Solve 6/(x+1)(x-1)-3/x-1=1

$$\frac { 6 } { ( x + 1 ) ( x - 1 ) } - \frac { 3 } { x - 1 } = 1$$

Solve for x

$$x=-4$$

Steps Using the Quadratic Formula

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), the least common multiple of \left(x+1\right)\left(x-1\right),x-1.

$$6-\left(x+1\right)\times 3=\left(x-1\right)\left(x+1\right)$$

Use the distributive property to multiply x+1 by 3.

$$6-\left(3x+3\right)=\left(x-1\right)\left(x+1\right)$$

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

$$6-3x-3=\left(x-1\right)\left(x+1\right)$$

Subtract 3 from 6 to get 3.

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

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.

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

Subtract x^{2} from both sides.

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

Add 1 to both sides.

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

Add 3 and 1 to get 4.

$$4-3x-x^{2}=0$$

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

$$-x^{2}-3x+4=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute -1 for a, -3 for b, and 4 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-\left(-3\right)±\sqrt{\left(-3\right)^{2}-4\left(-1\right)\times 4}}{2\left(-1\right)}$$

Square -3.

$$x=\frac{-\left(-3\right)±\sqrt{9-4\left(-1\right)\times 4}}{2\left(-1\right)}$$

Multiply -4 times -1.

$$x=\frac{-\left(-3\right)±\sqrt{9+4\times 4}}{2\left(-1\right)}$$

Multiply 4 times 4.

$$x=\frac{-\left(-3\right)±\sqrt{9+16}}{2\left(-1\right)}$$

Add 9 to 16.

$$x=\frac{-\left(-3\right)±\sqrt{25}}{2\left(-1\right)}$$

Take the square root of 25.

$$x=\frac{-\left(-3\right)±5}{2\left(-1\right)}$$

The opposite of -3 is 3.

$$x=\frac{3±5}{2\left(-1\right)}$$

Multiply 2 times -1.

$$x=\frac{3±5}{-2}$$

Now solve the equation x=\frac{3±5}{-2} when ± is plus. Add 3 to 5.

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

Divide 8 by -2.

$$x=-4$$

Now solve the equation x=\frac{3±5}{-2} when ± is minus. Subtract 5 from 3.

$$x=-\frac{2}{-2}$$

Divide -2 by -2.

$$x=1$$

The equation is now solved.

$$x=-4 x=1$$

Variable x cannot be equal to 1.

$$x=-4$$

Steps for Completing the Square

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), the least common multiple of \left(x+1\right)\left(x-1\right),x-1.

$$6-\left(x+1\right)\times 3=\left(x-1\right)\left(x+1\right)$$

Use the distributive property to multiply x+1 by 3.

$$6-\left(3x+3\right)=\left(x-1\right)\left(x+1\right)$$

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

$$6-3x-3=\left(x-1\right)\left(x+1\right)$$

Subtract 3 from 6 to get 3.

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

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.

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

Subtract x^{2} from both sides.

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

Subtract 3 from both sides.

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

Subtract 3 from -1 to get -4.

$$-3x-x^{2}=-4$$

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.

$$-x^{2}-3x=-4$$

Divide both sides by -1.

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

Dividing by -1 undoes the multiplication by -1.

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

Divide -3 by -1.

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

Divide -4 by -1.

$$x^{2}+3x=4$$

Divide 3, the coefficient of the x term, by 2 to get \frac{3}{2}. Then add the square of \frac{3}{2} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}+3x+\left(\frac{3}{2}\right)^{2}=4+\left(\frac{3}{2}\right)^{2}$$

Square \frac{3}{2} by squaring both the numerator and the denominator of the fraction.

$$x^{2}+3x+\frac{9}{4}=4+\frac{9}{4}$$

Add 4 to \frac{9}{4}.

$$x^{2}+3x+\frac{9}{4}=\frac{25}{4}$$

Factor x^{2}+3x+\frac{9}{4}. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(x+\frac{3}{2}\right)^{2}=\frac{25}{4}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x+\frac{3}{2}\right)^{2}}=\sqrt{\frac{25}{4}}$$

Simplify.

$$x+\frac{3}{2}=\frac{5}{2} x+\frac{3}{2}=-\frac{5}{2}$$

Subtract \frac{3}{2} from both sides of the equation.

$$x=1 x=-4$$

Variable x cannot be equal to 1.

$$x=-4$$