Solve 3x/2x=5x/2

$$\frac { 3 x } { 2 x } = \frac { 5 x } { 2 }$$

Solve for x

$$x=\frac{3}{5}=0.6$$

Steps Using Factoring

Variable x cannot be equal to 0 since division by zero is not defined. Multiply both sides of the equation by 2x, the least common multiple of 2x,2.

$$3x=x\times 5x$$

Multiply x and x to get x^{2}.

$$3x=x^{2}\times 5$$

Subtract x^{2}\times 5 from both sides.

$$3x-x^{2}\times 5=0$$

Multiply -1 and 5 to get -5.

$$3x-5x^{2}=0$$

Factor out x.

$$x\left(3-5x\right)=0$$

To find equation solutions, solve x=0 and 3-5x=0.

$$x=0 x=\frac{3}{5}$$

Variable x cannot be equal to 0.

$$x=\frac{3}{5}$$

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 2x, the least common multiple of 2x,2.

$$3x=x\times 5x$$

Multiply x and x to get x^{2}.

$$3x=x^{2}\times 5$$

Subtract x^{2}\times 5 from both sides.

$$3x-x^{2}\times 5=0$$

Multiply -1 and 5 to get -5.

$$3x-5x^{2}=0$$

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.

$$-5x^{2}+3x=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute -5 for a, 3 for b, and 0 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-3±\sqrt{3^{2}}}{2\left(-5\right)}$$

Take the square root of 3^{2}.

$$x=\frac{-3±3}{2\left(-5\right)}$$

Multiply 2 times -5.

$$x=\frac{-3±3}{-10}$$

Now solve the equation x=\frac{-3±3}{-10} when ± is plus. Add -3 to 3.

$$x=\frac{0}{-10}$$

Divide 0 by -10.

$$x=0$$

Now solve the equation x=\frac{-3±3}{-10} when ± is minus. Subtract 3 from -3.

$$x=-\frac{6}{-10}$$

Reduce the fraction \frac{-6}{-10} to lowest terms by extracting and canceling out 2.

$$x=\frac{3}{5}$$

The equation is now solved.

$$x=0 x=\frac{3}{5}$$

Variable x cannot be equal to 0.

$$x=\frac{3}{5}$$

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 2x, the least common multiple of 2x,2.

$$3x=x\times 5x$$

Multiply x and x to get x^{2}.

$$3x=x^{2}\times 5$$

Subtract x^{2}\times 5 from both sides.

$$3x-x^{2}\times 5=0$$

Multiply -1 and 5 to get -5.

$$3x-5x^{2}=0$$

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.

$$-5x^{2}+3x=0$$

Divide both sides by -5.

$$\frac{-5x^{2}+3x}{-5}=\frac{0}{-5}$$

Dividing by -5 undoes the multiplication by -5.

$$x^{2}+\frac{3}{-5}x=\frac{0}{-5}$$

Divide 3 by -5.

$$x^{2}-\frac{3}{5}x=\frac{0}{-5}$$

Divide 0 by -5.

$$x^{2}-\frac{3}{5}x=0$$

Divide -\frac{3}{5}, the coefficient of the x term, by 2 to get -\frac{3}{10}. Then add the square of -\frac{3}{10} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}-\frac{3}{5}x+\left(-\frac{3}{10}\right)^{2}=\left(-\frac{3}{10}\right)^{2}$$

Square -\frac{3}{10} by squaring both the numerator and the denominator of the fraction.

$$x^{2}-\frac{3}{5}x+\frac{9}{100}=\frac{9}{100}$$

Factor x^{2}-\frac{3}{5}x+\frac{9}{100}. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(x-\frac{3}{10}\right)^{2}=\frac{9}{100}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-\frac{3}{10}\right)^{2}}=\sqrt{\frac{9}{100}}$$

Simplify.

$$x-\frac{3}{10}=\frac{3}{10} x-\frac{3}{10}=-\frac{3}{10}$$

Add \frac{3}{10} to both sides of the equation.

$$x=\frac{3}{5} x=0$$

Variable x cannot be equal to 0.

$$x=\frac{3}{5}$$