How to solve y2+(1)/(3)y=2

1. Factor the equation (x + 2)(x + 3) = 0. 2. Set each factor equal to zero: x + 2 = 0 and x + 3 = 0. 3. Solve for x: x = -2 and x = -3.

Solve y2+(1)/(3)y=2 | Equatix

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$$y2+ \frac{ 1 }{ 3 } y=2$$

$y=6-3y_{2}$

Subtract $y_{2}$ from both sides.

$$\frac{1}{3}y=2-y_{2}$$

Multiply both sides by $3$.

$$\frac{\frac{1}{3}y}{\frac{1}{3}}=\frac{2-y_{2}}{\frac{1}{3}}$$

Dividing by $\frac{1}{3}$ undoes the multiplication by $\frac{1}{3}$.

$$y=\frac{2-y_{2}}{\frac{1}{3}}$$

Divide $2-y_{2}$ by $\frac{1}{3}$ by multiplying $2-y_{2}$ by the reciprocal of $\frac{1}{3}$.

$$y=6-3y_{2}$$

$y_{2}=-\frac{y}{3}+2$