follow these steps:

Expand the brackets, so #4(a-1)# becomes #4a-4#
Rearrange so you have the variables on the same side #-> -4 -2 = 7a-4a# (change signs when moving across the equals sign)
Simplify #-> -6 = 3a#
Divide both sides by #3# so #a# is by itself #-> -6/3=a#
#-6/3 = -2#, therefore, #a# must equal #-2#

Answer 2 · 2018-04-21T10:34:03

a = -2

#4(a-1) = 7a + 2#

On the left hand side, multiply each term inside the brackets by 4

#4a - 4 = 7a + 2#

Subtract 2 from both sides

#4a - 6 = 7a#

Subtract 4a from both sides

#-6 = 7a - 4a#
#-6 = 3a#

Divide both sides by 3

#-2 = a#
#a = -2#