Solve 1/frac{1{x-10}-1/x}=720

$$\frac{ 1 }{ \frac{ 1 }{ x-10 } - \frac{ 1 }{ x } } = 720$$

Solve for x

$$x=-80$$

$$x=90$$

Steps Using the Quadratic Formula

To add or subtract expressions, expand them to make their denominators the same. Least common multiple of x-10 and x is x\left(x-10\right). Multiply \frac{1}{x-10} times \frac{x}{x}. Multiply \frac{1}{x} times \frac{x-10}{x-10}.

$$\frac{1}{\frac{x}{x\left(x-10\right)}-\frac{x-10}{x\left(x-10\right)}}=720$$

Since \frac{x}{x\left(x-10\right)} and \frac{x-10}{x\left(x-10\right)} have the same denominator, subtract them by subtracting their numerators.

$$\frac{1}{\frac{x-\left(x-10\right)}{x\left(x-10\right)}}=720$$

Do the multiplications in x-\left(x-10\right).

$$\frac{1}{\frac{x-x+10}{x\left(x-10\right)}}=720$$

Combine like terms in x-x+10.

$$\frac{1}{\frac{10}{x\left(x-10\right)}}=720$$

Variable x cannot be equal to any of the values 0,10 since division by zero is not defined. Divide 1 by \frac{10}{x\left(x-10\right)} by multiplying 1 by the reciprocal of \frac{10}{x\left(x-10\right)}.

$$\frac{x\left(x-10\right)}{10}=720$$

Use the distributive property to multiply x by x-10.

$$\frac{x^{2}-10x}{10}=720$$

Divide each term of x^{2}-10x by 10 to get \frac{1}{10}x^{2}-x.

$$\frac{1}{10}x^{2}-x=720$$

Subtract 720 from both sides.

$$\frac{1}{10}x^{2}-x-720=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute \frac{1}{10} for a, -1 for b, and -720 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-\left(-1\right)±\sqrt{1-4\times \frac{1}{10}\left(-720\right)}}{2\times \frac{1}{10}}$$

Multiply -4 times \frac{1}{10}.

$$x=\frac{-\left(-1\right)±\sqrt{1-\frac{2}{5}\left(-720\right)}}{2\times \frac{1}{10}}$$

Multiply -\frac{2}{5} times -720.

$$x=\frac{-\left(-1\right)±\sqrt{1+288}}{2\times \frac{1}{10}}$$

Add 1 to 288.

$$x=\frac{-\left(-1\right)±\sqrt{289}}{2\times \frac{1}{10}}$$

Take the square root of 289.

$$x=\frac{-\left(-1\right)±17}{2\times \frac{1}{10}}$$

The opposite of -1 is 1.

$$x=\frac{1±17}{2\times \frac{1}{10}}$$

Multiply 2 times \frac{1}{10}.

$$x=\frac{1±17}{\frac{1}{5}}$$

Now solve the equation x=\frac{1±17}{\frac{1}{5}} when ± is plus. Add 1 to 17.

$$x=\frac{18}{\frac{1}{5}}$$

Divide 18 by \frac{1}{5} by multiplying 18 by the reciprocal of \frac{1}{5}.

$$x=90$$

Now solve the equation x=\frac{1±17}{\frac{1}{5}} when ± is minus. Subtract 17 from 1.

$$x=-\frac{16}{\frac{1}{5}}$$

Divide -16 by \frac{1}{5} by multiplying -16 by the reciprocal of \frac{1}{5}.

$$x=-80$$

The equation is now solved.

$$x=90 x=-80$$

Steps for Completing the Square

To add or subtract expressions, expand them to make their denominators the same. Least common multiple of x-10 and x is x\left(x-10\right). Multiply \frac{1}{x-10} times \frac{x}{x}. Multiply \frac{1}{x} times \frac{x-10}{x-10}.

$$\frac{1}{\frac{x}{x\left(x-10\right)}-\frac{x-10}{x\left(x-10\right)}}=720$$

Since \frac{x}{x\left(x-10\right)} and \frac{x-10}{x\left(x-10\right)} have the same denominator, subtract them by subtracting their numerators.

$$\frac{1}{\frac{x-\left(x-10\right)}{x\left(x-10\right)}}=720$$

Do the multiplications in x-\left(x-10\right).

$$\frac{1}{\frac{x-x+10}{x\left(x-10\right)}}=720$$

Combine like terms in x-x+10.

$$\frac{1}{\frac{10}{x\left(x-10\right)}}=720$$

Variable x cannot be equal to any of the values 0,10 since division by zero is not defined. Divide 1 by \frac{10}{x\left(x-10\right)} by multiplying 1 by the reciprocal of \frac{10}{x\left(x-10\right)}.

$$\frac{x\left(x-10\right)}{10}=720$$

Use the distributive property to multiply x by x-10.

$$\frac{x^{2}-10x}{10}=720$$

Divide each term of x^{2}-10x by 10 to get \frac{1}{10}x^{2}-x.

$$\frac{1}{10}x^{2}-x=720$$

Multiply both sides by 10.

$$\frac{\frac{1}{10}x^{2}-x}{\frac{1}{10}}=\frac{720}{\frac{1}{10}}$$

Dividing by \frac{1}{10} undoes the multiplication by \frac{1}{10}.

$$x^{2}+\left(-\frac{1}{\frac{1}{10}}\right)x=\frac{720}{\frac{1}{10}}$$

Divide -1 by \frac{1}{10} by multiplying -1 by the reciprocal of \frac{1}{10}.

$$x^{2}-10x=\frac{720}{\frac{1}{10}}$$

Divide 720 by \frac{1}{10} by multiplying 720 by the reciprocal of \frac{1}{10}.

$$x^{2}-10x=7200$$

Divide -10, the coefficient of the x term, by 2 to get -5. Then add the square of -5 to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}-10x+\left(-5\right)^{2}=7200+\left(-5\right)^{2}$$

Square -5.

$$x^{2}-10x+25=7200+25$$

Add 7200 to 25.

$$x^{2}-10x+25=7225$$

Factor x^{2}-10x+25. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(x-5\right)^{2}=7225$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-5\right)^{2}}=\sqrt{7225}$$

Simplify.

$$x-5=85 x-5=-85$$

Add 5 to both sides of the equation.

$$x=90 x=-80$$