Solve x/15-5x=2x/15

$$\frac { x } { 15 - 5 x } = \frac { 2 x } { 15 }$$

Solve for x

$$x = \frac{3}{2} = 1\frac{1}{2} = 1.5$$

$$x=0$$

Steps Using Factoring

Variable x cannot be equal to 3 since division by zero is not defined. Multiply both sides of the equation by 15\left(x-3\right), the least common multiple of 15-5x,15.

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

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

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

Use the distributive property to multiply 2x-6 by x.

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

Subtract 2x^{2} from both sides.

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

Add 6x to both sides.

$$-3x-2x^{2}+6x=0$$

Combine -3x and 6x to get 3x.

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

Factor out x.

$$x\left(3-2x\right)=0$$

To find equation solutions, solve x=0 and 3-2x=0.

$$x=0 x=\frac{3}{2}$$

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 15\left(x-3\right), the least common multiple of 15-5x,15.

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

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

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

Use the distributive property to multiply 2x-6 by x.

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

Subtract 2x^{2} from both sides.

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

Add 6x to both sides.

$$-3x-2x^{2}+6x=0$$

Combine -3x and 6x to get 3x.

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

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.

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

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

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

Take the square root of 3^{2}.

$$x=\frac{-3±3}{2\left(-2\right)}$$

Multiply 2 times -2.

$$x=\frac{-3±3}{-4}$$

Now solve the equation x=\frac{-3±3}{-4} when ± is plus. Add -3 to 3.

$$x=\frac{0}{-4}$$

Divide 0 by -4.

$$x=0$$

Now solve the equation x=\frac{-3±3}{-4} when ± is minus. Subtract 3 from -3.

$$x=-\frac{6}{-4}$$

Reduce the fraction \frac{-6}{-4} to lowest terms by extracting and canceling out 2.

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

The equation is now solved.

$$x=0 x=\frac{3}{2}$$

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 15\left(x-3\right), the least common multiple of 15-5x,15.

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

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

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

Use the distributive property to multiply 2x-6 by x.

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

Subtract 2x^{2} from both sides.

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

Add 6x to both sides.

$$-3x-2x^{2}+6x=0$$

Combine -3x and 6x to get 3x.

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

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.

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

Divide both sides by -2.

$$\frac{-2x^{2}+3x}{-2}=\frac{0}{-2}$$

Dividing by -2 undoes the multiplication by -2.

$$x^{2}+\frac{3}{-2}x=\frac{0}{-2}$$

Divide 3 by -2.

$$x^{2}-\frac{3}{2}x=\frac{0}{-2}$$

Divide 0 by -2.

$$x^{2}-\frac{3}{2}x=0$$

Divide -\frac{3}{2}, the coefficient of the x term, by 2 to get -\frac{3}{4}. Then add the square of -\frac{3}{4} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

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

Square -\frac{3}{4} by squaring both the numerator and the denominator of the fraction.

$$x^{2}-\frac{3}{2}x+\frac{9}{16}=\frac{9}{16}$$

Factor x^{2}-\frac{3}{2}x+\frac{9}{16}. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(x-\frac{3}{4}\right)^{2}=\frac{9}{16}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-\frac{3}{4}\right)^{2}}=\sqrt{\frac{9}{16}}$$

Simplify.

$$x-\frac{3}{4}=\frac{3}{4} x-\frac{3}{4}=-\frac{3}{4}$$

Add \frac{3}{4} to both sides of the equation.

$$x=\frac{3}{2} x=0$$