Solve y/sqrt{9+y^2}-2/3

$$\frac { y } { \sqrt { 9 } + y ^ { 2 } } - \frac { 2 } { 3 }$$

Evaluate

$$\frac{-2y^{2}+3y-6}{3\left(y^{2}+3\right)}$$

Short Solution Steps

Calculate the square root of 9 and get 3.

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

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

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

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

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

Do the multiplications in 3y-2\left(y^{2}+3\right).

$$\frac{3y-2y^{2}-6}{3\left(y^{2}+3\right)}$$

Expand 3\left(y^{2}+3\right).

$$\frac{3y-2y^{2}-6}{3y^{2}+9}$$

Factor

$$\frac{-2y^{2}+3y-6}{3\left(y^{2}+3\right)}$$