Solve -y^2/49y^2-81+3y/14y+18=

$$\frac { - y ^ { 2 } } { 49 y ^ { 2 } - 81 } + \frac { 3 y } { 14 y + 18 } =$$

Evaluate

$$\frac{y\left(19y-27\right)}{2\left(49y^{2}-81\right)}$$

Short Solution Steps

Factor 49y^{2}-81. Factor 14y+18.

$$\frac{-y^{2}}{\left(7y-9\right)\left(7y+9\right)}+\frac{3y}{2\left(7y+9\right)}$$

To add or subtract expressions, expand them to make their denominators the same. Least common multiple of \left(7y-9\right)\left(7y+9\right) and 2\left(7y+9\right) is 2\left(7y-9\right)\left(7y+9\right). Multiply \frac{-y^{2}}{\left(7y-9\right)\left(7y+9\right)} times \frac{2}{2}. Multiply \frac{3y}{2\left(7y+9\right)} times \frac{7y-9}{7y-9}.

$$\frac{2\left(-y^{2}\right)}{2\left(7y-9\right)\left(7y+9\right)}+\frac{3y\left(7y-9\right)}{2\left(7y-9\right)\left(7y+9\right)}$$

Since \frac{2\left(-y^{2}\right)}{2\left(7y-9\right)\left(7y+9\right)} and \frac{3y\left(7y-9\right)}{2\left(7y-9\right)\left(7y+9\right)} have the same denominator, add them by adding their numerators.

$$\frac{2\left(-y^{2}\right)+3y\left(7y-9\right)}{2\left(7y-9\right)\left(7y+9\right)}$$

Do the multiplications in 2\left(-y^{2}\right)+3y\left(7y-9\right).

$$\frac{-2y^{2}+21y^{2}-27y}{2\left(7y-9\right)\left(7y+9\right)}$$

Combine like terms in -2y^{2}+21y^{2}-27y.

$$\frac{19y^{2}-27y}{2\left(7y-9\right)\left(7y+9\right)}$$

Expand 2\left(7y-9\right)\left(7y+9\right).

$$\frac{19y^{2}-27y}{98y^{2}-162}$$

Factor

$$\frac{y\left(19y-27\right)}{2\left(49y^{2}-81\right)}$$