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

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

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 \left(x-5\right)\left(x+1\right),x+1,x-5.

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

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

$$10+x^{2}-5x=\left(x+1\right)\times 3$$

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

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

Subtract 3x from both sides.

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

Combine -5x and -3x to get -8x.

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

Subtract 3 from both sides.

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

Subtract 3 from 10 to get 7.

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

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}.

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

Square -8.

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

Multiply -4 times 7.

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

Add 64 to -28.

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

Take the square root of 36.

$$x=\frac{-\left(-8\right)±6}{2}$$

The opposite of -8 is 8.

$$x=\frac{8±6}{2}$$

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

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

Divide 14 by 2.

$$x=7$$

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

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

Divide 2 by 2.

$$x=1$$

The equation is now solved.

$$x=7 x=1$$

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 \left(x-5\right)\left(x+1\right),x+1,x-5.

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

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

$$10+x^{2}-5x=\left(x+1\right)\times 3$$

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

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

Subtract 3x from both sides.

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

Combine -5x and -3x to get -8x.

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

Subtract 10 from both sides.

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

Subtract 10 from 3 to get -7.

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

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.

$$x=7 x=1$$