Solve 120/x=140/x-2-3

$$\frac { 120 } { x } = \frac { 140 } { x - 2 } - 3$$

Solve for x

$$x = \frac{\sqrt{889} + 13}{3} \approx 14.272034344$$

$$x=\frac{13-\sqrt{889}}{3}\approx -5.605367677$$

Steps Using the Quadratic Formula

Variable x cannot be equal to any of the values 0,2 since division by zero is not defined. Multiply both sides of the equation by x\left(x-2\right), the least common multiple of x,x-2.

$$\left(x-2\right)\times 120=x\times 140+x\left(x-2\right)\left(-3\right)$$

Use the distributive property to multiply x-2 by 120.

$$120x-240=x\times 140+x\left(x-2\right)\left(-3\right)$$

Use the distributive property to multiply x by x-2.

$$120x-240=x\times 140+\left(x^{2}-2x\right)\left(-3\right)$$

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

$$120x-240=x\times 140-3x^{2}+6x$$

Combine x\times 140 and 6x to get 146x.

$$120x-240=146x-3x^{2}$$

Subtract 146x from both sides.

$$120x-240-146x=-3x^{2}$$

Combine 120x and -146x to get -26x.

$$-26x-240=-3x^{2}$$

Add 3x^{2} to both sides.

$$-26x-240+3x^{2}=0$$

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.

$$3x^{2}-26x-240=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 3 for a, -26 for b, and -240 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-\left(-26\right)±\sqrt{\left(-26\right)^{2}-4\times 3\left(-240\right)}}{2\times 3}$$

Square -26.

$$x=\frac{-\left(-26\right)±\sqrt{676-4\times 3\left(-240\right)}}{2\times 3}$$

Multiply -4 times 3.

$$x=\frac{-\left(-26\right)±\sqrt{676-12\left(-240\right)}}{2\times 3}$$

Multiply -12 times -240.

$$x=\frac{-\left(-26\right)±\sqrt{676+2880}}{2\times 3}$$

Add 676 to 2880.

$$x=\frac{-\left(-26\right)±\sqrt{3556}}{2\times 3}$$

Take the square root of 3556.

$$x=\frac{-\left(-26\right)±2\sqrt{889}}{2\times 3}$$

The opposite of -26 is 26.

$$x=\frac{26±2\sqrt{889}}{2\times 3}$$

Multiply 2 times 3.

$$x=\frac{26±2\sqrt{889}}{6}$$

Now solve the equation x=\frac{26±2\sqrt{889}}{6} when ± is plus. Add 26 to 2\sqrt{889}.

$$x=\frac{2\sqrt{889}+26}{6}$$

Divide 26+2\sqrt{889} by 6.

$$x=\frac{\sqrt{889}+13}{3}$$

Now solve the equation x=\frac{26±2\sqrt{889}}{6} when ± is minus. Subtract 2\sqrt{889} from 26.

$$x=\frac{26-2\sqrt{889}}{6}$$

Divide 26-2\sqrt{889} by 6.

$$x=\frac{13-\sqrt{889}}{3}$$

The equation is now solved.

$$x=\frac{\sqrt{889}+13}{3} x=\frac{13-\sqrt{889}}{3}$$

Steps for Completing the Square

Variable x cannot be equal to any of the values 0,2 since division by zero is not defined. Multiply both sides of the equation by x\left(x-2\right), the least common multiple of x,x-2.

$$\left(x-2\right)\times 120=x\times 140+x\left(x-2\right)\left(-3\right)$$

Use the distributive property to multiply x-2 by 120.

$$120x-240=x\times 140+x\left(x-2\right)\left(-3\right)$$

Use the distributive property to multiply x by x-2.

$$120x-240=x\times 140+\left(x^{2}-2x\right)\left(-3\right)$$

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

$$120x-240=x\times 140-3x^{2}+6x$$

Combine x\times 140 and 6x to get 146x.

$$120x-240=146x-3x^{2}$$

Subtract 146x from both sides.

$$120x-240-146x=-3x^{2}$$

Combine 120x and -146x to get -26x.

$$-26x-240=-3x^{2}$$

Add 3x^{2} to both sides.

$$-26x-240+3x^{2}=0$$

Add 240 to both sides. Anything plus zero gives itself.

$$-26x+3x^{2}=240$$

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.

$$3x^{2}-26x=240$$

Divide both sides by 3.

$$\frac{3x^{2}-26x}{3}=\frac{240}{3}$$

Dividing by 3 undoes the multiplication by 3.

$$x^{2}-\frac{26}{3}x=\frac{240}{3}$$

Divide 240 by 3.

$$x^{2}-\frac{26}{3}x=80$$

Divide -\frac{26}{3}, the coefficient of the x term, by 2 to get -\frac{13}{3}. Then add the square of -\frac{13}{3} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}-\frac{26}{3}x+\left(-\frac{13}{3}\right)^{2}=80+\left(-\frac{13}{3}\right)^{2}$$

Square -\frac{13}{3} by squaring both the numerator and the denominator of the fraction.

$$x^{2}-\frac{26}{3}x+\frac{169}{9}=80+\frac{169}{9}$$

Add 80 to \frac{169}{9}.

$$x^{2}-\frac{26}{3}x+\frac{169}{9}=\frac{889}{9}$$

Factor x^{2}-\frac{26}{3}x+\frac{169}{9}. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(x-\frac{13}{3}\right)^{2}=\frac{889}{9}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-\frac{13}{3}\right)^{2}}=\sqrt{\frac{889}{9}}$$

Simplify.

$$x-\frac{13}{3}=\frac{\sqrt{889}}{3} x-\frac{13}{3}=-\frac{\sqrt{889}}{3}$$

Add \frac{13}{3} to both sides of the equation.

$$x=\frac{\sqrt{889}+13}{3} x=\frac{13-\sqrt{889}}{3}$$