Solve 5/2x-3x-3/6x-9

$$\frac{ 5 }{ 2x } - \frac{ 3x-3 }{ 6x-9 }$$

Evaluate

$$\frac{-2x^{2}+12x-15}{2x\left(2x-3\right)}$$

Short Solution Steps

Factor the expressions that are not already factored in \frac{3x-3}{6x-9}.

$$\frac{5}{2x}-\frac{3\left(x-1\right)}{3\left(2x-3\right)}$$

Cancel out 3 in both numerator and denominator.

$$\frac{5}{2x}-\frac{x-1}{2x-3}$$

To add or subtract expressions, expand them to make their denominators the same. Least common multiple of 2x and 2x-3 is 2x\left(2x-3\right). Multiply \frac{5}{2x} times \frac{2x-3}{2x-3}. Multiply \frac{x-1}{2x-3} times \frac{2x}{2x}.

$$\frac{5\left(2x-3\right)}{2x\left(2x-3\right)}-\frac{\left(x-1\right)\times 2x}{2x\left(2x-3\right)}$$

Since \frac{5\left(2x-3\right)}{2x\left(2x-3\right)} and \frac{\left(x-1\right)\times 2x}{2x\left(2x-3\right)} have the same denominator, subtract them by subtracting their numerators.

$$\frac{5\left(2x-3\right)-\left(x-1\right)\times 2x}{2x\left(2x-3\right)}$$

Do the multiplications in 5\left(2x-3\right)-\left(x-1\right)\times 2x.

$$\frac{10x-15-2x^{2}+2x}{2x\left(2x-3\right)}$$

Combine like terms in 10x-15-2x^{2}+2x.

$$\frac{12x-15-2x^{2}}{2x\left(2x-3\right)}$$

Factor the expressions that are not already factored in \frac{12x-15-2x^{2}}{2x\left(2x-3\right)}.

$$\frac{-2\left(x-\left(-\frac{1}{2}\sqrt{6}+3\right)\right)\left(x-\left(\frac{1}{2}\sqrt{6}+3\right)\right)}{2x\left(2x-3\right)}$$

Cancel out 2 in both numerator and denominator.

$$\frac{-\left(x-\left(-\frac{1}{2}\sqrt{6}+3\right)\right)\left(x-\left(\frac{1}{2}\sqrt{6}+3\right)\right)}{x\left(2x-3\right)}$$

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

$$\frac{-\left(x-\left(-\frac{1}{2}\sqrt{6}+3\right)\right)\left(x-\left(\frac{1}{2}\sqrt{6}+3\right)\right)}{2x^{2}-3x}$$

To find the opposite of -\frac{1}{2}\sqrt{6}+3, find the opposite of each term.

$$\frac{-\left(x-\left(-\frac{1}{2}\sqrt{6}\right)-3\right)\left(x-\left(\frac{1}{2}\sqrt{6}+3\right)\right)}{2x^{2}-3x}$$

The opposite of -\frac{1}{2}\sqrt{6} is \frac{1}{2}\sqrt{6}.

$$\frac{-\left(x+\frac{1}{2}\sqrt{6}-3\right)\left(x-\left(\frac{1}{2}\sqrt{6}+3\right)\right)}{2x^{2}-3x}$$

To find the opposite of \frac{1}{2}\sqrt{6}+3, find the opposite of each term.

$$\frac{-\left(x+\frac{1}{2}\sqrt{6}-3\right)\left(x-\frac{1}{2}\sqrt{6}-3\right)}{2x^{2}-3x}$$

Use the distributive property to multiply -1 by x+\frac{1}{2}\sqrt{6}-3.

$$\frac{\left(-x-\frac{1}{2}\sqrt{6}+3\right)\left(x-\frac{1}{2}\sqrt{6}-3\right)}{2x^{2}-3x}$$

Apply the distributive property by multiplying each term of -x-\frac{1}{2}\sqrt{6}+3 by each term of x-\frac{1}{2}\sqrt{6}-3.

$$\frac{-x^{2}-x\left(-\frac{1}{2}\right)\sqrt{6}+3x-\frac{1}{2}\sqrt{6}x-\frac{1}{2}\sqrt{6}\left(-\frac{1}{2}\right)\sqrt{6}-\frac{1}{2}\sqrt{6}\left(-3\right)+3x+3\left(-\frac{1}{2}\right)\sqrt{6}-9}{2x^{2}-3x}$$

Multiply \sqrt{6} and \sqrt{6} to get 6.

$$\frac{-x^{2}-x\left(-\frac{1}{2}\right)\sqrt{6}+3x-\frac{1}{2}\sqrt{6}x-\frac{1}{2}\times 6\left(-\frac{1}{2}\right)-\frac{1}{2}\sqrt{6}\left(-3\right)+3x+3\left(-\frac{1}{2}\right)\sqrt{6}-9}{2x^{2}-3x}$$

Multiply -1 and -\frac{1}{2} to get \frac{1}{2}.

$$\frac{-x^{2}+\frac{1}{2}x\sqrt{6}+3x-\frac{1}{2}\sqrt{6}x-\frac{1}{2}\times 6\left(-\frac{1}{2}\right)-\frac{1}{2}\sqrt{6}\left(-3\right)+3x+3\left(-\frac{1}{2}\right)\sqrt{6}-9}{2x^{2}-3x}$$

Combine \frac{1}{2}x\sqrt{6} and -\frac{1}{2}\sqrt{6}x to get 0.

$$\frac{-x^{2}+3x-\frac{1}{2}\times 6\left(-\frac{1}{2}\right)-\frac{1}{2}\sqrt{6}\left(-3\right)+3x+3\left(-\frac{1}{2}\right)\sqrt{6}-9}{2x^{2}-3x}$$

Express -\frac{1}{2}\times 6 as a single fraction.

$$\frac{-x^{2}+3x+\frac{-6}{2}\left(-\frac{1}{2}\right)-\frac{1}{2}\sqrt{6}\left(-3\right)+3x+3\left(-\frac{1}{2}\right)\sqrt{6}-9}{2x^{2}-3x}$$

Divide -6 by 2 to get -3.

$$\frac{-x^{2}+3x-3\left(-\frac{1}{2}\right)-\frac{1}{2}\sqrt{6}\left(-3\right)+3x+3\left(-\frac{1}{2}\right)\sqrt{6}-9}{2x^{2}-3x}$$

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

$$\frac{-x^{2}+3x+\frac{-3\left(-1\right)}{2}-\frac{1}{2}\sqrt{6}\left(-3\right)+3x+3\left(-\frac{1}{2}\right)\sqrt{6}-9}{2x^{2}-3x}$$

Multiply -3 and -1 to get 3.

$$\frac{-x^{2}+3x+\frac{3}{2}-\frac{1}{2}\sqrt{6}\left(-3\right)+3x+3\left(-\frac{1}{2}\right)\sqrt{6}-9}{2x^{2}-3x}$$

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

$$\frac{-x^{2}+3x+\frac{3}{2}+\frac{-\left(-3\right)}{2}\sqrt{6}+3x+3\left(-\frac{1}{2}\right)\sqrt{6}-9}{2x^{2}-3x}$$

Multiply -1 and -3 to get 3.

$$\frac{-x^{2}+3x+\frac{3}{2}+\frac{3}{2}\sqrt{6}+3x+3\left(-\frac{1}{2}\right)\sqrt{6}-9}{2x^{2}-3x}$$

Combine 3x and 3x to get 6x.

$$\frac{-x^{2}+6x+\frac{3}{2}+\frac{3}{2}\sqrt{6}+3\left(-\frac{1}{2}\right)\sqrt{6}-9}{2x^{2}-3x}$$

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

$$\frac{-x^{2}+6x+\frac{3}{2}+\frac{3}{2}\sqrt{6}+\frac{3\left(-1\right)}{2}\sqrt{6}-9}{2x^{2}-3x}$$

Multiply 3 and -1 to get -3.

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

Fraction \frac{-3}{2} can be rewritten as -\frac{3}{2} by extracting the negative sign.

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

Combine \frac{3}{2}\sqrt{6} and -\frac{3}{2}\sqrt{6} to get 0.

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

Convert 9 to fraction \frac{18}{2}.

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

Since \frac{3}{2} and \frac{18}{2} have the same denominator, subtract them by subtracting their numerators.

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

Subtract 18 from 3 to get -15.

$$\frac{-x^{2}+6x-\frac{15}{2}}{2x^{2}-3x}$$

Expand

$$-\frac{2x^{2}-12x+15}{2x\left(2x-3\right)}$$

Short Solution Steps

Factor the expressions that are not already factored in \frac{3x-3}{6x-9}.

$$\frac{5}{2x}-\frac{3\left(x-1\right)}{3\left(2x-3\right)}$$

Cancel out 3 in both numerator and denominator.

$$\frac{5}{2x}-\frac{x-1}{2x-3}$$

To add or subtract expressions, expand them to make their denominators the same. Least common multiple of 2x and 2x-3 is 2x\left(2x-3\right). Multiply \frac{5}{2x} times \frac{2x-3}{2x-3}. Multiply \frac{x-1}{2x-3} times \frac{2x}{2x}.

$$\frac{5\left(2x-3\right)}{2x\left(2x-3\right)}-\frac{\left(x-1\right)\times 2x}{2x\left(2x-3\right)}$$

Since \frac{5\left(2x-3\right)}{2x\left(2x-3\right)} and \frac{\left(x-1\right)\times 2x}{2x\left(2x-3\right)} have the same denominator, subtract them by subtracting their numerators.

$$\frac{5\left(2x-3\right)-\left(x-1\right)\times 2x}{2x\left(2x-3\right)}$$

Do the multiplications in 5\left(2x-3\right)-\left(x-1\right)\times 2x.

$$\frac{10x-15-2x^{2}+2x}{2x\left(2x-3\right)}$$

Combine like terms in 10x-15-2x^{2}+2x.

$$\frac{12x-15-2x^{2}}{2x\left(2x-3\right)}$$

Factor the expressions that are not already factored in \frac{12x-15-2x^{2}}{2x\left(2x-3\right)}.

$$\frac{-2\left(x-\left(-\frac{1}{2}\sqrt{6}+3\right)\right)\left(x-\left(\frac{1}{2}\sqrt{6}+3\right)\right)}{2x\left(2x-3\right)}$$

Cancel out 2 in both numerator and denominator.

$$\frac{-\left(x-\left(-\frac{1}{2}\sqrt{6}+3\right)\right)\left(x-\left(\frac{1}{2}\sqrt{6}+3\right)\right)}{x\left(2x-3\right)}$$

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

$$\frac{-\left(x-\left(-\frac{1}{2}\sqrt{6}+3\right)\right)\left(x-\left(\frac{1}{2}\sqrt{6}+3\right)\right)}{2x^{2}-3x}$$

To find the opposite of -\frac{1}{2}\sqrt{6}+3, find the opposite of each term.

$$\frac{-\left(x-\left(-\frac{1}{2}\sqrt{6}\right)-3\right)\left(x-\left(\frac{1}{2}\sqrt{6}+3\right)\right)}{2x^{2}-3x}$$

The opposite of -\frac{1}{2}\sqrt{6} is \frac{1}{2}\sqrt{6}.

$$\frac{-\left(x+\frac{1}{2}\sqrt{6}-3\right)\left(x-\left(\frac{1}{2}\sqrt{6}+3\right)\right)}{2x^{2}-3x}$$

To find the opposite of \frac{1}{2}\sqrt{6}+3, find the opposite of each term.

$$\frac{-\left(x+\frac{1}{2}\sqrt{6}-3\right)\left(x-\frac{1}{2}\sqrt{6}-3\right)}{2x^{2}-3x}$$

Use the distributive property to multiply -1 by x+\frac{1}{2}\sqrt{6}-3.

$$\frac{\left(-x-\frac{1}{2}\sqrt{6}+3\right)\left(x-\frac{1}{2}\sqrt{6}-3\right)}{2x^{2}-3x}$$

Apply the distributive property by multiplying each term of -x-\frac{1}{2}\sqrt{6}+3 by each term of x-\frac{1}{2}\sqrt{6}-3.

$$\frac{-x^{2}-x\left(-\frac{1}{2}\right)\sqrt{6}+3x-\frac{1}{2}\sqrt{6}x-\frac{1}{2}\sqrt{6}\left(-\frac{1}{2}\right)\sqrt{6}-\frac{1}{2}\sqrt{6}\left(-3\right)+3x+3\left(-\frac{1}{2}\right)\sqrt{6}-9}{2x^{2}-3x}$$

Multiply \sqrt{6} and \sqrt{6} to get 6.

$$\frac{-x^{2}-x\left(-\frac{1}{2}\right)\sqrt{6}+3x-\frac{1}{2}\sqrt{6}x-\frac{1}{2}\times 6\left(-\frac{1}{2}\right)-\frac{1}{2}\sqrt{6}\left(-3\right)+3x+3\left(-\frac{1}{2}\right)\sqrt{6}-9}{2x^{2}-3x}$$

Multiply -1 and -\frac{1}{2} to get \frac{1}{2}.

$$\frac{-x^{2}+\frac{1}{2}x\sqrt{6}+3x-\frac{1}{2}\sqrt{6}x-\frac{1}{2}\times 6\left(-\frac{1}{2}\right)-\frac{1}{2}\sqrt{6}\left(-3\right)+3x+3\left(-\frac{1}{2}\right)\sqrt{6}-9}{2x^{2}-3x}$$

Combine \frac{1}{2}x\sqrt{6} and -\frac{1}{2}\sqrt{6}x to get 0.

$$\frac{-x^{2}+3x-\frac{1}{2}\times 6\left(-\frac{1}{2}\right)-\frac{1}{2}\sqrt{6}\left(-3\right)+3x+3\left(-\frac{1}{2}\right)\sqrt{6}-9}{2x^{2}-3x}$$

Express -\frac{1}{2}\times 6 as a single fraction.

$$\frac{-x^{2}+3x+\frac{-6}{2}\left(-\frac{1}{2}\right)-\frac{1}{2}\sqrt{6}\left(-3\right)+3x+3\left(-\frac{1}{2}\right)\sqrt{6}-9}{2x^{2}-3x}$$

Divide -6 by 2 to get -3.

$$\frac{-x^{2}+3x-3\left(-\frac{1}{2}\right)-\frac{1}{2}\sqrt{6}\left(-3\right)+3x+3\left(-\frac{1}{2}\right)\sqrt{6}-9}{2x^{2}-3x}$$

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

$$\frac{-x^{2}+3x+\frac{-3\left(-1\right)}{2}-\frac{1}{2}\sqrt{6}\left(-3\right)+3x+3\left(-\frac{1}{2}\right)\sqrt{6}-9}{2x^{2}-3x}$$

Multiply -3 and -1 to get 3.

$$\frac{-x^{2}+3x+\frac{3}{2}-\frac{1}{2}\sqrt{6}\left(-3\right)+3x+3\left(-\frac{1}{2}\right)\sqrt{6}-9}{2x^{2}-3x}$$

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

$$\frac{-x^{2}+3x+\frac{3}{2}+\frac{-\left(-3\right)}{2}\sqrt{6}+3x+3\left(-\frac{1}{2}\right)\sqrt{6}-9}{2x^{2}-3x}$$

Multiply -1 and -3 to get 3.

$$\frac{-x^{2}+3x+\frac{3}{2}+\frac{3}{2}\sqrt{6}+3x+3\left(-\frac{1}{2}\right)\sqrt{6}-9}{2x^{2}-3x}$$

Combine 3x and 3x to get 6x.

$$\frac{-x^{2}+6x+\frac{3}{2}+\frac{3}{2}\sqrt{6}+3\left(-\frac{1}{2}\right)\sqrt{6}-9}{2x^{2}-3x}$$

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

$$\frac{-x^{2}+6x+\frac{3}{2}+\frac{3}{2}\sqrt{6}+\frac{3\left(-1\right)}{2}\sqrt{6}-9}{2x^{2}-3x}$$

Multiply 3 and -1 to get -3.

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

Fraction \frac{-3}{2} can be rewritten as -\frac{3}{2} by extracting the negative sign.

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

Combine \frac{3}{2}\sqrt{6} and -\frac{3}{2}\sqrt{6} to get 0.

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

Convert 9 to fraction \frac{18}{2}.

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

Since \frac{3}{2} and \frac{18}{2} have the same denominator, subtract them by subtracting their numerators.

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

Subtract 18 from 3 to get -15.

$$\frac{-x^{2}+6x-\frac{15}{2}}{2x^{2}-3x}$$