Solve 3/x-5-x/x+1=10/(x+1)(x-5)

$$\frac{ 3 }{ x-5 } - \frac{ x }{ x+1 } = \frac{ 10 }{ (x+1)(x-5) }$$

Solve for x

$$x=1$$

$$x=7$$

Steps Using the Quadratic Formula

Variable x cannot be equal to any of the values -1,5 since division by zero is not defined. Multiply both sides of the equation by \left(x-5\right)\left(x+1\right), the least common multiple of x-5,x+1,\left(x+1\right)\left(x-5\right).

$$\left(x+1\right)\times 3-\left(x-5\right)x=10$$

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

$$3x+3-\left(x-5\right)x=10$$

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

$$3x+3-\left(x^{2}-5x\right)=10$$

To find the opposite of x^{2}-5x, find the opposite of each term.

$$3x+3-x^{2}+5x=10$$

Combine 3x and 5x to get 8x.

$$8x+3-x^{2}=10$$

Subtract 10 from both sides.

$$8x+3-x^{2}-10=0$$

Subtract 10 from 3 to get -7.

$$8x-7-x^{2}=0$$

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.

$$-x^{2}+8x-7=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute -1 for a, 8 for b, and -7 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-8±\sqrt{8^{2}-4\left(-1\right)\left(-7\right)}}{2\left(-1\right)}$$

Square 8.

$$x=\frac{-8±\sqrt{64-4\left(-1\right)\left(-7\right)}}{2\left(-1\right)}$$

Multiply -4 times -1.

$$x=\frac{-8±\sqrt{64+4\left(-7\right)}}{2\left(-1\right)}$$

Multiply 4 times -7.

$$x=\frac{-8±\sqrt{64-28}}{2\left(-1\right)}$$

Add 64 to -28.

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

Take the square root of 36.

$$x=\frac{-8±6}{2\left(-1\right)}$$

Multiply 2 times -1.

$$x=\frac{-8±6}{-2}$$

Now solve the equation x=\frac{-8±6}{-2} when ± is plus. Add -8 to 6.

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

Divide -2 by -2.

$$x=1$$

Now solve the equation x=\frac{-8±6}{-2} when ± is minus. Subtract 6 from -8.

$$x=-\frac{14}{-2}$$

Divide -14 by -2.

$$x=7$$

The equation is now solved.

$$x=1 x=7$$

Steps for Completing the Square

Variable x cannot be equal to any of the values -1,5 since division by zero is not defined. Multiply both sides of the equation by \left(x-5\right)\left(x+1\right), the least common multiple of x-5,x+1,\left(x+1\right)\left(x-5\right).

$$\left(x+1\right)\times 3-\left(x-5\right)x=10$$

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

$$3x+3-\left(x-5\right)x=10$$

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

$$3x+3-\left(x^{2}-5x\right)=10$$

To find the opposite of x^{2}-5x, find the opposite of each term.

$$3x+3-x^{2}+5x=10$$

Combine 3x and 5x to get 8x.

$$8x+3-x^{2}=10$$

Subtract 3 from both sides.

$$8x-x^{2}=10-3$$

Subtract 3 from 10 to get 7.

$$8x-x^{2}=7$$

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.

$$-x^{2}+8x=7$$

Divide both sides by -1.

$$\frac{-x^{2}+8x}{-1}=\frac{7}{-1}$$

Dividing by -1 undoes the multiplication by -1.

$$x^{2}+\frac{8}{-1}x=\frac{7}{-1}$$

Divide 8 by -1.

$$x^{2}-8x=\frac{7}{-1}$$

Divide 7 by -1.

$$x^{2}-8x=-7$$

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.

$$x^{2}-8x+\left(-4\right)^{2}=-7+\left(-4\right)^{2}$$

Square -4.

$$x^{2}-8x+16=-7+16$$

Add -7 to 16.

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

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}.

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

Take the square root of both sides of the equation.

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

Simplify.

$$x-4=3 x-4=-3$$

Add 4 to both sides of the equation.

$$x=7 x=1$$