Solve (2)-a^2-25/a^2+10a+25diva^2-1/a+5*a^2+2a+1/a-5

$$( 2 ) - \frac { a ^ { 2 } - 25 } { a ^ { 2 } + 10 a + 25 } \div \frac { a ^ { 2 } - 1 } { a + 5 } \cdot \frac { a ^ { 2 } + 2 a + 1 } { a - 5 }$$

Evaluate

$$\frac{a-3}{a-1}$$

Short Solution Steps

Divide \frac{a^{2}-25}{a^{2}+10a+25} by \frac{a^{2}-1}{a+5} by multiplying \frac{a^{2}-25}{a^{2}+10a+25} by the reciprocal of \frac{a^{2}-1}{a+5}.

$$2-\frac{\left(a^{2}-25\right)\left(a+5\right)}{\left(a^{2}+10a+25\right)\left(a^{2}-1\right)}\times \frac{a^{2}+2a+1}{a-5}$$

Factor the expressions that are not already factored in \frac{\left(a^{2}-25\right)\left(a+5\right)}{\left(a^{2}+10a+25\right)\left(a^{2}-1\right)}.

$$2-\frac{\left(a-5\right)\left(a+5\right)^{2}}{\left(a-1\right)\left(a+1\right)\left(a+5\right)^{2}}\times \frac{a^{2}+2a+1}{a-5}$$

Cancel out \left(a+5\right)^{2} in both numerator and denominator.

$$2-\frac{a-5}{\left(a-1\right)\left(a+1\right)}\times \frac{a^{2}+2a+1}{a-5}$$

Multiply \frac{a-5}{\left(a-1\right)\left(a+1\right)} times \frac{a^{2}+2a+1}{a-5} by multiplying numerator times numerator and denominator times denominator.

$$2-\frac{\left(a-5\right)\left(a^{2}+2a+1\right)}{\left(a-1\right)\left(a+1\right)\left(a-5\right)}$$

Cancel out a-5 in both numerator and denominator.

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

Factor the expressions that are not already factored in \frac{a^{2}+2a+1}{\left(a-1\right)\left(a+1\right)}.

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

Cancel out a+1 in both numerator and denominator.

$$2-\frac{a+1}{a-1}$$

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

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

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

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

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

$$\frac{2a-2-a-1}{a-1}$$

Combine like terms in 2a-2-a-1.

$$\frac{a-3}{a-1}$$

Expand

$$\frac{a-3}{a-1}$$

Short Solution Steps

Divide \frac{a^{2}-25}{a^{2}+10a+25} by \frac{a^{2}-1}{a+5} by multiplying \frac{a^{2}-25}{a^{2}+10a+25} by the reciprocal of \frac{a^{2}-1}{a+5}.

$$2-\frac{\left(a^{2}-25\right)\left(a+5\right)}{\left(a^{2}+10a+25\right)\left(a^{2}-1\right)}\times \frac{a^{2}+2a+1}{a-5}$$

Factor the expressions that are not already factored in \frac{\left(a^{2}-25\right)\left(a+5\right)}{\left(a^{2}+10a+25\right)\left(a^{2}-1\right)}.

$$2-\frac{\left(a-5\right)\left(a+5\right)^{2}}{\left(a-1\right)\left(a+1\right)\left(a+5\right)^{2}}\times \frac{a^{2}+2a+1}{a-5}$$

Cancel out \left(a+5\right)^{2} in both numerator and denominator.

$$2-\frac{a-5}{\left(a-1\right)\left(a+1\right)}\times \frac{a^{2}+2a+1}{a-5}$$

Multiply \frac{a-5}{\left(a-1\right)\left(a+1\right)} times \frac{a^{2}+2a+1}{a-5} by multiplying numerator times numerator and denominator times denominator.

$$2-\frac{\left(a-5\right)\left(a^{2}+2a+1\right)}{\left(a-1\right)\left(a+1\right)\left(a-5\right)}$$

Cancel out a-5 in both numerator and denominator.

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

Factor the expressions that are not already factored in \frac{a^{2}+2a+1}{\left(a-1\right)\left(a+1\right)}.

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

Cancel out a+1 in both numerator and denominator.

$$2-\frac{a+1}{a-1}$$

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

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

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

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

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

$$\frac{2a-2-a-1}{a-1}$$

Combine like terms in 2a-2-a-1.

$$\frac{a-3}{a-1}$$