Solve =(2x+1/1-3x)+2/(frac{2x+1{1-3x})-3}

$$= \frac { ( \frac { 2 x + 1 } { 1 - 3 x } ) + 2 } { ( \frac { 2 x + 1 } { 1 - 3 x } ) - 3 }$$

Evaluate

$$\frac{3-4x}{11x-2}$$

Short Solution Steps

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

$$\frac{\frac{2x+1}{1-3x}+\frac{2\left(1-3x\right)}{1-3x}}{\frac{2x+1}{1-3x}-3}$$

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

$$\frac{\frac{2x+1+2\left(1-3x\right)}{1-3x}}{\frac{2x+1}{1-3x}-3}$$

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

$$\frac{\frac{2x+1+2-6x}{1-3x}}{\frac{2x+1}{1-3x}-3}$$

Combine like terms in 2x+1+2-6x.

$$\frac{\frac{-4x+3}{1-3x}}{\frac{2x+1}{1-3x}-3}$$

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

$$\frac{\frac{-4x+3}{1-3x}}{\frac{2x+1}{1-3x}-\frac{3\left(1-3x\right)}{1-3x}}$$

Since \frac{2x+1}{1-3x} and \frac{3\left(1-3x\right)}{1-3x} have the same denominator, subtract them by subtracting their numerators.

$$\frac{\frac{-4x+3}{1-3x}}{\frac{2x+1-3\left(1-3x\right)}{1-3x}}$$

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

$$\frac{\frac{-4x+3}{1-3x}}{\frac{2x+1-3+9x}{1-3x}}$$

Combine like terms in 2x+1-3+9x.

$$\frac{\frac{-4x+3}{1-3x}}{\frac{11x-2}{1-3x}}$$

Divide \frac{-4x+3}{1-3x} by \frac{11x-2}{1-3x} by multiplying \frac{-4x+3}{1-3x} by the reciprocal of \frac{11x-2}{1-3x}.

$$\frac{\left(-4x+3\right)\left(1-3x\right)}{\left(1-3x\right)\left(11x-2\right)}$$

Cancel out -3x+1 in both numerator and denominator.

$$\frac{-4x+3}{11x-2}$$

Expand

$$\frac{3-4x}{11x-2}$$

Short Solution Steps

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

$$\frac{\frac{2x+1}{1-3x}+\frac{2\left(1-3x\right)}{1-3x}}{\frac{2x+1}{1-3x}-3}$$

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

$$\frac{\frac{2x+1+2\left(1-3x\right)}{1-3x}}{\frac{2x+1}{1-3x}-3}$$

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

$$\frac{\frac{2x+1+2-6x}{1-3x}}{\frac{2x+1}{1-3x}-3}$$

Combine like terms in 2x+1+2-6x.

$$\frac{\frac{-4x+3}{1-3x}}{\frac{2x+1}{1-3x}-3}$$

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

$$\frac{\frac{-4x+3}{1-3x}}{\frac{2x+1}{1-3x}-\frac{3\left(1-3x\right)}{1-3x}}$$

Since \frac{2x+1}{1-3x} and \frac{3\left(1-3x\right)}{1-3x} have the same denominator, subtract them by subtracting their numerators.

$$\frac{\frac{-4x+3}{1-3x}}{\frac{2x+1-3\left(1-3x\right)}{1-3x}}$$

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

$$\frac{\frac{-4x+3}{1-3x}}{\frac{2x+1-3+9x}{1-3x}}$$

Combine like terms in 2x+1-3+9x.

$$\frac{\frac{-4x+3}{1-3x}}{\frac{11x-2}{1-3x}}$$

Divide \frac{-4x+3}{1-3x} by \frac{11x-2}{1-3x} by multiplying \frac{-4x+3}{1-3x} by the reciprocal of \frac{11x-2}{1-3x}.

$$\frac{\left(-4x+3\right)\left(1-3x\right)}{\left(1-3x\right)\left(11x-2\right)}$$

Cancel out -3x+1 in both numerator and denominator.

$$\frac{-4x+3}{11x-2}$$