Solve t^2-3/1-t^2+t+1/t-1=4/1+t

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

Solve for t

$$t=4$$

Steps for Solving Linear Equation

Variable t cannot be equal to any of the values -1,1 since division by zero is not defined. Multiply both sides of the equation by \left(t-1\right)\left(t+1\right), the least common multiple of 1-t^{2},t-1,1+t.

$$-\left(t^{2}-3\right)+\left(t+1\right)\left(t+1\right)=\left(t-1\right)\times 4$$

Multiply t+1 and t+1 to get \left(t+1\right)^{2}.

$$-\left(t^{2}-3\right)+\left(t+1\right)^{2}=\left(t-1\right)\times 4$$

To find the opposite of t^{2}-3, find the opposite of each term.

$$-t^{2}+3+\left(t+1\right)^{2}=\left(t-1\right)\times 4$$

Use binomial theorem \left(a+b\right)^{2}=a^{2}+2ab+b^{2} to expand \left(t+1\right)^{2}.

$$-t^{2}+3+t^{2}+2t+1=\left(t-1\right)\times 4$$

Combine -t^{2} and t^{2} to get 0.

$$3+2t+1=\left(t-1\right)\times 4$$

Add 3 and 1 to get 4.

$$4+2t=\left(t-1\right)\times 4$$

Use the distributive property to multiply t-1 by 4.

$$4+2t=4t-4$$

Subtract 4t from both sides.

$$4+2t-4t=-4$$

Combine 2t and -4t to get -2t.

$$4-2t=-4$$

Subtract 4 from both sides.

$$-2t=-4-4$$

Subtract 4 from -4 to get -8.

$$-2t=-8$$

Divide both sides by -2.

$$t=\frac{-8}{-2}$$

Divide -8 by -2 to get 4.

$$t=4$$