Solve =(a-1-4a-1/a+1)diva^2-8a+16/a+1

$$= ( a - 1 - \frac { 4 a - 1 } { a + 1 } ) \div \frac { a ^ { 2 } - 8 a + 16 } { a + 1 }$$

Evaluate

$$\frac{a}{a-4}$$

Short Solution Steps

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

$$\frac{\frac{\left(a-1\right)\left(a+1\right)}{a+1}-\frac{4a-1}{a+1}}{\frac{a^{2}-8a+16}{a+1}}$$

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

$$\frac{\frac{\left(a-1\right)\left(a+1\right)-\left(4a-1\right)}{a+1}}{\frac{a^{2}-8a+16}{a+1}}$$

Do the multiplications in \left(a-1\right)\left(a+1\right)-\left(4a-1\right).

$$\frac{\frac{a^{2}+a-a-1-4a+1}{a+1}}{\frac{a^{2}-8a+16}{a+1}}$$

Combine like terms in a^{2}+a-a-1-4a+1.

$$\frac{\frac{a^{2}-4a}{a+1}}{\frac{a^{2}-8a+16}{a+1}}$$

Divide \frac{a^{2}-4a}{a+1} by \frac{a^{2}-8a+16}{a+1} by multiplying \frac{a^{2}-4a}{a+1} by the reciprocal of \frac{a^{2}-8a+16}{a+1}.

$$\frac{\left(a^{2}-4a\right)\left(a+1\right)}{\left(a+1\right)\left(a^{2}-8a+16\right)}$$

Cancel out a+1 in both numerator and denominator.

$$\frac{a^{2}-4a}{a^{2}-8a+16}$$

Factor the expressions that are not already factored.

$$\frac{a\left(a-4\right)}{\left(a-4\right)^{2}}$$

Cancel out a-4 in both numerator and denominator.

$$\frac{a}{a-4}$$

Expand

$$\frac{a}{a-4}$$

Short Solution Steps

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

$$\frac{\frac{\left(a-1\right)\left(a+1\right)}{a+1}-\frac{4a-1}{a+1}}{\frac{a^{2}-8a+16}{a+1}}$$

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

$$\frac{\frac{\left(a-1\right)\left(a+1\right)-\left(4a-1\right)}{a+1}}{\frac{a^{2}-8a+16}{a+1}}$$

Do the multiplications in \left(a-1\right)\left(a+1\right)-\left(4a-1\right).

$$\frac{\frac{a^{2}+a-a-1-4a+1}{a+1}}{\frac{a^{2}-8a+16}{a+1}}$$

Combine like terms in a^{2}+a-a-1-4a+1.

$$\frac{\frac{a^{2}-4a}{a+1}}{\frac{a^{2}-8a+16}{a+1}}$$

Divide \frac{a^{2}-4a}{a+1} by \frac{a^{2}-8a+16}{a+1} by multiplying \frac{a^{2}-4a}{a+1} by the reciprocal of \frac{a^{2}-8a+16}{a+1}.

$$\frac{\left(a^{2}-4a\right)\left(a+1\right)}{\left(a+1\right)\left(a^{2}-8a+16\right)}$$

Cancel out a+1 in both numerator and denominator.

$$\frac{a^{2}-4a}{a^{2}-8a+16}$$

Factor the expressions that are not already factored.

$$\frac{a\left(a-4\right)}{\left(a-4\right)^{2}}$$

Cancel out a-4 in both numerator and denominator.

$$\frac{a}{a-4}$$