Solve 2+1/2+1{1+frac{1/2+frac{1{A}}}}=64/27

$$2 + \frac { 1 } { 2 + \frac { 1 } { 1 + \frac { 1 } { 2 + \frac { 1 } { A } } } } = \frac { 64 } { 27 }$$

Solve for A

$$A=3$$

Steps for Solving Linear Equation

To add or subtract expressions, expand them to make their denominators the same. Multiply 2 times \frac{A}{A}.

$$2+\frac{1}{2+\frac{1}{1+\frac{1}{\frac{2A}{A}+\frac{1}{A}}}}=\frac{64}{27}$$

Since \frac{2A}{A} and \frac{1}{A} have the same denominator, add them by adding their numerators.

$$2+\frac{1}{2+\frac{1}{1+\frac{1}{\frac{2A+1}{A}}}}=\frac{64}{27}$$

Variable A cannot be equal to 0 since division by zero is not defined. Divide 1 by \frac{2A+1}{A} by multiplying 1 by the reciprocal of \frac{2A+1}{A}.

$$2+\frac{1}{2+\frac{1}{1+\frac{A}{2A+1}}}=\frac{64}{27}$$

To add or subtract expressions, expand them to make their denominators the same. Multiply 1 times \frac{2A+1}{2A+1}.

$$2+\frac{1}{2+\frac{1}{\frac{2A+1}{2A+1}+\frac{A}{2A+1}}}=\frac{64}{27}$$

Since \frac{2A+1}{2A+1} and \frac{A}{2A+1} have the same denominator, add them by adding their numerators.

$$2+\frac{1}{2+\frac{1}{\frac{2A+1+A}{2A+1}}}=\frac{64}{27}$$

Combine like terms in 2A+1+A.

$$2+\frac{1}{2+\frac{1}{\frac{3A+1}{2A+1}}}=\frac{64}{27}$$

Variable A cannot be equal to -\frac{1}{2} since division by zero is not defined. Divide 1 by \frac{3A+1}{2A+1} by multiplying 1 by the reciprocal of \frac{3A+1}{2A+1}.

$$2+\frac{1}{2+\frac{2A+1}{3A+1}}=\frac{64}{27}$$

To add or subtract expressions, expand them to make their denominators the same. Multiply 2 times \frac{3A+1}{3A+1}.

$$2+\frac{1}{\frac{2\left(3A+1\right)}{3A+1}+\frac{2A+1}{3A+1}}=\frac{64}{27}$$

Since \frac{2\left(3A+1\right)}{3A+1} and \frac{2A+1}{3A+1} have the same denominator, add them by adding their numerators.

$$2+\frac{1}{\frac{2\left(3A+1\right)+2A+1}{3A+1}}=\frac{64}{27}$$

Do the multiplications in 2\left(3A+1\right)+2A+1.

$$2+\frac{1}{\frac{6A+2+2A+1}{3A+1}}=\frac{64}{27}$$

Combine like terms in 6A+2+2A+1.

$$2+\frac{1}{\frac{8A+3}{3A+1}}=\frac{64}{27}$$

Variable A cannot be equal to -\frac{1}{3} since division by zero is not defined. Divide 1 by \frac{8A+3}{3A+1} by multiplying 1 by the reciprocal of \frac{8A+3}{3A+1}.

$$2+\frac{3A+1}{8A+3}=\frac{64}{27}$$

To add or subtract expressions, expand them to make their denominators the same. Multiply 2 times \frac{8A+3}{8A+3}.

$$\frac{2\left(8A+3\right)}{8A+3}+\frac{3A+1}{8A+3}=\frac{64}{27}$$

Since \frac{2\left(8A+3\right)}{8A+3} and \frac{3A+1}{8A+3} have the same denominator, add them by adding their numerators.

$$\frac{2\left(8A+3\right)+3A+1}{8A+3}=\frac{64}{27}$$

Do the multiplications in 2\left(8A+3\right)+3A+1.

$$\frac{16A+6+3A+1}{8A+3}=\frac{64}{27}$$

Combine like terms in 16A+6+3A+1.

$$\frac{19A+7}{8A+3}=\frac{64}{27}$$

Variable A cannot be equal to -\frac{3}{8} since division by zero is not defined. Multiply both sides of the equation by 27\left(8A+3\right), the least common multiple of 8A+3,27.

$$27\left(19A+7\right)=64\left(8A+3\right)$$

Use the distributive property to multiply 27 by 19A+7.

$$513A+189=64\left(8A+3\right)$$

Use the distributive property to multiply 64 by 8A+3.

$$513A+189=512A+192$$

Subtract 512A from both sides.

$$513A+189-512A=192$$

Combine 513A and -512A to get A.

$$A+189=192$$

Subtract 189 from both sides.

$$A=192-189$$

Subtract 189 from 192 to get 3.

$$A=3$$