Solve 6x/x^2-1=1+5/1-x+3/x+1

$$\frac { 6 x } { x ^ { 2 } - 1 } = 1 + \frac { 5 } { 1 - x } + \frac { 3 } { x + 1 }$$

Solve for x

$$x=9$$

Steps Using Factoring By Grouping

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

$$6x=\left(x-1\right)\left(x+1\right)-\left(1+x\right)\times 5+\left(x-1\right)\times 3$$

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.

$$6x=x^{2}-1-\left(1+x\right)\times 5+\left(x-1\right)\times 3$$

Multiply -1 and 5 to get -5.

$$6x=x^{2}-1-5\left(1+x\right)+\left(x-1\right)\times 3$$

Use the distributive property to multiply -5 by 1+x.

$$6x=x^{2}-1-5-5x+\left(x-1\right)\times 3$$

Subtract 5 from -1 to get -6.

$$6x=x^{2}-6-5x+\left(x-1\right)\times 3$$

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

$$6x=x^{2}-6-5x+3x-3$$

Combine -5x and 3x to get -2x.

$$6x=x^{2}-6-2x-3$$

Subtract 3 from -6 to get -9.

$$6x=x^{2}-9-2x$$

Subtract x^{2} from both sides.

$$6x-x^{2}=-9-2x$$

Subtract -9 from both sides.

$$6x-x^{2}-\left(-9\right)=-2x$$

The opposite of -9 is 9.

$$6x-x^{2}+9=-2x$$

Add 2x to both sides.

$$6x-x^{2}+9+2x=0$$

Combine 6x and 2x to get 8x.

$$8x-x^{2}+9=0$$

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

$$-x^{2}+8x+9=0$$

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as -x^{2}+ax+bx+9. To find a and b, set up a system to be solved.

$$a+b=8 ab=-9=-9$$

Since ab is negative, a and b have the opposite signs. Since a+b is positive, the positive number has greater absolute value than the negative. List all such integer pairs that give product -9.

$$-1,9 -3,3$$

Calculate the sum for each pair.

$$-1+9=8 -3+3=0$$

The solution is the pair that gives sum 8.

$$a=9 b=-1$$

Rewrite -x^{2}+8x+9 as \left(-x^{2}+9x\right)+\left(-x+9\right).

$$\left(-x^{2}+9x\right)+\left(-x+9\right)$$

Factor out -x in the first and -1 in the second group.

$$-x\left(x-9\right)-\left(x-9\right)$$

Factor out common term x-9 by using distributive property.

$$\left(x-9\right)\left(-x-1\right)$$

To find equation solutions, solve x-9=0 and -x-1=0.

$$x=9 x=-1$$

Variable x cannot be equal to -1.

$$x=9$$

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

$$6x=\left(x-1\right)\left(x+1\right)-\left(1+x\right)\times 5+\left(x-1\right)\times 3$$

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.

$$6x=x^{2}-1-\left(1+x\right)\times 5+\left(x-1\right)\times 3$$

Multiply -1 and 5 to get -5.

$$6x=x^{2}-1-5\left(1+x\right)+\left(x-1\right)\times 3$$

Use the distributive property to multiply -5 by 1+x.

$$6x=x^{2}-1-5-5x+\left(x-1\right)\times 3$$

Subtract 5 from -1 to get -6.

$$6x=x^{2}-6-5x+\left(x-1\right)\times 3$$

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

$$6x=x^{2}-6-5x+3x-3$$

Combine -5x and 3x to get -2x.

$$6x=x^{2}-6-2x-3$$

Subtract 3 from -6 to get -9.

$$6x=x^{2}-9-2x$$

Subtract x^{2} from both sides.

$$6x-x^{2}=-9-2x$$

Subtract -9 from both sides.

$$6x-x^{2}-\left(-9\right)=-2x$$

The opposite of -9 is 9.

$$6x-x^{2}+9=-2x$$

Add 2x to both sides.

$$6x-x^{2}+9+2x=0$$

Combine 6x and 2x to get 8x.

$$8x-x^{2}+9=0$$

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.

$$-x^{2}+8x+9=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute -1 for a, 8 for b, and 9 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-8±\sqrt{8^{2}-4\left(-1\right)\times 9}}{2\left(-1\right)}$$

Square 8.

$$x=\frac{-8±\sqrt{64-4\left(-1\right)\times 9}}{2\left(-1\right)}$$

Multiply -4 times -1.

$$x=\frac{-8±\sqrt{64+4\times 9}}{2\left(-1\right)}$$

Multiply 4 times 9.

$$x=\frac{-8±\sqrt{64+36}}{2\left(-1\right)}$$

Add 64 to 36.

$$x=\frac{-8±\sqrt{100}}{2\left(-1\right)}$$

Take the square root of 100.

$$x=\frac{-8±10}{2\left(-1\right)}$$

Multiply 2 times -1.

$$x=\frac{-8±10}{-2}$$

Now solve the equation x=\frac{-8±10}{-2} when ± is plus. Add -8 to 10.

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

Divide 2 by -2.

$$x=-1$$

Now solve the equation x=\frac{-8±10}{-2} when ± is minus. Subtract 10 from -8.

$$x=-\frac{18}{-2}$$

Divide -18 by -2.

$$x=9$$

The equation is now solved.

$$x=-1 x=9$$

Variable x cannot be equal to -1.

$$x=9$$

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

$$6x=\left(x-1\right)\left(x+1\right)-\left(1+x\right)\times 5+\left(x-1\right)\times 3$$

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.

$$6x=x^{2}-1-\left(1+x\right)\times 5+\left(x-1\right)\times 3$$

Multiply -1 and 5 to get -5.

$$6x=x^{2}-1-5\left(1+x\right)+\left(x-1\right)\times 3$$

Use the distributive property to multiply -5 by 1+x.

$$6x=x^{2}-1-5-5x+\left(x-1\right)\times 3$$

Subtract 5 from -1 to get -6.

$$6x=x^{2}-6-5x+\left(x-1\right)\times 3$$

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

$$6x=x^{2}-6-5x+3x-3$$

Combine -5x and 3x to get -2x.

$$6x=x^{2}-6-2x-3$$

Subtract 3 from -6 to get -9.

$$6x=x^{2}-9-2x$$

Subtract x^{2} from both sides.

$$6x-x^{2}=-9-2x$$

Add 2x to both sides.

$$6x-x^{2}+2x=-9$$

Combine 6x and 2x to get 8x.

$$8x-x^{2}=-9$$

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.

$$-x^{2}+8x=-9$$

Divide both sides by -1.

$$\frac{-x^{2}+8x}{-1}=-\frac{9}{-1}$$

Dividing by -1 undoes the multiplication by -1.

$$x^{2}+\frac{8}{-1}x=-\frac{9}{-1}$$

Divide 8 by -1.

$$x^{2}-8x=-\frac{9}{-1}$$

Divide -9 by -1.

$$x^{2}-8x=9$$

Divide -8, the coefficient of the x term, by 2 to get -4. Then add the square of -4 to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}-8x+\left(-4\right)^{2}=9+\left(-4\right)^{2}$$

Square -4.

$$x^{2}-8x+16=9+16$$

Add 9 to 16.

$$x^{2}-8x+16=25$$

Factor x^{2}-8x+16. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(x-4\right)^{2}=25$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-4\right)^{2}}=\sqrt{25}$$

Simplify.

$$x-4=5 x-4=-5$$

Add 4 to both sides of the equation.

$$x=9 x=-1$$

Variable x cannot be equal to -1.

$$x=9$$