How to solve y = (8 - 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 = (8 - 3x)/2 | Equatix

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$$y=\frac{8-3x}{2}$$

$x=\frac{8-2y}{3}$

Divide each term of $8-3x$ by $2$ to get $4-\frac{3}{2}x$.

$$y=4-\frac{3}{2}x$$

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

$$4-\frac{3}{2}x=y$$

Subtract $4$ from both sides.

$$-\frac{3}{2}x=y-4$$

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

$$\frac{-\frac{3}{2}x}{-\frac{3}{2}}=\frac{y-4}{-\frac{3}{2}}$$

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

$$x=\frac{y-4}{-\frac{3}{2}}$$

Divide $y-4$ by $-\frac{3}{2}$ by multiplying $y-4$ by the reciprocal of $-\frac{3}{2}$.

$$x=\frac{8-2y}{3}$$

$y=-\frac{3x}{2}+4$