Solve 3x/1+i=j+1

$$\frac { 3 x } { 1 + i } = j + 1$$

Solve for j

$$j=\left(\frac{3}{2}-\frac{3}{2}i\right)x-1$$

Solution Steps

Divide 3x by 1+i to get \left(\frac{3}{2}-\frac{3}{2}i\right)x.

$$\left(\frac{3}{2}-\frac{3}{2}i\right)x=j+1$$

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

$$j+1=\left(\frac{3}{2}-\frac{3}{2}i\right)x$$

Subtract 1 from both sides.

$$j=\left(\frac{3}{2}-\frac{3}{2}i\right)x-1$$

Solve for x

$$x=\left(\frac{1}{3}+\frac{1}{3}i\right)\left(j+1\right)$$

Steps for Solving Linear Equation

Divide 3x by 1+i to get \left(\frac{3}{2}-\frac{3}{2}i\right)x.

$$\left(\frac{3}{2}-\frac{3}{2}i\right)x=j+1$$

Divide both sides by \frac{3}{2}-\frac{3}{2}i.

$$\frac{\left(\frac{3}{2}-\frac{3}{2}i\right)x}{\frac{3}{2}-\frac{3}{2}i}=\frac{j+1}{\frac{3}{2}-\frac{3}{2}i}$$

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

$$x=\frac{j+1}{\frac{3}{2}-\frac{3}{2}i}$$

Divide j+1 by \frac{3}{2}-\frac{3}{2}i.

$$x=\left(\frac{1}{3}+\frac{1}{3}i\right)j+\left(\frac{1}{3}+\frac{1}{3}i\right)$$