Solve 5/x+10/x-10=1

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

Solve for x

$$x = \frac{5 \sqrt{17} + 25}{2} \approx 22.807764064$$

$$x = \frac{25 - 5 \sqrt{17}}{2} \approx 2.192235936$$

Steps Using the Quadratic Formula

Variable x cannot be equal to any of the values 0,10 since division by zero is not defined. Multiply both sides of the equation by x\left(x-10\right), the least common multiple of x,x-10.

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

Use the distributive property to multiply x-10 by 5.

$$5x-50+x\times 10=x\left(x-10\right)$$

Combine 5x and x\times 10 to get 15x.

$$15x-50=x\left(x-10\right)$$

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

$$15x-50=x^{2}-10x$$

Subtract x^{2} from both sides.

$$15x-50-x^{2}=-10x$$

Add 10x to both sides.

$$15x-50-x^{2}+10x=0$$

Combine 15x and 10x to get 25x.

$$25x-50-x^{2}=0$$

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.

$$-x^{2}+25x-50=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute -1 for a, 25 for b, and -50 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-25±\sqrt{25^{2}-4\left(-1\right)\left(-50\right)}}{2\left(-1\right)}$$

Square 25.

$$x=\frac{-25±\sqrt{625-4\left(-1\right)\left(-50\right)}}{2\left(-1\right)}$$

Multiply -4 times -1.

$$x=\frac{-25±\sqrt{625+4\left(-50\right)}}{2\left(-1\right)}$$

Multiply 4 times -50.

$$x=\frac{-25±\sqrt{625-200}}{2\left(-1\right)}$$

Add 625 to -200.

$$x=\frac{-25±\sqrt{425}}{2\left(-1\right)}$$

Take the square root of 425.

$$x=\frac{-25±5\sqrt{17}}{2\left(-1\right)}$$

Multiply 2 times -1.

$$x=\frac{-25±5\sqrt{17}}{-2}$$

Now solve the equation x=\frac{-25±5\sqrt{17}}{-2} when ± is plus. Add -25 to 5\sqrt{17}.

$$x=\frac{5\sqrt{17}-25}{-2}$$

Divide -25+5\sqrt{17} by -2.

$$x=\frac{25-5\sqrt{17}}{2}$$

Now solve the equation x=\frac{-25±5\sqrt{17}}{-2} when ± is minus. Subtract 5\sqrt{17} from -25.

$$x=\frac{-5\sqrt{17}-25}{-2}$$

Divide -25-5\sqrt{17} by -2.

$$x=\frac{5\sqrt{17}+25}{2}$$

The equation is now solved.

$$x=\frac{25-5\sqrt{17}}{2} x=\frac{5\sqrt{17}+25}{2}$$

Steps for Completing the Square

Variable x cannot be equal to any of the values 0,10 since division by zero is not defined. Multiply both sides of the equation by x\left(x-10\right), the least common multiple of x,x-10.

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

Use the distributive property to multiply x-10 by 5.

$$5x-50+x\times 10=x\left(x-10\right)$$

Combine 5x and x\times 10 to get 15x.

$$15x-50=x\left(x-10\right)$$

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

$$15x-50=x^{2}-10x$$

Subtract x^{2} from both sides.

$$15x-50-x^{2}=-10x$$

Add 10x to both sides.

$$15x-50-x^{2}+10x=0$$

Combine 15x and 10x to get 25x.

$$25x-50-x^{2}=0$$

Add 50 to both sides. Anything plus zero gives itself.

$$25x-x^{2}=50$$

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.

$$-x^{2}+25x=50$$

Divide both sides by -1.

$$\frac{-x^{2}+25x}{-1}=\frac{50}{-1}$$

Dividing by -1 undoes the multiplication by -1.

$$x^{2}+\frac{25}{-1}x=\frac{50}{-1}$$

Divide 25 by -1.

$$x^{2}-25x=\frac{50}{-1}$$

Divide 50 by -1.

$$x^{2}-25x=-50$$

Divide -25, the coefficient of the x term, by 2 to get -\frac{25}{2}. Then add the square of -\frac{25}{2} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}-25x+\left(-\frac{25}{2}\right)^{2}=-50+\left(-\frac{25}{2}\right)^{2}$$

Square -\frac{25}{2} by squaring both the numerator and the denominator of the fraction.

$$x^{2}-25x+\frac{625}{4}=-50+\frac{625}{4}$$

Add -50 to \frac{625}{4}.

$$x^{2}-25x+\frac{625}{4}=\frac{425}{4}$$

Factor x^{2}-25x+\frac{625}{4}. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(x-\frac{25}{2}\right)^{2}=\frac{425}{4}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-\frac{25}{2}\right)^{2}}=\sqrt{\frac{425}{4}}$$

Simplify.

$$x-\frac{25}{2}=\frac{5\sqrt{17}}{2} x-\frac{25}{2}=-\frac{5\sqrt{17}}{2}$$

Add \frac{25}{2} to both sides of the equation.

$$x=\frac{5\sqrt{17}+25}{2} x=\frac{25-5\sqrt{17}}{2}$$