Solve 2-(1-bc)/a+3(cx+d)/e=f

$$\frac { 2 - ( 1 - b c ) } { a } + \frac { 3 ( c x + d ) } { e } = f$$

Solve for b (complex solution)

$$\left\{\begin{matrix}b=-\frac{3acx-eaf+3ad+e}{ec}\text{, }&c\neq 0\text{ and }a\neq 0\\b\in \mathrm{C}\text{, }&c=0\text{ and }d\neq \frac{ef}{3}\text{ and }a=-\frac{e}{3d-ef}\end{matrix}\right.$$

Steps for Solving Linear Equation

Multiply both sides of the equation by a.

$$2-\left(1-bc\right)+ae^{-1}\times 3\left(cx+d\right)=fa$$

To find the opposite of 1-bc, find the opposite of each term.

$$2-1+bc+ae^{-1}\times 3\left(cx+d\right)=fa$$

Subtract 1 from 2 to get 1.

$$1+bc+ae^{-1}\times 3\left(cx+d\right)=fa$$

Use the distributive property to multiply ae^{-1}\times 3 by cx+d.

$$1+bc+3ae^{-1}cx+3ae^{-1}d=fa$$

Subtract 1 from both sides.

$$bc+3ae^{-1}cx+3ae^{-1}d=fa-1$$

Subtract 3ae^{-1}cx from both sides.

$$bc+3ae^{-1}d=fa-1-3ae^{-1}cx$$

Subtract 3ae^{-1}d from both sides.

$$bc=fa-1-3ae^{-1}cx-3ae^{-1}d$$

Reorder the terms.

$$bc=af-1-3\times \frac{1}{e}acx-3\times \frac{1}{e}ad$$

Express -3\times \frac{1}{e} as a single fraction.

$$bc=af-1+\frac{-3}{e}acx-3\times \frac{1}{e}ad$$

Express \frac{-3}{e}a as a single fraction.

$$bc=af-1+\frac{-3a}{e}cx-3\times \frac{1}{e}ad$$

Express \frac{-3a}{e}c as a single fraction.

$$bc=af-1+\frac{-3ac}{e}x-3\times \frac{1}{e}ad$$

Express \frac{-3ac}{e}x as a single fraction.

$$bc=af-1+\frac{-3acx}{e}-3\times \frac{1}{e}ad$$

Express -3\times \frac{1}{e} as a single fraction.

$$bc=af-1+\frac{-3acx}{e}+\frac{-3}{e}ad$$

Express \frac{-3}{e}a as a single fraction.

$$bc=af-1+\frac{-3acx}{e}+\frac{-3a}{e}d$$

Express \frac{-3a}{e}d as a single fraction.

$$bc=af-1+\frac{-3acx}{e}+\frac{-3ad}{e}$$

To add or subtract expressions, expand them to make their denominators the same. Multiply af-1 times \frac{e}{e}.

$$bc=\frac{\left(af-1\right)e}{e}+\frac{-3acx}{e}+\frac{-3ad}{e}$$

Since \frac{\left(af-1\right)e}{e} and \frac{-3acx}{e} have the same denominator, add them by adding their numerators.

$$bc=\frac{\left(af-1\right)e-3acx}{e}+\frac{-3ad}{e}$$

Do the multiplications in \left(af-1\right)e-3acx.

$$bc=\frac{afe-e-3xac}{e}+\frac{-3ad}{e}$$

Since \frac{afe-e-3xac}{e} and \frac{-3ad}{e} have the same denominator, add them by adding their numerators.

$$bc=\frac{afe-e-3xac-3ad}{e}$$

The equation is in standard form.

$$cb=\frac{-3acx+eaf-3ad-e}{e}$$

Divide both sides by c.

$$\frac{cb}{c}=\frac{-3acx+eaf-3ad-e}{ec}$$

Dividing by c undoes the multiplication by c.

$$b=\frac{-3acx+eaf-3ad-e}{ec}$$

Solve for b

$$\left\{\begin{matrix}b=-\frac{3acx-eaf+3ad+e}{ec}\text{, }&c\neq 0\text{ and }a\neq 0\\b\in \mathrm{R}\text{, }&c=0\text{ and }d\neq \frac{ef}{3}\text{ and }a=-\frac{e}{3d-ef}\end{matrix}\right.$$

Steps for Solving Linear Equation

Multiply both sides of the equation by a.

$$2-\left(1-bc\right)+ae^{-1}\times 3\left(cx+d\right)=fa$$

To find the opposite of 1-bc, find the opposite of each term.

$$2-1+bc+ae^{-1}\times 3\left(cx+d\right)=fa$$

Subtract 1 from 2 to get 1.

$$1+bc+ae^{-1}\times 3\left(cx+d\right)=fa$$

Use the distributive property to multiply ae^{-1}\times 3 by cx+d.

$$1+bc+3ae^{-1}cx+3ae^{-1}d=fa$$

Subtract 1 from both sides.

$$bc+3ae^{-1}cx+3ae^{-1}d=fa-1$$

Subtract 3ae^{-1}cx from both sides.

$$bc+3ae^{-1}d=fa-1-3ae^{-1}cx$$

Subtract 3ae^{-1}d from both sides.

$$bc=fa-1-3ae^{-1}cx-3ae^{-1}d$$

Reorder the terms.

$$bc=af-1-3\times \frac{1}{e}acx-3\times \frac{1}{e}ad$$

Express -3\times \frac{1}{e} as a single fraction.

$$bc=af-1+\frac{-3}{e}acx-3\times \frac{1}{e}ad$$

Express \frac{-3}{e}a as a single fraction.

$$bc=af-1+\frac{-3a}{e}cx-3\times \frac{1}{e}ad$$

Express \frac{-3a}{e}c as a single fraction.

$$bc=af-1+\frac{-3ac}{e}x-3\times \frac{1}{e}ad$$

Express \frac{-3ac}{e}x as a single fraction.

$$bc=af-1+\frac{-3acx}{e}-3\times \frac{1}{e}ad$$

Express -3\times \frac{1}{e} as a single fraction.

$$bc=af-1+\frac{-3acx}{e}+\frac{-3}{e}ad$$

Express \frac{-3}{e}a as a single fraction.

$$bc=af-1+\frac{-3acx}{e}+\frac{-3a}{e}d$$

Express \frac{-3a}{e}d as a single fraction.

$$bc=af-1+\frac{-3acx}{e}+\frac{-3ad}{e}$$

To add or subtract expressions, expand them to make their denominators the same. Multiply af-1 times \frac{e}{e}.

$$bc=\frac{\left(af-1\right)e}{e}+\frac{-3acx}{e}+\frac{-3ad}{e}$$

Since \frac{\left(af-1\right)e}{e} and \frac{-3acx}{e} have the same denominator, add them by adding their numerators.

$$bc=\frac{\left(af-1\right)e-3acx}{e}+\frac{-3ad}{e}$$

Do the multiplications in \left(af-1\right)e-3acx.

$$bc=\frac{afe-e-3xac}{e}+\frac{-3ad}{e}$$

Since \frac{afe-e-3xac}{e} and \frac{-3ad}{e} have the same denominator, add them by adding their numerators.

$$bc=\frac{afe-e-3xac-3ad}{e}$$

The equation is in standard form.

$$cb=\frac{-3acx+eaf-3ad-e}{e}$$

Divide both sides by c.

$$\frac{cb}{c}=\frac{-3acx+eaf-3ad-e}{ec}$$

Dividing by c undoes the multiplication by c.

$$b=\frac{-3acx+eaf-3ad-e}{ec}$$

Solve for a

$$\left\{\begin{matrix}a=-\frac{e\left(bc+1\right)}{3cx-ef+3d}\text{, }&\left(c=0\text{ and }d\neq \frac{ef}{3}\right)\text{ or }\left(b\neq -\frac{1}{c}\text{ and }f\neq \frac{3\left(cx+d\right)}{e}\text{ and }c\neq 0\right)\\a\neq 0\text{, }&d=-cx+\frac{ef}{3}\text{ and }b=-\frac{1}{c}\text{ and }c\neq 0\end{matrix}\right.$$