Solve (x-1/2x^2+3x-5-x+1/3x^2+4x+1)*(6x^2+17x+5)

$$( \frac { x - 1 } { 2 x ^ { 2 } + 3 x - 5 } - \frac { x + 1 } { 3 x ^ { 2 } + 4 x + 1 } ) \cdot ( 6 x ^ { 2 } + 17 x + 5 )$$

Evaluate

$$x-4$$

Short Solution Steps

Factor the expressions that are not already factored in \frac{x-1}{2x^{2}+3x-5}.

$$\left(\frac{x-1}{\left(x-1\right)\left(2x+5\right)}-\frac{x+1}{3x^{2}+4x+1}\right)\left(6x^{2}+17x+5\right)$$

Cancel out x-1 in both numerator and denominator.

$$\left(\frac{1}{2x+5}-\frac{x+1}{3x^{2}+4x+1}\right)\left(6x^{2}+17x+5\right)$$

Factor the expressions that are not already factored in \frac{x+1}{3x^{2}+4x+1}.

$$\left(\frac{1}{2x+5}-\frac{x+1}{\left(x+1\right)\left(3x+1\right)}\right)\left(6x^{2}+17x+5\right)$$

Cancel out x+1 in both numerator and denominator.

$$\left(\frac{1}{2x+5}-\frac{1}{3x+1}\right)\left(6x^{2}+17x+5\right)$$

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

$$\left(\frac{3x+1}{\left(2x+5\right)\left(3x+1\right)}-\frac{2x+5}{\left(2x+5\right)\left(3x+1\right)}\right)\left(6x^{2}+17x+5\right)$$

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

$$\frac{3x+1-\left(2x+5\right)}{\left(2x+5\right)\left(3x+1\right)}\left(6x^{2}+17x+5\right)$$

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

$$\frac{3x+1-2x-5}{\left(2x+5\right)\left(3x+1\right)}\left(6x^{2}+17x+5\right)$$

Combine like terms in 3x+1-2x-5.

$$\frac{x-4}{\left(2x+5\right)\left(3x+1\right)}\left(6x^{2}+17x+5\right)$$

Express \frac{x-4}{\left(2x+5\right)\left(3x+1\right)}\left(6x^{2}+17x+5\right) as a single fraction.

$$\frac{\left(x-4\right)\left(6x^{2}+17x+5\right)}{\left(2x+5\right)\left(3x+1\right)}$$

Factor the expressions that are not already factored.

$$\frac{\left(x-4\right)\left(2x+5\right)\left(3x+1\right)}{\left(2x+5\right)\left(3x+1\right)}$$

Cancel out \left(2x+5\right)\left(3x+1\right) in both numerator and denominator.

$$x-4$$

Expand

$$x-4$$

Short Solution Steps

Factor the expressions that are not already factored in \frac{x-1}{2x^{2}+3x-5}.

$$\left(\frac{x-1}{\left(x-1\right)\left(2x+5\right)}-\frac{x+1}{3x^{2}+4x+1}\right)\left(6x^{2}+17x+5\right)$$

Cancel out x-1 in both numerator and denominator.

$$\left(\frac{1}{2x+5}-\frac{x+1}{3x^{2}+4x+1}\right)\left(6x^{2}+17x+5\right)$$

Factor the expressions that are not already factored in \frac{x+1}{3x^{2}+4x+1}.

$$\left(\frac{1}{2x+5}-\frac{x+1}{\left(x+1\right)\left(3x+1\right)}\right)\left(6x^{2}+17x+5\right)$$

Cancel out x+1 in both numerator and denominator.

$$\left(\frac{1}{2x+5}-\frac{1}{3x+1}\right)\left(6x^{2}+17x+5\right)$$

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

$$\left(\frac{3x+1}{\left(2x+5\right)\left(3x+1\right)}-\frac{2x+5}{\left(2x+5\right)\left(3x+1\right)}\right)\left(6x^{2}+17x+5\right)$$

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

$$\frac{3x+1-\left(2x+5\right)}{\left(2x+5\right)\left(3x+1\right)}\left(6x^{2}+17x+5\right)$$

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

$$\frac{3x+1-2x-5}{\left(2x+5\right)\left(3x+1\right)}\left(6x^{2}+17x+5\right)$$

Combine like terms in 3x+1-2x-5.

$$\frac{x-4}{\left(2x+5\right)\left(3x+1\right)}\left(6x^{2}+17x+5\right)$$

Express \frac{x-4}{\left(2x+5\right)\left(3x+1\right)}\left(6x^{2}+17x+5\right) as a single fraction.

$$\frac{\left(x-4\right)\left(6x^{2}+17x+5\right)}{\left(2x+5\right)\left(3x+1\right)}$$

Factor the expressions that are not already factored.

$$\frac{\left(x-4\right)\left(2x+5\right)\left(3x+1\right)}{\left(2x+5\right)\left(3x+1\right)}$$

Cancel out \left(2x+5\right)\left(3x+1\right) in both numerator and denominator.

$$x-4$$