Solve 3-x/3=2x/6-2x

$$\frac{ 3-x }{ 3 } = \frac{ 2x }{ 6-2x }$$

Solve for x

$$x = \frac{3 \sqrt{5} + 9}{2} \approx 7.854101966$$

$$x = \frac{9 - 3 \sqrt{5}}{2} \approx 1.145898034$$

Steps Using the Quadratic Formula

Variable x cannot be equal to 3 since division by zero is not defined. Multiply both sides of the equation by 6\left(x-3\right), the least common multiple of 3,6-2x.

$$\left(2x-6\right)\left(3-x\right)=-3\times 2x$$

Use the distributive property to multiply 2x-6 by 3-x and combine like terms.

$$12x-2x^{2}-18=-3\times 2x$$

Multiply -3 and 2 to get -6.

$$12x-2x^{2}-18=-6x$$

Add 6x to both sides.

$$12x-2x^{2}-18+6x=0$$

Combine 12x and 6x to get 18x.

$$18x-2x^{2}-18=0$$

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.

$$-2x^{2}+18x-18=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute -2 for a, 18 for b, and -18 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-18±\sqrt{18^{2}-4\left(-2\right)\left(-18\right)}}{2\left(-2\right)}$$

Square 18.

$$x=\frac{-18±\sqrt{324-4\left(-2\right)\left(-18\right)}}{2\left(-2\right)}$$

Multiply -4 times -2.

$$x=\frac{-18±\sqrt{324+8\left(-18\right)}}{2\left(-2\right)}$$

Multiply 8 times -18.

$$x=\frac{-18±\sqrt{324-144}}{2\left(-2\right)}$$

Add 324 to -144.

$$x=\frac{-18±\sqrt{180}}{2\left(-2\right)}$$

Take the square root of 180.

$$x=\frac{-18±6\sqrt{5}}{2\left(-2\right)}$$

Multiply 2 times -2.

$$x=\frac{-18±6\sqrt{5}}{-4}$$

Now solve the equation x=\frac{-18±6\sqrt{5}}{-4} when ± is plus. Add -18 to 6\sqrt{5}.

$$x=\frac{6\sqrt{5}-18}{-4}$$

Divide -18+6\sqrt{5} by -4.

$$x=\frac{9-3\sqrt{5}}{2}$$

Now solve the equation x=\frac{-18±6\sqrt{5}}{-4} when ± is minus. Subtract 6\sqrt{5} from -18.

$$x=\frac{-6\sqrt{5}-18}{-4}$$

Divide -18-6\sqrt{5} by -4.

$$x=\frac{3\sqrt{5}+9}{2}$$

The equation is now solved.

$$x=\frac{9-3\sqrt{5}}{2} x=\frac{3\sqrt{5}+9}{2}$$

Steps for Completing the Square

Variable x cannot be equal to 3 since division by zero is not defined. Multiply both sides of the equation by 6\left(x-3\right), the least common multiple of 3,6-2x.

$$\left(2x-6\right)\left(3-x\right)=-3\times 2x$$

Use the distributive property to multiply 2x-6 by 3-x and combine like terms.

$$12x-2x^{2}-18=-3\times 2x$$

Multiply -3 and 2 to get -6.

$$12x-2x^{2}-18=-6x$$

Add 6x to both sides.

$$12x-2x^{2}-18+6x=0$$

Combine 12x and 6x to get 18x.

$$18x-2x^{2}-18=0$$

Add 18 to both sides. Anything plus zero gives itself.

$$18x-2x^{2}=18$$

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.

$$-2x^{2}+18x=18$$

Divide both sides by -2.

$$\frac{-2x^{2}+18x}{-2}=\frac{18}{-2}$$

Dividing by -2 undoes the multiplication by -2.

$$x^{2}+\frac{18}{-2}x=\frac{18}{-2}$$

Divide 18 by -2.

$$x^{2}-9x=\frac{18}{-2}$$

Divide 18 by -2.

$$x^{2}-9x=-9$$

Divide -9, the coefficient of the x term, by 2 to get -\frac{9}{2}. Then add the square of -\frac{9}{2} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}-9x+\left(-\frac{9}{2}\right)^{2}=-9+\left(-\frac{9}{2}\right)^{2}$$

Square -\frac{9}{2} by squaring both the numerator and the denominator of the fraction.

$$x^{2}-9x+\frac{81}{4}=-9+\frac{81}{4}$$

Add -9 to \frac{81}{4}.

$$x^{2}-9x+\frac{81}{4}=\frac{45}{4}$$

Factor x^{2}-9x+\frac{81}{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}.

$$\left(x-\frac{9}{2}\right)^{2}=\frac{45}{4}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-\frac{9}{2}\right)^{2}}=\sqrt{\frac{45}{4}}$$

Simplify.

$$x-\frac{9}{2}=\frac{3\sqrt{5}}{2} x-\frac{9}{2}=-\frac{3\sqrt{5}}{2}$$

Add \frac{9}{2} to both sides of the equation.

$$x=\frac{3\sqrt{5}+9}{2} x=\frac{9-3\sqrt{5}}{2}$$