How to solve y = (1 + 3x)/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 y = (1 + 3x)/2 | Equatix

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

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

Divide each term of $1+3x$ by $2$ to get $\frac{1}{2}+\frac{3}{2}x$.

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

Swap sides so that all variable terms are on the left hand side.

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

Subtract $\frac{1}{2}$ from both sides.

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

Divide both sides of the equation by $\frac{3}{2}$, which is the same as multiplying both sides by the reciprocal of the fraction.

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

Dividing by $\frac{3}{2}$ undoes the multiplication by $\frac{3}{2}$.

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

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

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

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