Solve 4z+3=6+2zquad2

$$4 z + 3 = 6 + 2 z \quad 2$$

Solve for z

$$z=\frac{z_{2}}{2}+\frac{3}{4}$$

Steps for Solving Linear Equation

Subtract 3 from both sides.

$$4z=6+2z_{2}-3$$

Subtract 3 from 6 to get 3.

$$4z=3+2z_{2}$$

The equation is in standard form.

$$4z=2z_{2}+3$$

Divide both sides by 4.

$$\frac{4z}{4}=\frac{2z_{2}+3}{4}$$

Dividing by 4 undoes the multiplication by 4.

$$z=\frac{2z_{2}+3}{4}$$

Divide 3+2z_{2} by 4.

$$z=\frac{z_{2}}{2}+\frac{3}{4}$$

Solve for z_2

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

Steps for Solving Linear Equation

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

$$6+2z_{2}=4z+3$$

Subtract 6 from both sides.

$$2z_{2}=4z+3-6$$

Subtract 6 from 3 to get -3.

$$2z_{2}=4z-3$$

Divide both sides by 2.

$$\frac{2z_{2}}{2}=\frac{4z-3}{2}$$

Dividing by 2 undoes the multiplication by 2.

$$z_{2}=\frac{4z-3}{2}$$

Divide 4z-3 by 2.

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