How do you solve for b in $d = 3 a + 3 b$ $d = 3a + 3b$ ?

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Answer 1 · 2018-04-23T21:40:48

$\Rightarrow b = \frac{d}{3} - a$ $=>b = d/3 - a$

We are given

$d = 3 a + 3 b$ $d = 3a + 3b$

Subtract $3 a$ $3a$ from both sides

$d - 3 a = 3 a + 3 b - 3 a$ $d - 3a = 3a + 3b - 3a$

$d - 3a = cancel(3a)+3b cancel(-3a)$

$d - 3a = 3b$

Divide both sides by $3$

$(d-3a)/3 = (3b)/3$

$d/3 - (cancel(3)a)/cancel3 = (cancel(3)b)/cancel3$

$d/3 - a = b$

How do you solve for b in d = 3a + 3b d=3a+3bd = 3a + 3b ?