Solve 1/u-3+6/u^2-9=3/u+3

$$\frac { 1 } { u - 3 } + \frac { 6 } { u ^ { 2 } - 9 } = \frac { 3 } { u + 3 }$$

Solve for u

$$u=9$$

Steps for Solving Linear Equation

Variable u cannot be equal to any of the values -3,3 since division by zero is not defined. Multiply both sides of the equation by \left(u-3\right)\left(u+3\right), the least common multiple of u-3,u^{2}-9,u+3.

$$u+3+6=\left(u-3\right)\times 3$$

Add 3 and 6 to get 9.

$$u+9=\left(u-3\right)\times 3$$

Use the distributive property to multiply u-3 by 3.

$$u+9=3u-9$$

Subtract 3u from both sides.

$$u+9-3u=-9$$

Combine u and -3u to get -2u.

$$-2u+9=-9$$

Subtract 9 from both sides.

$$-2u=-9-9$$

Subtract 9 from -9 to get -18.

$$-2u=-18$$

Divide both sides by -2.

$$u=\frac{-18}{-2}$$

Divide -18 by -2 to get 9.

$$u=9$$