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

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

Solve for t

$$t=2$$

$$t=0$$

Steps Using Factoring

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 t^{2}-1,t+1.

$$t=\left(t-1\right)t$$

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

$$t=t^{2}-t$$

Subtract t^{2} from both sides.

$$t-t^{2}=-t$$

Add t to both sides.

$$t-t^{2}+t=0$$

Combine t and t to get 2t.

$$2t-t^{2}=0$$

Factor out t.

$$t\left(2-t\right)=0$$

To find equation solutions, solve t=0 and 2-t=0.

$$t=0 t=2$$

Steps Using the Quadratic Formula

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 t^{2}-1,t+1.

$$t=\left(t-1\right)t$$

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

$$t=t^{2}-t$$

Subtract t^{2} from both sides.

$$t-t^{2}=-t$$

Add t to both sides.

$$t-t^{2}+t=0$$

Combine t and t to get 2t.

$$2t-t^{2}=0$$

All equations of the form ax^{2}+bx+c=0 can be solved using the quadratic formula: \frac{-b±\sqrt{b^{2}-4ac}}{2a}. The quadratic formula gives two solutions, one when ± is addition and one when it is subtraction.

$$-t^{2}+2t=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute -1 for a, 2 for b, and 0 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$t=\frac{-2±\sqrt{2^{2}}}{2\left(-1\right)}$$

Take the square root of 2^{2}.

$$t=\frac{-2±2}{2\left(-1\right)}$$

Multiply 2 times -1.

$$t=\frac{-2±2}{-2}$$

Now solve the equation t=\frac{-2±2}{-2} when ± is plus. Add -2 to 2.

$$t=\frac{0}{-2}$$

Divide 0 by -2.

$$t=0$$

Now solve the equation t=\frac{-2±2}{-2} when ± is minus. Subtract 2 from -2.

$$t=-\frac{4}{-2}$$

Divide -4 by -2.

$$t=2$$

The equation is now solved.

$$t=0 t=2$$

Steps for Completing the Square

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 t^{2}-1,t+1.

$$t=\left(t-1\right)t$$

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

$$t=t^{2}-t$$

Subtract t^{2} from both sides.

$$t-t^{2}=-t$$

Add t to both sides.

$$t-t^{2}+t=0$$

Combine t and t to get 2t.

$$2t-t^{2}=0$$

Quadratic equations such as this one can be solved by completing the square. In order to complete the square, the equation must first be in the form x^{2}+bx=c.

$$-t^{2}+2t=0$$

Divide both sides by -1.

$$\frac{-t^{2}+2t}{-1}=\frac{0}{-1}$$

Dividing by -1 undoes the multiplication by -1.

$$t^{2}+\frac{2}{-1}t=\frac{0}{-1}$$

Divide 2 by -1.

$$t^{2}-2t=\frac{0}{-1}$$

Divide 0 by -1.

$$t^{2}-2t=0$$

Divide -2, the coefficient of the x term, by 2 to get -1. Then add the square of -1 to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$t^{2}-2t+1=1$$

Factor t^{2}-2t+1. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(t-1\right)^{2}=1$$

Take the square root of both sides of the equation.

$$\sqrt{\left(t-1\right)^{2}}=\sqrt{1}$$

Simplify.

$$t-1=1 t-1=-1$$

Add 1 to both sides of the equation.

$$t=2 t=0$$