Solve 9-3x/9=15-9x/9x

$$\frac{ 9-3x }{ 9 } = \frac{ 15-9x }{ 9x }$$

Solve for x

$$x=1$$

$$x=5$$

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 9x, the least common multiple of 9,9x.

$$x\left(9-3x\right)=15-9x$$

Use the distributive property to multiply x by 9-3x.

$$9x-3x^{2}=15-9x$$

Subtract 15 from both sides.

$$9x-3x^{2}-15=-9x$$

Add 9x to both sides.

$$9x-3x^{2}-15+9x=0$$

Combine 9x and 9x to get 18x.

$$18x-3x^{2}-15=0$$

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.

$$-3x^{2}+18x-15=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute -3 for a, 18 for b, and -15 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-18±\sqrt{18^{2}-4\left(-3\right)\left(-15\right)}}{2\left(-3\right)}$$

Square 18.

$$x=\frac{-18±\sqrt{324-4\left(-3\right)\left(-15\right)}}{2\left(-3\right)}$$

Multiply -4 times -3.

$$x=\frac{-18±\sqrt{324+12\left(-15\right)}}{2\left(-3\right)}$$

Multiply 12 times -15.

$$x=\frac{-18±\sqrt{324-180}}{2\left(-3\right)}$$

Add 324 to -180.

$$x=\frac{-18±\sqrt{144}}{2\left(-3\right)}$$

Take the square root of 144.

$$x=\frac{-18±12}{2\left(-3\right)}$$

Multiply 2 times -3.

$$x=\frac{-18±12}{-6}$$

Now solve the equation x=\frac{-18±12}{-6} when ± is plus. Add -18 to 12.

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

Divide -6 by -6.

$$x=1$$

Now solve the equation x=\frac{-18±12}{-6} when ± is minus. Subtract 12 from -18.

$$x=-\frac{30}{-6}$$

Divide -30 by -6.

$$x=5$$

The equation is now solved.

$$x=1 x=5$$

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 9x, the least common multiple of 9,9x.

$$x\left(9-3x\right)=15-9x$$

Use the distributive property to multiply x by 9-3x.

$$9x-3x^{2}=15-9x$$

Add 9x to both sides.

$$9x-3x^{2}+9x=15$$

Combine 9x and 9x to get 18x.

$$18x-3x^{2}=15$$

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.

$$-3x^{2}+18x=15$$

Divide both sides by -3.

$$\frac{-3x^{2}+18x}{-3}=\frac{15}{-3}$$

Dividing by -3 undoes the multiplication by -3.

$$x^{2}+\frac{18}{-3}x=\frac{15}{-3}$$

Divide 18 by -3.

$$x^{2}-6x=\frac{15}{-3}$$

Divide 15 by -3.

$$x^{2}-6x=-5$$

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.

$$x^{2}-6x+\left(-3\right)^{2}=-5+\left(-3\right)^{2}$$

Square -3.

$$x^{2}-6x+9=-5+9$$

Add -5 to 9.

$$x^{2}-6x+9=4$$

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}.

$$\left(x-3\right)^{2}=4$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-3\right)^{2}}=\sqrt{4}$$

Simplify.

$$x-3=2 x-3=-2$$

Add 3 to both sides of the equation.

$$x=5 x=1$$