How do you solve for a in $3 a - 2 b = 4 c$ $3a-2b=4c$ ?

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Answer 1 · 2016-02-21T16:54:50

$a = \frac{2 (2 c + b)}{3}$ $color(blue)(a = (2(2c +b))/3$

$3 a - 2 b = 4 c$ $3a -2b = 4c$

$3 a = 4 c + 2 b$ $3a = 4c +2b$

$2$ $2$ is common to both terms of the R.H.S:

$3 a = 2 (2 c + b)$ $3a = 2(2c +b)$

$a = \frac{2 (2 c + b)}{3}$ $a = (2(2c +b))/3$

How do you solve for a in 3a-2b=4c3a−2b=4c3a-2b=4c?