Solve 6/a^2-1+4/a+1*3/a-1=

$$\frac { 6 } { a ^ { 2 } - 1 } + \frac { 4 } { a + 1 } \cdot \frac { 3 } { a - 1 } =$$

Evaluate

$$\frac{18}{a^{2}-1}$$

Short Solution Steps

Multiply \frac{4}{a+1} times \frac{3}{a-1} by multiplying numerator times numerator and denominator times denominator.

$$\frac{6}{a^{2}-1}+\frac{4\times 3}{\left(a+1\right)\left(a-1\right)}$$

Factor a^{2}-1.

$$\frac{6}{\left(a-1\right)\left(a+1\right)}+\frac{4\times 3}{\left(a+1\right)\left(a-1\right)}$$

Since \frac{6}{\left(a-1\right)\left(a+1\right)} and \frac{4\times 3}{\left(a+1\right)\left(a-1\right)} have the same denominator, add them by adding their numerators.

$$\frac{6+4\times 3}{\left(a-1\right)\left(a+1\right)}$$

Do the multiplications in 6+4\times 3.

$$\frac{6+12}{\left(a-1\right)\left(a+1\right)}$$

Do the calculations in 6+12.

$$\frac{18}{\left(a-1\right)\left(a+1\right)}$$

Expand \left(a-1\right)\left(a+1\right).

$$\frac{18}{a^{2}-1}$$

Factor

$$\frac{18}{a^{2}-1}$$