Solve x+1/x-5=7

$$x + \frac { 1 } { x - 5 } = 7$$

Solve for x

$$x=6$$

Steps Using the Quadratic Formula

Variable x cannot be equal to 5 since division by zero is not defined. Multiply both sides of the equation by x-5.

$$\left(x-5\right)x+1=7\left(x-5\right)$$

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

$$x^{2}-5x+1=7\left(x-5\right)$$

Use the distributive property to multiply 7 by x-5.

$$x^{2}-5x+1=7x-35$$

Subtract 7x from both sides.

$$x^{2}-5x+1-7x=-35$$

Combine -5x and -7x to get -12x.

$$x^{2}-12x+1=-35$$

Add 35 to both sides.

$$x^{2}-12x+1+35=0$$

Add 1 and 35 to get 36.

$$x^{2}-12x+36=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 1 for a, -12 for b, and 36 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-\left(-12\right)±\sqrt{\left(-12\right)^{2}-4\times 36}}{2}$$

Square -12.

$$x=\frac{-\left(-12\right)±\sqrt{144-4\times 36}}{2}$$

Multiply -4 times 36.

$$x=\frac{-\left(-12\right)±\sqrt{144-144}}{2}$$

Add 144 to -144.

$$x=\frac{-\left(-12\right)±\sqrt{0}}{2}$$

Take the square root of 0.

$$x=-\frac{-12}{2}$$

The opposite of -12 is 12.

$$x=\frac{12}{2}$$

Divide 12 by 2.

$$x=6$$

Steps for Completing the Square

Variable x cannot be equal to 5 since division by zero is not defined. Multiply both sides of the equation by x-5.

$$\left(x-5\right)x+1=7\left(x-5\right)$$

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

$$x^{2}-5x+1=7\left(x-5\right)$$

Use the distributive property to multiply 7 by x-5.

$$x^{2}-5x+1=7x-35$$

Subtract 7x from both sides.

$$x^{2}-5x+1-7x=-35$$

Combine -5x and -7x to get -12x.

$$x^{2}-12x+1=-35$$

Subtract 1 from both sides.

$$x^{2}-12x=-35-1$$

Subtract 1 from -35 to get -36.

$$x^{2}-12x=-36$$

Divide -12, the coefficient of the x term, by 2 to get -6. Then add the square of -6 to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}-12x+\left(-6\right)^{2}=-36+\left(-6\right)^{2}$$

Square -6.

$$x^{2}-12x+36=-36+36$$

Add -36 to 36.

$$x^{2}-12x+36=0$$

Factor x^{2}-12x+36. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(x-6\right)^{2}=0$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-6\right)^{2}}=\sqrt{0}$$

Simplify.

$$x-6=0 x-6=0$$

Add 6 to both sides of the equation.

$$x=6 x=6$$

The equation is now solved. Solutions are the same.

$$x=6$$