Solve 2/3-5x=bx+1/3

$$\frac { 2 } { 3 } - 5 x = b x + \frac { 1 } { 3 }$$

Solve for b

$$b=-5+\frac{1}{3x},x\neq 0$$

Steps for Solving Linear Equation

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

$$bx+\frac{1}{3}=\frac{2}{3}-5x$$

Subtract \frac{1}{3} from both sides.

$$bx=\frac{2}{3}-5x-\frac{1}{3}$$

Subtract \frac{1}{3} from \frac{2}{3} to get \frac{1}{3}.

$$bx=\frac{1}{3}-5x$$

The equation is in standard form.

$$xb=\frac{1}{3}-5x$$

Divide both sides by x.

$$\frac{xb}{x}=\frac{\frac{1}{3}-5x}{x}$$

Dividing by x undoes the multiplication by x.

$$b=\frac{\frac{1}{3}-5x}{x}$$

Divide \frac{1}{3}-5x by x.

$$b=-5+\frac{1}{3x}$$

Solve for x

$$x=\frac{1}{3\left(b+5\right)},b\neq -5$$

Steps for Solving Linear Equation

Subtract bx from both sides.

$$\frac{2}{3}-5x-bx=\frac{1}{3}$$

Subtract \frac{2}{3} from both sides.

$$-5x-bx=\frac{1}{3}-\frac{2}{3}$$

Subtract \frac{2}{3} from \frac{1}{3} to get -\frac{1}{3}.

$$-5x-bx=-\frac{1}{3}$$

Combine all terms containing x.

$$\left(-5-b\right)x=-\frac{1}{3}$$

The equation is in standard form.

$$\left(-b-5\right)x=-\frac{1}{3}$$

Divide both sides by -5-b.

$$\frac{\left(-b-5\right)x}{-b-5}=\frac{-\frac{1}{3}}{-b-5}$$

Dividing by -5-b undoes the multiplication by -5-b.

$$x=\frac{-\frac{1}{3}}{-b-5}$$

Divide -\frac{1}{3} by -5-b.

$$x=\frac{1}{3\left(b+5\right)}$$