Solve 1/3-t+4/3+t+13/9-t^2=0

$$\frac { 1 } { 3 - t } + \frac { 4 } { 3 + t } + \frac { 13 } { 9 - t ^ { 2 } } = 0$$

Solve for t

$$t = \frac{28}{3} = 9\frac{1}{3} \approx 9.333333333$$

Steps for Solving Linear Equation

Variable t 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(t-3\right)\left(t+3\right), the least common multiple of 3-t,3+t,9-t^{2}.

$$-\left(3+t\right)+\left(t-3\right)\times 4-13=0$$

To find the opposite of 3+t, find the opposite of each term.

$$-3-t+\left(t-3\right)\times 4-13=0$$

Use the distributive property to multiply t-3 by 4.

$$-3-t+4t-12-13=0$$

Combine -t and 4t to get 3t.

$$-3+3t-12-13=0$$

Subtract 12 from -3 to get -15.

$$-15+3t-13=0$$

Subtract 13 from -15 to get -28.

$$-28+3t=0$$

Add 28 to both sides. Anything plus zero gives itself.

$$3t=28$$

Divide both sides by 3.

$$t=\frac{28}{3}$$