Solve 3x/3x-1quad+1/xquad=1/3x-1

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

Solve for x

$$x=-1$$

Steps Using Factoring By Grouping

Variable x cannot be equal to any of the values 0,\frac{1}{3} since division by zero is not defined. Multiply both sides of the equation by x\left(3x-1\right), the least common multiple of 3x-1,x.

$$x\times 3x+3x-1=x$$

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

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

Subtract x from both sides.

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

Combine 3x and -x to get 2x.

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

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.

$$a+b=2 ab=3\left(-1\right)=-3$$

Since ab is negative, a and b have the opposite signs. Since a+b is positive, the positive number has greater absolute value than the negative. 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 3x^{2}-x.

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

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.

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

Variable x cannot be equal to \frac{1}{3}.

$$x=-1$$

Steps Using the Quadratic Formula

Variable x cannot be equal to any of the values 0,\frac{1}{3} since division by zero is not defined. Multiply both sides of the equation by x\left(3x-1\right), the least common multiple of 3x-1,x.

$$x\times 3x+3x-1=x$$

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

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

Subtract x from both sides.

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

Combine 3x and -x to get 2x.

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

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}.

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

Square 2.

$$x=\frac{-2±\sqrt{4-4\times 3\left(-1\right)}}{2\times 3}$$

Multiply -4 times 3.

$$x=\frac{-2±\sqrt{4-12\left(-1\right)}}{2\times 3}$$

Multiply -12 times -1.

$$x=\frac{-2±\sqrt{4+12}}{2\times 3}$$

Add 4 to 12.

$$x=\frac{-2±\sqrt{16}}{2\times 3}$$

Take the square root of 16.

$$x=\frac{-2±4}{2\times 3}$$

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.

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

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

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

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

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

Divide -6 by 6.

$$x=-1$$

The equation is now solved.

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

Variable x cannot be equal to \frac{1}{3}.

$$x=-1$$

Steps for Completing the Square

Variable x cannot be equal to any of the values 0,\frac{1}{3} since division by zero is not defined. Multiply both sides of the equation by x\left(3x-1\right), the least common multiple of 3x-1,x.

$$x\times 3x+3x-1=x$$

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

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

Subtract x from both sides.

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

Combine 3x and -x to get 2x.

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

Add 1 to both sides. Anything plus zero gives itself.

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

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.

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

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.

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

Variable x cannot be equal to \frac{1}{3}.

$$x=-1$$