Solve 6-x/6=3/x

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

Solve for x (complex solution)

$$x=3+3i$$

$$x=3-3i$$

Steps Using the Quadratic Formula

Variable x cannot be equal to 0 since division by zero is not defined. Multiply both sides of the equation by 6x, the least common multiple of 6,x.

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

Use the distributive property to multiply x by 6-x.

$$6x-x^{2}=6\times 3$$

Multiply 6 and 3 to get 18.

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

Subtract 18 from both sides.

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

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.

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

This equation is in standard form: ax^{2}+bx+c=0. Substitute -1 for a, 6 for b, and -18 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-6±\sqrt{6^{2}-4\left(-1\right)\left(-18\right)}}{2\left(-1\right)}$$

Square 6.

$$x=\frac{-6±\sqrt{36-4\left(-1\right)\left(-18\right)}}{2\left(-1\right)}$$

Multiply -4 times -1.

$$x=\frac{-6±\sqrt{36+4\left(-18\right)}}{2\left(-1\right)}$$

Multiply 4 times -18.

$$x=\frac{-6±\sqrt{36-72}}{2\left(-1\right)}$$

Add 36 to -72.

$$x=\frac{-6±\sqrt{-36}}{2\left(-1\right)}$$

Take the square root of -36.

$$x=\frac{-6±6i}{2\left(-1\right)}$$

Multiply 2 times -1.

$$x=\frac{-6±6i}{-2}$$

Now solve the equation x=\frac{-6±6i}{-2} when ± is plus. Add -6 to 6i.

$$x=\frac{-6+6i}{-2}$$

Divide -6+6i by -2.

$$x=3-3i$$

Now solve the equation x=\frac{-6±6i}{-2} when ± is minus. Subtract 6i from -6.

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

Divide -6-6i by -2.

$$x=3+3i$$

The equation is now solved.

$$x=3-3i x=3+3i$$

Steps for Completing the Square

Variable x cannot be equal to 0 since division by zero is not defined. Multiply both sides of the equation by 6x, the least common multiple of 6,x.

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

Use the distributive property to multiply x by 6-x.

$$6x-x^{2}=6\times 3$$

Multiply 6 and 3 to get 18.

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

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.

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

Divide both sides by -1.

$$\frac{-x^{2}+6x}{-1}=\frac{18}{-1}$$

Dividing by -1 undoes the multiplication by -1.

$$x^{2}+\frac{6}{-1}x=\frac{18}{-1}$$

Divide 6 by -1.

$$x^{2}-6x=\frac{18}{-1}$$

Divide 18 by -1.

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

Divide -6, the coefficient of the x term, by 2 to get -3. Then add the square of -3 to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}-6x+\left(-3\right)^{2}=-18+\left(-3\right)^{2}$$

Square -3.

$$x^{2}-6x+9=-18+9$$

Add -18 to 9.

$$x^{2}-6x+9=-9$$

Factor x^{2}-6x+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}.

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

Take the square root of both sides of the equation.

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

Simplify.

$$x-3=3i x-3=-3i$$

Add 3 to both sides of the equation.

$$x=3+3i x=3-3i$$