Solve 27/x=20/x-3-10

$$\frac{ 27 }{ x } = \frac{ 20 }{ x-3 } -10$$

Solve for x

$$x = \frac{\sqrt{3769} + 23}{20} \approx 4.219609096$$

$$x=\frac{23-\sqrt{3769}}{20}\approx -1.919609096$$

Steps Using the Quadratic Formula

Variable x cannot be equal to any of the values 0,3 since division by zero is not defined. Multiply both sides of the equation by x\left(x-3\right), the least common multiple of x,x-3.

$$\left(x-3\right)\times 27=x\times 20+x\left(x-3\right)\left(-10\right)$$

Use the distributive property to multiply x-3 by 27.

$$27x-81=x\times 20+x\left(x-3\right)\left(-10\right)$$

Use the distributive property to multiply x by x-3.

$$27x-81=x\times 20+\left(x^{2}-3x\right)\left(-10\right)$$

Use the distributive property to multiply x^{2}-3x by -10.

$$27x-81=x\times 20-10x^{2}+30x$$

Combine x\times 20 and 30x to get 50x.

$$27x-81=50x-10x^{2}$$

Subtract 50x from both sides.

$$27x-81-50x=-10x^{2}$$

Combine 27x and -50x to get -23x.

$$-23x-81=-10x^{2}$$

Add 10x^{2} to both sides.

$$-23x-81+10x^{2}=0$$

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.

$$10x^{2}-23x-81=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 10 for a, -23 for b, and -81 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-\left(-23\right)±\sqrt{\left(-23\right)^{2}-4\times 10\left(-81\right)}}{2\times 10}$$

Square -23.

$$x=\frac{-\left(-23\right)±\sqrt{529-4\times 10\left(-81\right)}}{2\times 10}$$

Multiply -4 times 10.

$$x=\frac{-\left(-23\right)±\sqrt{529-40\left(-81\right)}}{2\times 10}$$

Multiply -40 times -81.

$$x=\frac{-\left(-23\right)±\sqrt{529+3240}}{2\times 10}$$

Add 529 to 3240.

$$x=\frac{-\left(-23\right)±\sqrt{3769}}{2\times 10}$$

The opposite of -23 is 23.

$$x=\frac{23±\sqrt{3769}}{2\times 10}$$

Multiply 2 times 10.

$$x=\frac{23±\sqrt{3769}}{20}$$

Now solve the equation x=\frac{23±\sqrt{3769}}{20} when ± is plus. Add 23 to \sqrt{3769}.

$$x=\frac{\sqrt{3769}+23}{20}$$

Now solve the equation x=\frac{23±\sqrt{3769}}{20} when ± is minus. Subtract \sqrt{3769} from 23.

$$x=\frac{23-\sqrt{3769}}{20}$$

The equation is now solved.

$$x=\frac{\sqrt{3769}+23}{20} x=\frac{23-\sqrt{3769}}{20}$$

Steps for Completing the Square

Variable x cannot be equal to any of the values 0,3 since division by zero is not defined. Multiply both sides of the equation by x\left(x-3\right), the least common multiple of x,x-3.

$$\left(x-3\right)\times 27=x\times 20+x\left(x-3\right)\left(-10\right)$$

Use the distributive property to multiply x-3 by 27.

$$27x-81=x\times 20+x\left(x-3\right)\left(-10\right)$$

Use the distributive property to multiply x by x-3.

$$27x-81=x\times 20+\left(x^{2}-3x\right)\left(-10\right)$$

Use the distributive property to multiply x^{2}-3x by -10.

$$27x-81=x\times 20-10x^{2}+30x$$

Combine x\times 20 and 30x to get 50x.

$$27x-81=50x-10x^{2}$$

Subtract 50x from both sides.

$$27x-81-50x=-10x^{2}$$

Combine 27x and -50x to get -23x.

$$-23x-81=-10x^{2}$$

Add 10x^{2} to both sides.

$$-23x-81+10x^{2}=0$$

Add 81 to both sides. Anything plus zero gives itself.

$$-23x+10x^{2}=81$$

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.

$$10x^{2}-23x=81$$

Divide both sides by 10.

$$\frac{10x^{2}-23x}{10}=\frac{81}{10}$$

Dividing by 10 undoes the multiplication by 10.

$$x^{2}-\frac{23}{10}x=\frac{81}{10}$$

Divide -\frac{23}{10}, the coefficient of the x term, by 2 to get -\frac{23}{20}. Then add the square of -\frac{23}{20} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}-\frac{23}{10}x+\left(-\frac{23}{20}\right)^{2}=\frac{81}{10}+\left(-\frac{23}{20}\right)^{2}$$

Square -\frac{23}{20} by squaring both the numerator and the denominator of the fraction.

$$x^{2}-\frac{23}{10}x+\frac{529}{400}=\frac{81}{10}+\frac{529}{400}$$

Add \frac{81}{10} to \frac{529}{400} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$x^{2}-\frac{23}{10}x+\frac{529}{400}=\frac{3769}{400}$$

Factor x^{2}-\frac{23}{10}x+\frac{529}{400}. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(x-\frac{23}{20}\right)^{2}=\frac{3769}{400}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-\frac{23}{20}\right)^{2}}=\sqrt{\frac{3769}{400}}$$

Simplify.

$$x-\frac{23}{20}=\frac{\sqrt{3769}}{20} x-\frac{23}{20}=-\frac{\sqrt{3769}}{20}$$

Add \frac{23}{20} to both sides of the equation.

$$x=\frac{\sqrt{3769}+23}{20} x=\frac{23-\sqrt{3769}}{20}$$