How to solve x=(5+z-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 x=(5+z-y)/(2) | Equatix

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$$x= \frac{ 5+z-y }{ 2 }$$

$y=5+z-2x$

Divide each term of $5+z-y$ by $2$ to get $\frac{5}{2}+\frac{1}{2}z-\frac{1}{2}y$.

$$x=\frac{5}{2}+\frac{1}{2}z-\frac{1}{2}y$$

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

$$\frac{5}{2}+\frac{1}{2}z-\frac{1}{2}y=x$$

Subtract $\frac{5}{2}$ from both sides.

$$\frac{1}{2}z-\frac{1}{2}y=x-\frac{5}{2}$$

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

$$-\frac{1}{2}y=x-\frac{5}{2}-\frac{1}{2}z$$

The equation is in standard form.

$$-\frac{1}{2}y=-\frac{z}{2}+x-\frac{5}{2}$$

Multiply both sides by $-2$.

$$\frac{-\frac{1}{2}y}{-\frac{1}{2}}=\frac{-\frac{z}{2}+x-\frac{5}{2}}{-\frac{1}{2}}$$

Dividing by $-\frac{1}{2}$ undoes the multiplication by $-\frac{1}{2}$.

$$y=\frac{-\frac{z}{2}+x-\frac{5}{2}}{-\frac{1}{2}}$$

Divide $x-\frac{5}{2}-\frac{z}{2}$ by $-\frac{1}{2}$ by multiplying $x-\frac{5}{2}-\frac{z}{2}$ by the reciprocal of $-\frac{1}{2}$.

$$y=5+z-2x$$

$x=\frac{5+z-y}{2}$