Solve (x-1)(x+1)/x^2-1=-x+3

$$\frac { ( x - 1 ) ( x + 1 ) } { x ^ { 2 } - 1 } = - x + 3$$

Solve for x

$$x=2$$

Solution Steps

Variable x 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(x-1\right)\left(x+1\right).

$$\left(x-1\right)\left(x+1\right)=-\left(x-1\right)\left(x+1\right)x+\left(x-1\right)\left(x+1\right)\times 3$$

Consider \left(x-1\right)\left(x+1\right). Multiplication can be transformed into difference of squares using the rule: \left(a-b\right)\left(a+b\right)=a^{2}-b^{2}. Square 1.

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

Use the distributive property to multiply x-1 by x+1 and combine like terms.

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

Use the distributive property to multiply x^{2}-1 by x.

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

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

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

Use the distributive property to multiply x-1 by x+1 and combine like terms.

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

Use the distributive property to multiply x^{2}-1 by 3.

$$x^{2}-1=-x^{3}+x+3x^{2}-3$$

Add x^{3} to both sides.

$$x^{2}-1+x^{3}=x+3x^{2}-3$$

Subtract x from both sides.

$$x^{2}-1+x^{3}-x=3x^{2}-3$$

Subtract 3x^{2} from both sides.

$$x^{2}-1+x^{3}-x-3x^{2}=-3$$

Combine x^{2} and -3x^{2} to get -2x^{2}.

$$-2x^{2}-1+x^{3}-x=-3$$

Add 3 to both sides.

$$-2x^{2}-1+x^{3}-x+3=0$$

Add -1 and 3 to get 2.

$$-2x^{2}+2+x^{3}-x=0$$

Rearrange the equation to put it in standard form. Place the terms in order from highest to lowest power.

$$x^{3}-2x^{2}-x+2=0$$

By Rational Root Theorem, all rational roots of a polynomial are in the form \frac{p}{q}, where p divides the constant term 2 and q divides the leading coefficient 1. List all candidates \frac{p}{q}.

$$±2,±1$$

Find one such root by trying out all the integer values, starting from the smallest by absolute value. If no integer roots are found, try out fractions.

$$x=1$$

By Factor theorem, x-k is a factor of the polynomial for each root k. Divide x^{3}-2x^{2}-x+2 by x-1 to get x^{2}-x-2. Solve the equation where the result equals to 0.

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

All equations of the form ax^{2}+bx+c=0 can be solved using the quadratic formula: \frac{-b±\sqrt{b^{2}-4ac}}{2a}. Substitute 1 for a, -1 for b, and -2 for c in the quadratic formula.

$$x=\frac{-\left(-1\right)±\sqrt{\left(-1\right)^{2}-4\times 1\left(-2\right)}}{2}$$

Do the calculations.

$$x=\frac{1±3}{2}$$

Solve the equation x^{2}-x-2=0 when ± is plus and when ± is minus.

$$x=-1 x=2$$

Remove the values that the variable cannot be equal to.

$$x=2$$

List all found solutions.

$$x=1 x=-1 x=2$$

Variable x cannot be equal to any of the values 1,-1.

$$x=2$$