Solve x=4-(3+2z-2y/3)

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

Solve for x

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

Solution Steps

Divide each term of 3+2z-2y by 3 to get 1+\frac{2}{3}z-\frac{2}{3}y.

$$x=4-\left(1+\frac{2}{3}z-\frac{2}{3}y\right)$$

To find the opposite of 1+\frac{2}{3}z-\frac{2}{3}y, find the opposite of each term.

$$x=4-1-\frac{2}{3}z+\frac{2}{3}y$$

Subtract 1 from 4 to get 3.

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

Solve for y

$$y=\frac{3x}{2}+z-\frac{9}{2}$$

Steps for Solving Linear Equation

Divide each term of 3+2z-2y by 3 to get 1+\frac{2}{3}z-\frac{2}{3}y.

$$x=4-\left(1+\frac{2}{3}z-\frac{2}{3}y\right)$$

To find the opposite of 1+\frac{2}{3}z-\frac{2}{3}y, find the opposite of each term.

$$x=4-1-\frac{2}{3}z+\frac{2}{3}y$$

Subtract 1 from 4 to get 3.

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

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

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

Subtract 3 from both sides.

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

Add \frac{2}{3}z to both sides.

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

The equation is in standard form.

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

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

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

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

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

Divide x-3+\frac{2z}{3} by \frac{2}{3} by multiplying x-3+\frac{2z}{3} by the reciprocal of \frac{2}{3}.

$$y=\frac{3x}{2}+z-\frac{9}{2}$$