Solve varepsilonA/d_{1}+eA/d_{2}=frac{(d_{1}+d_{2})varepsilon_{A}}{d_{1}d_{2}}

$$\frac { \varepsilon A } { d _ { 1 } } + \frac { e A } { d _ { 2 } } = \frac { ( d _ { 1 } + d _ { 2 } ) \varepsilon _ { A } } { d _ { 1 } d _ { 2 } }$$

Solve for A (complex solution)

$$\left\{\begin{matrix}A=\frac{\epsilon _{A}\left(d_{1}+d_{2}\right)}{ed_{1}+d_{2}\epsilon }\text{, }&d_{1}\neq -\frac{d_{2}\epsilon }{e}\text{ and }d_{1}\neq 0\text{ and }d_{2}\neq 0\\A\in \mathrm{C}\text{, }&\left(\epsilon _{A}=0\text{ and }d_{1}=-\frac{d_{2}\epsilon }{e}\text{ and }\epsilon \neq 0\text{ and }d_{2}\neq 0\right)\text{ or }\left(d_{1}=-d_{2}\text{ and }\epsilon =e\text{ and }d_{2}\neq 0\right)\end{matrix}\right.$$

Steps for Solving Linear Equation

Multiply both sides of the equation by d_{1}d_{2}, the least common multiple of d_{1},d_{2},d_{1}d_{2}.

$$d_{2}\epsilon A+d_{1}eA=\left(d_{1}+d_{2}\right)\epsilon _{A}$$

Use the distributive property to multiply d_{1}+d_{2} by \epsilon _{A}.

$$d_{2}\epsilon A+d_{1}eA=d_{1}\epsilon _{A}+d_{2}\epsilon _{A}$$

Combine all terms containing A.

$$\left(d_{2}\epsilon +d_{1}e\right)A=d_{1}\epsilon _{A}+d_{2}\epsilon _{A}$$

The equation is in standard form.

$$\left(ed_{1}+d_{2}\epsilon \right)A=d_{1}\epsilon _{A}+d_{2}\epsilon _{A}$$

Divide both sides by ed_{1}+d_{2}\epsilon .

$$\frac{\left(ed_{1}+d_{2}\epsilon \right)A}{ed_{1}+d_{2}\epsilon }=\frac{\epsilon _{A}\left(d_{1}+d_{2}\right)}{ed_{1}+d_{2}\epsilon }$$

Dividing by ed_{1}+d_{2}\epsilon undoes the multiplication by ed_{1}+d_{2}\epsilon .

$$A=\frac{\epsilon _{A}\left(d_{1}+d_{2}\right)}{ed_{1}+d_{2}\epsilon }$$

Solve for A

$$\left\{\begin{matrix}A=\frac{\epsilon _{A}\left(d_{1}+d_{2}\right)}{ed_{1}+d_{2}\epsilon }\text{, }&d_{1}\neq -\frac{d_{2}\epsilon }{e}\text{ and }d_{1}\neq 0\text{ and }d_{2}\neq 0\\A\in \mathrm{R}\text{, }&\left(\epsilon _{A}=0\text{ and }d_{1}=-\frac{d_{2}\epsilon }{e}\text{ and }\epsilon \neq 0\text{ and }d_{2}\neq 0\right)\text{ or }\left(d_{1}=-d_{2}\text{ and }\epsilon =e\text{ and }d_{2}\neq 0\right)\end{matrix}\right.$$

Steps for Solving Linear Equation

Multiply both sides of the equation by d_{1}d_{2}, the least common multiple of d_{1},d_{2},d_{1}d_{2}.

$$d_{2}\epsilon A+d_{1}eA=\left(d_{1}+d_{2}\right)\epsilon _{A}$$

Use the distributive property to multiply d_{1}+d_{2} by \epsilon _{A}.

$$d_{2}\epsilon A+d_{1}eA=d_{1}\epsilon _{A}+d_{2}\epsilon _{A}$$

Combine all terms containing A.

$$\left(d_{2}\epsilon +d_{1}e\right)A=d_{1}\epsilon _{A}+d_{2}\epsilon _{A}$$

The equation is in standard form.

$$\left(ed_{1}+d_{2}\epsilon \right)A=d_{1}\epsilon _{A}+d_{2}\epsilon _{A}$$

Divide both sides by ed_{1}+d_{2}\epsilon .

$$\frac{\left(ed_{1}+d_{2}\epsilon \right)A}{ed_{1}+d_{2}\epsilon }=\frac{\epsilon _{A}\left(d_{1}+d_{2}\right)}{ed_{1}+d_{2}\epsilon }$$

Dividing by ed_{1}+d_{2}\epsilon undoes the multiplication by ed_{1}+d_{2}\epsilon .

$$A=\frac{\epsilon _{A}\left(d_{1}+d_{2}\right)}{ed_{1}+d_{2}\epsilon }$$

Solve for d_1

$$\left\{\begin{matrix}d_{1}=-\frac{d_{2}\left(A\epsilon -\epsilon _{A}\right)}{eA-\epsilon _{A}}\text{, }&\left(d_{2}\neq 0\text{ and }A=0\text{ and }\epsilon _{A}\neq 0\right)\text{ or }\left(d_{2}\neq 0\text{ and }A\neq 0\text{ and }\epsilon \neq \frac{\epsilon _{A}}{A}\text{ and }A\neq \frac{\epsilon _{A}}{e}\right)\\d_{1}\neq 0\text{, }&\left(A=0\text{ and }\epsilon _{A}=0\text{ and }d_{2}\neq 0\right)\text{ or }\left(\epsilon =e\text{ and }\epsilon _{A}\neq 0\text{ and }A=\frac{\epsilon _{A}}{e}\text{ and }d_{2}\neq 0\right)\end{matrix}\right.$$

Steps for Solving Linear Equation

Variable d_{1} cannot be equal to 0 since division by zero is not defined. Multiply both sides of the equation by d_{1}d_{2}, the least common multiple of d_{1},d_{2},d_{1}d_{2}.

$$d_{2}\epsilon A+d_{1}eA=\left(d_{1}+d_{2}\right)\epsilon _{A}$$

Use the distributive property to multiply d_{1}+d_{2} by \epsilon _{A}.

$$d_{2}\epsilon A+d_{1}eA=d_{1}\epsilon _{A}+d_{2}\epsilon _{A}$$

Subtract d_{1}\epsilon _{A} from both sides.

$$d_{2}\epsilon A+d_{1}eA-d_{1}\epsilon _{A}=d_{2}\epsilon _{A}$$

Subtract d_{2}\epsilon A from both sides.

$$d_{1}eA-d_{1}\epsilon _{A}=d_{2}\epsilon _{A}-d_{2}\epsilon A$$

Reorder the terms.

$$eAd_{1}-d_{1}\epsilon _{A}=-Ad_{2}\epsilon +d_{2}\epsilon _{A}$$

Combine all terms containing d_{1}.

$$\left(eA-\epsilon _{A}\right)d_{1}=-Ad_{2}\epsilon +d_{2}\epsilon _{A}$$

The equation is in standard form.

$$\left(eA-\epsilon _{A}\right)d_{1}=d_{2}\epsilon _{A}-Ad_{2}\epsilon$$

Divide both sides by eA-\epsilon _{A}.

$$\frac{\left(eA-\epsilon _{A}\right)d_{1}}{eA-\epsilon _{A}}=\frac{d_{2}\left(\epsilon _{A}-A\epsilon \right)}{eA-\epsilon _{A}}$$

Dividing by eA-\epsilon _{A} undoes the multiplication by eA-\epsilon _{A}.

$$d_{1}=\frac{d_{2}\left(\epsilon _{A}-A\epsilon \right)}{eA-\epsilon _{A}}$$

Variable d_{1} cannot be equal to 0.

$$d_{1}=\frac{d_{2}\left(\epsilon _{A}-A\epsilon \right)}{eA-\epsilon _{A}}\text{, }d_{1}\neq 0$$