How do you solve #\frac { 1} { 2} ( 4b + 1) = 7- \frac { 1} { 4} ( 6b - 2)#?

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Answer 1 · 2016-11-01T02:24:33

#b = 2#

To solve this problem:
1. Expand the terms in parenthesis.
2. Group like terms.
3. Solve for #b# while keeping the equation balanced.

#(1/2)4b + 1/2 = 7 - 1/4(6b) + (1/4)2#

#2b + 1/2 = 7 1/2 - 6/4 b#

#2b + 1/2 - 1/2 + 6/4 b = 7 1/2 - 6/4 b - 1/2 + 6/4 b#

#2b + 6/4 b = 7 1/2 - 1/2#

#8/4 b + 6/4 b = 7#

#14/4 b = 7#

#4/14 14/4 b = 7 4/14#

#b = 2#