Solve 2/3/(2x-1)(x-4)+3(x-1/3)(1/3+x)]=2/3(5x^2-x)+frac{14}{9}

$$\frac { 2 } { 3 } / ( 2 x - 1 ) ( x - 4 ) + 3 ( x - \frac { 1 } { 3 } ) ( \frac { 1 } { 3 } + x ) ] = \frac { 2 } { 3 } ( 5 x ^ { 2 } - x ) + \frac { 14 } { 9 }$$

Evaluate

$$\frac{18x^{3}-9x^{2}-7}{3\left(2x-1\right)}$$

Short Solution Steps

Express \frac{\frac{2}{3}}{2x-1} as a single fraction.

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

Express \frac{2}{3\left(2x-1\right)}\left(x-4\right) as a single fraction.

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

Use the distributive property to multiply 3 by x-\frac{1}{3}.

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

Cancel out 3 and 3.

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

Apply the distributive property by multiplying each term of 3x-1 by each term of \frac{1}{3}+x.

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

Cancel out 3 and 3.

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

Combine x and -x to get 0.

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

To add or subtract expressions, expand them to make their denominators the same. Least common multiple of 3\left(2x-1\right) and 3 is 3\left(2x-1\right). Multiply \frac{1}{3} times \frac{2x-1}{2x-1}.

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

Since \frac{2\left(x-4\right)}{3\left(2x-1\right)} and \frac{2x-1}{3\left(2x-1\right)} have the same denominator, subtract them by subtracting their numerators.

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

Do the multiplications in 2\left(x-4\right)-\left(2x-1\right).

$$\frac{2x-8-2x+1}{3\left(2x-1\right)}+3x^{2}$$

Combine like terms in 2x-8-2x+1.

$$\frac{-7}{3\left(2x-1\right)}+3x^{2}$$

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

$$\frac{-7}{3\left(2x-1\right)}+\frac{3x^{2}\times 3\left(2x-1\right)}{3\left(2x-1\right)}$$

Since \frac{-7}{3\left(2x-1\right)} and \frac{3x^{2}\times 3\left(2x-1\right)}{3\left(2x-1\right)} have the same denominator, add them by adding their numerators.

$$\frac{-7+3x^{2}\times 3\left(2x-1\right)}{3\left(2x-1\right)}$$

Do the multiplications in -7+3x^{2}\times 3\left(2x-1\right).

$$\frac{-7+18x^{3}-9x^{2}}{3\left(2x-1\right)}$$

Expand 3\left(2x-1\right).

$$\frac{-7+18x^{3}-9x^{2}}{6x-3}$$

Expand

$$\frac{18x^{3}-9x^{2}-7}{3\left(2x-1\right)}$$

Short Solution Steps

Express \frac{\frac{2}{3}}{2x-1} as a single fraction.

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

Express \frac{2}{3\left(2x-1\right)}\left(x-4\right) as a single fraction.

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

Use the distributive property to multiply 3 by x-\frac{1}{3}.

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

Cancel out 3 and 3.

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

Apply the distributive property by multiplying each term of 3x-1 by each term of \frac{1}{3}+x.

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

Cancel out 3 and 3.

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

Combine x and -x to get 0.

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

To add or subtract expressions, expand them to make their denominators the same. Least common multiple of 3\left(2x-1\right) and 3 is 3\left(2x-1\right). Multiply \frac{1}{3} times \frac{2x-1}{2x-1}.

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

Since \frac{2\left(x-4\right)}{3\left(2x-1\right)} and \frac{2x-1}{3\left(2x-1\right)} have the same denominator, subtract them by subtracting their numerators.

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

Do the multiplications in 2\left(x-4\right)-\left(2x-1\right).

$$\frac{2x-8-2x+1}{3\left(2x-1\right)}+3x^{2}$$

Combine like terms in 2x-8-2x+1.

$$\frac{-7}{3\left(2x-1\right)}+3x^{2}$$

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

$$\frac{-7}{3\left(2x-1\right)}+\frac{3x^{2}\times 3\left(2x-1\right)}{3\left(2x-1\right)}$$

Since \frac{-7}{3\left(2x-1\right)} and \frac{3x^{2}\times 3\left(2x-1\right)}{3\left(2x-1\right)} have the same denominator, add them by adding their numerators.

$$\frac{-7+3x^{2}\times 3\left(2x-1\right)}{3\left(2x-1\right)}$$

Do the multiplications in -7+3x^{2}\times 3\left(2x-1\right).

$$\frac{-7+18x^{3}-9x^{2}}{3\left(2x-1\right)}$$

Expand 3\left(2x-1\right).

$$\frac{-7+18x^{3}-9x^{2}}{6x-3}$$