Solve (z_{1}+z_{2})=(z_{3}-z_{2})

$$( z _ { 1 } + z _ { 2 } ) = ( z _ { 3 } - z _ { 2 } )$$

Solve for z_1

$$z_{1}=z_{3}-2z_{2}$$

Solution Steps

Subtract z_{2} from both sides.

$$z_{1}=z_{3}-z_{2}-z_{2}$$

Combine -z_{2} and -z_{2} to get -2z_{2}.

$$z_{1}=z_{3}-2z_{2}$$

Solve for z_2

$$z_{2}=\frac{z_{3}-z_{1}}{2}$$

Steps for Solving Linear Equation

Add z_{2} to both sides.

$$z_{1}+z_{2}+z_{2}=z_{3}$$

Combine z_{2} and z_{2} to get 2z_{2}.

$$z_{1}+2z_{2}=z_{3}$$

Subtract z_{1} from both sides.

$$2z_{2}=z_{3}-z_{1}$$

Divide both sides by 2.

$$\frac{2z_{2}}{2}=\frac{z_{3}-z_{1}}{2}$$

Dividing by 2 undoes the multiplication by 2.

$$z_{2}=\frac{z_{3}-z_{1}}{2}$$