Solve x+4/x+1+5/1-x=6x/x^2-1

$$\frac { x + 4 } { x + 1 } + \frac { 5 } { 1 - x } = \frac { 6 x } { x ^ { 2 } - 1 }$$

Solve for x

$$x=9$$

Steps Using Factoring

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

$$\left(x-1\right)\left(x+4\right)-\left(1+x\right)\times 5=6x$$

Use the distributive property to multiply x-1 by x+4 and combine like terms.

$$x^{2}+3x-4-\left(1+x\right)\times 5=6x$$

Multiply -1 and 5 to get -5.

$$x^{2}+3x-4-5\left(1+x\right)=6x$$

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

$$x^{2}+3x-4-5-5x=6x$$

Subtract 5 from -4 to get -9.

$$x^{2}+3x-9-5x=6x$$

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

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

Subtract 6x from both sides.

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

Combine -2x and -6x to get -8x.

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

To solve the equation, factor x^{2}-8x-9 using formula x^{2}+\left(a+b\right)x+ab=\left(x+a\right)\left(x+b\right). To find a and b, set up a system to be solved.

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

Since ab is negative, a and b have the opposite signs. Since a+b is negative, the negative number has greater absolute value than the positive. 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 factored expression \left(x+a\right)\left(x+b\right) using the obtained values.

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

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

$$\left(x-1\right)\left(x+4\right)-\left(1+x\right)\times 5=6x$$

Use the distributive property to multiply x-1 by x+4 and combine like terms.

$$x^{2}+3x-4-\left(1+x\right)\times 5=6x$$

Multiply -1 and 5 to get -5.

$$x^{2}+3x-4-5\left(1+x\right)=6x$$

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

$$x^{2}+3x-4-5-5x=6x$$

Subtract 5 from -4 to get -9.

$$x^{2}+3x-9-5x=6x$$

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

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

Subtract 6x from both sides.

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

Combine -2x and -6x to get -8x.

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

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.

$$a+b=-8 ab=1\left(-9\right)=-9$$

Since ab is negative, a and b have the opposite signs. Since a+b is negative, the negative number has greater absolute value than the positive. 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 x^{2}-9x.

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

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

$$\left(x-1\right)\left(x+4\right)-\left(1+x\right)\times 5=6x$$

Use the distributive property to multiply x-1 by x+4 and combine like terms.

$$x^{2}+3x-4-\left(1+x\right)\times 5=6x$$

Multiply -1 and 5 to get -5.

$$x^{2}+3x-4-5\left(1+x\right)=6x$$

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

$$x^{2}+3x-4-5-5x=6x$$

Subtract 5 from -4 to get -9.

$$x^{2}+3x-9-5x=6x$$

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

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

Subtract 6x from both sides.

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

Combine -2x and -6x to get -8x.

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

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}.

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

Square -8.

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

Multiply -4 times -9.

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

Add 64 to 36.

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

Take the square root of 100.

$$x=\frac{-\left(-8\right)±10}{2}$$

The opposite of -8 is 8.

$$x=\frac{8±10}{2}$$

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

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

Divide 18 by 2.

$$x=9$$

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

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

Divide -2 by 2.

$$x=-1$$

The equation is now solved.

$$x=9 x=-1$$

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

$$\left(x-1\right)\left(x+4\right)-\left(1+x\right)\times 5=6x$$

Use the distributive property to multiply x-1 by x+4 and combine like terms.

$$x^{2}+3x-4-\left(1+x\right)\times 5=6x$$

Multiply -1 and 5 to get -5.

$$x^{2}+3x-4-5\left(1+x\right)=6x$$

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

$$x^{2}+3x-4-5-5x=6x$$

Subtract 5 from -4 to get -9.

$$x^{2}+3x-9-5x=6x$$

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

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

Subtract 6x from both sides.

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

Combine -2x and -6x to get -8x.

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

Add 9 to both sides. Anything plus zero gives itself.

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

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.

$$x=9$$