Question

find the value of Y if
$\frac{x}{2} + \frac{y}{3} = 1$

Answer 1

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find the value y if
$\frac{x}{2} + \frac{y}{3} = 1$

$\sf\frac{3x+2y}{6}=1$
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$\large\sf\:3x+2y\:=\:6$

2y=6-3x

$\large\bf\implies\:y\:=\frac{6-3x}{2}$

Answer 2

$\frac{x}{2} + \frac{y}{3} = 1$
$\frac{x}{2} + \frac{y}{3} = 1 \\ \frac{3x + 2y}{6} = 1 \\ 3x + 2y = 6 \\ 2y = 6 - 3x \\ y = \frac{6 - 3x}{2}$