Solve 1/{R_{2}}+1/{R_{3}}=1/5+1/1

$$\frac{ 1 }{ { R }_{ 2 } } + \frac{ 1 }{ { R }_{ 3 } } = \frac{ 1 }{ 5 } + \frac{ 1 }{ 1 }$$

Solve for R_2

$$R_{2}=-\frac{5R_{3}}{5-6R_{3}}$$

$$R_{3}\neq 0\text{ and }R_{3}\neq \frac{5}{6}$$

Steps for Solving Linear Equation

Variable R_{2} cannot be equal to 0 since division by zero is not defined. Multiply both sides of the equation by 5R_{2}R_{3}, the least common multiple of R_{2},R_{3},5.

$$5R_{3}+5R_{2}=5R_{2}R_{3}\times \frac{1}{5}+5R_{2}R_{3}$$

Multiply 5 and \frac{1}{5} to get 1.

$$5R_{3}+5R_{2}=R_{2}R_{3}+5R_{2}R_{3}$$

Combine R_{2}R_{3} and 5R_{2}R_{3} to get 6R_{2}R_{3}.

$$5R_{3}+5R_{2}=6R_{2}R_{3}$$

Subtract 6R_{2}R_{3} from both sides.

$$5R_{3}+5R_{2}-6R_{2}R_{3}=0$$

Subtract 5R_{3} from both sides. Anything subtracted from zero gives its negation.

$$5R_{2}-6R_{2}R_{3}=-5R_{3}$$

Combine all terms containing R_{2}.

$$\left(5-6R_{3}\right)R_{2}=-5R_{3}$$

Divide both sides by 5-6R_{3}.

$$\frac{\left(5-6R_{3}\right)R_{2}}{5-6R_{3}}=-\frac{5R_{3}}{5-6R_{3}}$$

Dividing by 5-6R_{3} undoes the multiplication by 5-6R_{3}.

$$R_{2}=-\frac{5R_{3}}{5-6R_{3}}$$

Variable R_{2} cannot be equal to 0.

$$R_{2}=-\frac{5R_{3}}{5-6R_{3}}\text{, }R_{2}\neq 0$$

Solve for R_3

$$R_{3}=-\frac{5R_{2}}{5-6R_{2}}$$

$$R_{2}\neq 0\text{ and }R_{2}\neq \frac{5}{6}$$

Steps for Solving Linear Equation

Variable R_{3} cannot be equal to 0 since division by zero is not defined. Multiply both sides of the equation by 5R_{2}R_{3}, the least common multiple of R_{2},R_{3},5.

$$5R_{3}+5R_{2}=5R_{2}R_{3}\times \frac{1}{5}+5R_{2}R_{3}$$

Multiply 5 and \frac{1}{5} to get 1.

$$5R_{3}+5R_{2}=R_{2}R_{3}+5R_{2}R_{3}$$

Combine R_{2}R_{3} and 5R_{2}R_{3} to get 6R_{2}R_{3}.

$$5R_{3}+5R_{2}=6R_{2}R_{3}$$

Subtract 6R_{2}R_{3} from both sides.

$$5R_{3}+5R_{2}-6R_{2}R_{3}=0$$

Subtract 5R_{2} from both sides. Anything subtracted from zero gives its negation.

$$5R_{3}-6R_{2}R_{3}=-5R_{2}$$

Combine all terms containing R_{3}.

$$\left(5-6R_{2}\right)R_{3}=-5R_{2}$$

Divide both sides by 5-6R_{2}.

$$\frac{\left(5-6R_{2}\right)R_{3}}{5-6R_{2}}=-\frac{5R_{2}}{5-6R_{2}}$$

Dividing by 5-6R_{2} undoes the multiplication by 5-6R_{2}.

$$R_{3}=-\frac{5R_{2}}{5-6R_{2}}$$

Variable R_{3} cannot be equal to 0.

$$R_{3}=-\frac{5R_{2}}{5-6R_{2}}\text{, }R_{3}\neq 0$$