Solve (1-5/a^2-4)*a+2/a^2-3a

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

Evaluate

$$\frac{a+3}{a\left(a-2\right)}$$

Short Solution Steps

Factor a^{2}-4.

$$\left(1-\frac{5}{\left(a-2\right)\left(a+2\right)}\right)\times \frac{a+2}{a^{2}-3a}$$

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

$$\left(\frac{\left(a-2\right)\left(a+2\right)}{\left(a-2\right)\left(a+2\right)}-\frac{5}{\left(a-2\right)\left(a+2\right)}\right)\times \frac{a+2}{a^{2}-3a}$$

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

$$\frac{\left(a-2\right)\left(a+2\right)-5}{\left(a-2\right)\left(a+2\right)}\times \frac{a+2}{a^{2}-3a}$$

Do the multiplications in \left(a-2\right)\left(a+2\right)-5.

$$\frac{a^{2}+2a-2a-4-5}{\left(a-2\right)\left(a+2\right)}\times \frac{a+2}{a^{2}-3a}$$

Combine like terms in a^{2}+2a-2a-4-5.

$$\frac{a^{2}-9}{\left(a-2\right)\left(a+2\right)}\times \frac{a+2}{a^{2}-3a}$$

Multiply \frac{a^{2}-9}{\left(a-2\right)\left(a+2\right)} times \frac{a+2}{a^{2}-3a} by multiplying numerator times numerator and denominator times denominator.

$$\frac{\left(a^{2}-9\right)\left(a+2\right)}{\left(a-2\right)\left(a+2\right)\left(a^{2}-3a\right)}$$

Cancel out a+2 in both numerator and denominator.

$$\frac{a^{2}-9}{\left(a-2\right)\left(a^{2}-3a\right)}$$

Factor the expressions that are not already factored.

$$\frac{\left(a-3\right)\left(a+3\right)}{a\left(a-3\right)\left(a-2\right)}$$

Cancel out a-3 in both numerator and denominator.

$$\frac{a+3}{a\left(a-2\right)}$$

Expand the expression.

$$\frac{a+3}{a^{2}-2a}$$

Expand

$$\frac{a+3}{a\left(a-2\right)}$$

Short Solution Steps

Factor a^{2}-4.

$$\left(1-\frac{5}{\left(a-2\right)\left(a+2\right)}\right)\times \frac{a+2}{a^{2}-3a}$$

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

$$\left(\frac{\left(a-2\right)\left(a+2\right)}{\left(a-2\right)\left(a+2\right)}-\frac{5}{\left(a-2\right)\left(a+2\right)}\right)\times \frac{a+2}{a^{2}-3a}$$

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

$$\frac{\left(a-2\right)\left(a+2\right)-5}{\left(a-2\right)\left(a+2\right)}\times \frac{a+2}{a^{2}-3a}$$

Do the multiplications in \left(a-2\right)\left(a+2\right)-5.

$$\frac{a^{2}+2a-2a-4-5}{\left(a-2\right)\left(a+2\right)}\times \frac{a+2}{a^{2}-3a}$$

Combine like terms in a^{2}+2a-2a-4-5.

$$\frac{a^{2}-9}{\left(a-2\right)\left(a+2\right)}\times \frac{a+2}{a^{2}-3a}$$

Multiply \frac{a^{2}-9}{\left(a-2\right)\left(a+2\right)} times \frac{a+2}{a^{2}-3a} by multiplying numerator times numerator and denominator times denominator.

$$\frac{\left(a^{2}-9\right)\left(a+2\right)}{\left(a-2\right)\left(a+2\right)\left(a^{2}-3a\right)}$$

Cancel out a+2 in both numerator and denominator.

$$\frac{a^{2}-9}{\left(a-2\right)\left(a^{2}-3a\right)}$$

Factor the expressions that are not already factored.

$$\frac{\left(a-3\right)\left(a+3\right)}{a\left(a-3\right)\left(a-2\right)}$$

Cancel out a-3 in both numerator and denominator.

$$\frac{a+3}{a\left(a-2\right)}$$

Expand the expression.

$$\frac{a+3}{a^{2}-2a}$$