Solve A=1+frac{1/1+frac{I{3}}}{1/1+frac{1{3}}}

$$A = \frac { 1 + \frac { 1 } { 1 + \frac { I } { 3 } } } { \frac { 1 } { 1 + \frac { 1 } { 3 } } }$$

Solve for I

$$I=-\frac{3\left(8-3A\right)}{4-3A}$$

$$A\neq \frac{4}{3}$$

Steps for Solving Linear Equation

Add 1 and \frac{1}{3} to get \frac{4}{3}.

$$A=\frac{1+\frac{1}{1+\frac{I}{3}}}{\frac{1}{\frac{4}{3}}}$$

Divide 1 by \frac{4}{3} by multiplying 1 by the reciprocal of \frac{4}{3}.

$$A=\frac{1+\frac{1}{1+\frac{I}{3}}}{1\times \frac{3}{4}}$$

Multiply 1 and \frac{3}{4} to get \frac{3}{4}.

$$A=\frac{1+\frac{1}{1+\frac{I}{3}}}{\frac{3}{4}}$$

Divide each term of 1+\frac{1}{1+\frac{I}{3}} by \frac{3}{4} to get \frac{1}{\frac{3}{4}}+\frac{\frac{1}{1+\frac{I}{3}}}{\frac{3}{4}}.

$$A=\frac{1}{\frac{3}{4}}+\frac{\frac{1}{1+\frac{I}{3}}}{\frac{3}{4}}$$

Divide 1 by \frac{3}{4} by multiplying 1 by the reciprocal of \frac{3}{4}.

$$A=1\times \frac{4}{3}+\frac{\frac{1}{1+\frac{I}{3}}}{\frac{3}{4}}$$

Multiply 1 and \frac{4}{3} to get \frac{4}{3}.

$$A=\frac{4}{3}+\frac{\frac{1}{1+\frac{I}{3}}}{\frac{3}{4}}$$

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

$$\frac{4}{3}+\frac{\frac{1}{1+\frac{I}{3}}}{\frac{3}{4}}=A$$

Subtract \frac{4}{3} from both sides.

$$\frac{\frac{1}{1+\frac{I}{3}}}{\frac{3}{4}}=A-\frac{4}{3}$$

Reorder the terms.

$$\frac{1}{\frac{3}{4}\left(\frac{I}{3}+1\right)}=A-\frac{4}{3}$$

Use the distributive property to multiply \frac{3}{4} by \frac{I}{3}+1.

$$\frac{1}{\frac{3}{4}\times \frac{I}{3}+\frac{3}{4}}=A-\frac{4}{3}$$

Multiply \frac{3}{4} times \frac{I}{3} by multiplying numerator times numerator and denominator times denominator.

$$\frac{1}{\frac{3I}{4\times 3}+\frac{3}{4}}=A-\frac{4}{3}$$

Cancel out 3 in both numerator and denominator.

$$\frac{1}{\frac{I}{4}+\frac{3}{4}}=A-\frac{4}{3}$$

Since \frac{I}{4} and \frac{3}{4} have the same denominator, add them by adding their numerators.

$$\frac{1}{\frac{I+3}{4}}=A-\frac{4}{3}$$

Divide 1 by \frac{I+3}{4} by multiplying 1 by the reciprocal of \frac{I+3}{4}.

$$\frac{4}{I+3}=A-\frac{4}{3}$$

Variable I cannot be equal to -3 since division by zero is not defined. Multiply both sides of the equation by 3\left(I+3\right), the least common multiple of I+3,3.

$$3\times 4=3\left(I+3\right)A+3\left(I+3\right)\left(-\frac{4}{3}\right)$$

Reorder the terms.

$$3\times 4=3A\left(I+3\right)-\frac{4}{3}\times 3\left(I+3\right)$$

Multiply 3 and 4 to get 12.

$$12=3A\left(I+3\right)-\frac{4}{3}\times 3\left(I+3\right)$$

Use the distributive property to multiply 3A by I+3.

$$12=3AI+9A-\frac{4}{3}\times 3\left(I+3\right)$$

Multiply -\frac{4}{3} and 3 to get -4.

$$12=3AI+9A-4\left(I+3\right)$$

Use the distributive property to multiply -4 by I+3.

$$12=3AI+9A-4I-12$$

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

$$3AI+9A-4I-12=12$$

Subtract 9A from both sides.

$$3AI-4I-12=12-9A$$

Add 12 to both sides.

$$3AI-4I=12-9A+12$$

Add 12 and 12 to get 24.

$$3AI-4I=24-9A$$

Combine all terms containing I.

$$\left(3A-4\right)I=24-9A$$

Divide both sides by 3A-4.

$$\frac{\left(3A-4\right)I}{3A-4}=\frac{24-9A}{3A-4}$$

Dividing by 3A-4 undoes the multiplication by 3A-4.

$$I=\frac{24-9A}{3A-4}$$

Divide 24-9A by 3A-4.

$$I=\frac{3\left(8-3A\right)}{3A-4}$$

Variable I cannot be equal to -3.

$$I=\frac{3\left(8-3A\right)}{3A-4}\text{, }I\neq -3$$

Solve for A

$$A=\frac{4\left(I+6\right)}{3\left(I+3\right)}$$

$$I\neq -3$$