Solve for y 3c-y=2

$3 c - y = 2$

Subtract $3 c$ from both sides of the equation.

$- y = 2 - 3 c$

Multiply each term in $- y = 2 - 3 c$ by $- 1$

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Multiply each term in $- y = 2 - 3 c$ by $- 1$ .

$(- y) \cdot - 1 = 2 \cdot - 1 + (- 3 c) \cdot - 1$

Multiply $(- y) \cdot - 1$ .

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Multiply $- 1$ by $- 1$ .

$1 y = 2 \cdot - 1 + (- 3 c) \cdot - 1$

Multiply $y$ by $1$ .

$y = 2 \cdot - 1 + (- 3 c) \cdot - 1$

Simplify each term.

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Multiply $2$ by $- 1$ .

$y = - 2 + (- 3 c) \cdot - 1$

Multiply $- 1$ by $- 3$ .

$y = - 2 + 3 c$

Solve for y 3c-y=2

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