How do you solve $2( 6+ 4b ) + 3b = 7b - 4( b - 3)$ ?

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Answer 1 · 2017-10-30T02:12:28

$therefore b=0$

$2( 6+ 4b ) + 3b = 7b - 4( b - 3)$

Use distributive property $a(b+c)=ab+ac$ to open the brackets:

$=> 12+8b + 3b = 7b - 4b+ 12$

Take all terms with variable $b$ on left hand side and rest of the terms on right hand side of the equation applying rules of transposition:

$=> 8b + 3b + 4b - 7b = +12-12$

$=> 8b = 0$

$therefore b=0$

How do you solve 2( 6+ 4b ) + 3b = 7b - 4( b - 3)2( 6+ 4b ) + 3b = 7b - 4( b - 3)?