Solve 2x-1/x-3x/2x-1=2

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

Solve for x

$$x=-1$$

$$x=\frac{1}{3}\approx 0.333333333$$

Steps Using Factoring By Grouping

Variable x cannot be equal to any of the values 0,\frac{1}{2} since division by zero is not defined. Multiply both sides of the equation by x\left(2x-1\right), the least common multiple of x,2x-1.

$$\left(2x-1\right)\left(2x-1\right)-x\times 3x=2x\left(2x-1\right)$$

Multiply 2x-1 and 2x-1 to get \left(2x-1\right)^{2}.

$$\left(2x-1\right)^{2}-x\times 3x=2x\left(2x-1\right)$$

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(2x-1\right)^{2}.

$$4x^{2}-4x+1-x\times 3x=2x\left(2x-1\right)$$

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

$$4x^{2}-4x+1-x^{2}\times 3=2x\left(2x-1\right)$$

Use the distributive property to multiply 2x by 2x-1.

$$4x^{2}-4x+1-x^{2}\times 3=4x^{2}-2x$$

Subtract 4x^{2} from both sides.

$$4x^{2}-4x+1-x^{2}\times 3-4x^{2}=-2x$$

Combine 4x^{2} and -4x^{2} to get 0.

$$-4x+1-x^{2}\times 3=-2x$$

Add 2x to both sides.

$$-4x+1-x^{2}\times 3+2x=0$$

Combine -4x and 2x to get -2x.

$$-2x+1-x^{2}\times 3=0$$

Multiply -1 and 3 to get -3.

$$-2x+1-3x^{2}=0$$

Rearrange the polynomial to put it in standard form. Place the terms in order from highest to lowest power.

$$-3x^{2}-2x+1=0$$

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as -3x^{2}+ax+bx+1. To find a and b, set up a system to be solved.

$$a+b=-2 ab=-3=-3$$

Since ab is negative, a and b have the opposite signs. Since a+b is negative, the negative number has greater absolute value than the positive. The only such pair is the system solution.

$$a=1 b=-3$$

Rewrite -3x^{2}-2x+1 as \left(-3x^{2}+x\right)+\left(-3x+1\right).

$$\left(-3x^{2}+x\right)+\left(-3x+1\right)$$

Factor out -x in the first and -1 in the second group.

$$-x\left(3x-1\right)-\left(3x-1\right)$$

Factor out common term 3x-1 by using distributive property.

$$\left(3x-1\right)\left(-x-1\right)$$

To find equation solutions, solve 3x-1=0 and -x-1=0.

$$x=\frac{1}{3} x=-1$$

Steps Using the Quadratic Formula

Variable x cannot be equal to any of the values 0,\frac{1}{2} since division by zero is not defined. Multiply both sides of the equation by x\left(2x-1\right), the least common multiple of x,2x-1.

$$\left(2x-1\right)\left(2x-1\right)-x\times 3x=2x\left(2x-1\right)$$

Multiply 2x-1 and 2x-1 to get \left(2x-1\right)^{2}.

$$\left(2x-1\right)^{2}-x\times 3x=2x\left(2x-1\right)$$

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(2x-1\right)^{2}.

$$4x^{2}-4x+1-x\times 3x=2x\left(2x-1\right)$$

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

$$4x^{2}-4x+1-x^{2}\times 3=2x\left(2x-1\right)$$

Use the distributive property to multiply 2x by 2x-1.

$$4x^{2}-4x+1-x^{2}\times 3=4x^{2}-2x$$

Subtract 4x^{2} from both sides.

$$4x^{2}-4x+1-x^{2}\times 3-4x^{2}=-2x$$

Combine 4x^{2} and -4x^{2} to get 0.

$$-4x+1-x^{2}\times 3=-2x$$

Add 2x to both sides.

$$-4x+1-x^{2}\times 3+2x=0$$

Combine -4x and 2x to get -2x.

$$-2x+1-x^{2}\times 3=0$$

Multiply -1 and 3 to get -3.

$$-2x+1-3x^{2}=0$$

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.

$$-3x^{2}-2x+1=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute -3 for a, -2 for b, and 1 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

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

Square -2.

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

Multiply -4 times -3.

$$x=\frac{-\left(-2\right)±\sqrt{4+12}}{2\left(-3\right)}$$

Add 4 to 12.

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

Take the square root of 16.

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

The opposite of -2 is 2.

$$x=\frac{2±4}{2\left(-3\right)}$$

Multiply 2 times -3.

$$x=\frac{2±4}{-6}$$

Now solve the equation x=\frac{2±4}{-6} when ± is plus. Add 2 to 4.

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

Divide 6 by -6.

$$x=-1$$

Now solve the equation x=\frac{2±4}{-6} when ± is minus. Subtract 4 from 2.

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

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

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

The equation is now solved.

$$x=-1 x=\frac{1}{3}$$

Steps for Completing the Square

Variable x cannot be equal to any of the values 0,\frac{1}{2} since division by zero is not defined. Multiply both sides of the equation by x\left(2x-1\right), the least common multiple of x,2x-1.

$$\left(2x-1\right)\left(2x-1\right)-x\times 3x=2x\left(2x-1\right)$$

Multiply 2x-1 and 2x-1 to get \left(2x-1\right)^{2}.

$$\left(2x-1\right)^{2}-x\times 3x=2x\left(2x-1\right)$$

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(2x-1\right)^{2}.

$$4x^{2}-4x+1-x\times 3x=2x\left(2x-1\right)$$

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

$$4x^{2}-4x+1-x^{2}\times 3=2x\left(2x-1\right)$$

Use the distributive property to multiply 2x by 2x-1.

$$4x^{2}-4x+1-x^{2}\times 3=4x^{2}-2x$$

Subtract 4x^{2} from both sides.

$$4x^{2}-4x+1-x^{2}\times 3-4x^{2}=-2x$$

Combine 4x^{2} and -4x^{2} to get 0.

$$-4x+1-x^{2}\times 3=-2x$$

Add 2x to both sides.

$$-4x+1-x^{2}\times 3+2x=0$$

Combine -4x and 2x to get -2x.

$$-2x+1-x^{2}\times 3=0$$

Subtract 1 from both sides. Anything subtracted from zero gives its negation.

$$-2x-x^{2}\times 3=-1$$

Multiply -1 and 3 to get -3.

$$-2x-3x^{2}=-1$$

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.

$$-3x^{2}-2x=-1$$

Divide both sides by -3.

$$\frac{-3x^{2}-2x}{-3}=-\frac{1}{-3}$$

Dividing by -3 undoes the multiplication by -3.

$$x^{2}+\left(-\frac{2}{-3}\right)x=-\frac{1}{-3}$$

Divide -2 by -3.

$$x^{2}+\frac{2}{3}x=-\frac{1}{-3}$$

Divide -1 by -3.

$$x^{2}+\frac{2}{3}x=\frac{1}{3}$$

Divide \frac{2}{3}, the coefficient of the x term, by 2 to get \frac{1}{3}. Then add the square of \frac{1}{3} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}+\frac{2}{3}x+\left(\frac{1}{3}\right)^{2}=\frac{1}{3}+\left(\frac{1}{3}\right)^{2}$$

Square \frac{1}{3} by squaring both the numerator and the denominator of the fraction.

$$x^{2}+\frac{2}{3}x+\frac{1}{9}=\frac{1}{3}+\frac{1}{9}$$

Add \frac{1}{3} to \frac{1}{9} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$x^{2}+\frac{2}{3}x+\frac{1}{9}=\frac{4}{9}$$

Factor x^{2}+\frac{2}{3}x+\frac{1}{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}.

$$\left(x+\frac{1}{3}\right)^{2}=\frac{4}{9}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x+\frac{1}{3}\right)^{2}}=\sqrt{\frac{4}{9}}$$

Simplify.

$$x+\frac{1}{3}=\frac{2}{3} x+\frac{1}{3}=-\frac{2}{3}$$

Subtract \frac{1}{3} from both sides of the equation.

$$x=\frac{1}{3} x=-1$$