Solve x+1/3x^2+x+2x-1/x-3x^2=1/9x^2-1

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

Solve for x

$$x=\frac{2}{3}\approx 0.666666667$$

Steps Using Factoring

Variable x cannot be equal to any of the values -\frac{1}{3},0,\frac{1}{3} since division by zero is not defined. Multiply both sides of the equation by x\left(3x-1\right)\left(3x+1\right), the least common multiple of 3x^{2}+x,x-3x^{2},9x^{2}-1.

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

Use the distributive property to multiply 3x-1 by x+1 and combine like terms.

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

Use the distributive property to multiply -1-3x by 2x-1 and combine like terms.

$$3x^{2}+2x-1+x+1-6x^{2}=x$$

Combine 2x and x to get 3x.

$$3x^{2}+3x-1+1-6x^{2}=x$$

Add -1 and 1 to get 0.

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

Combine 3x^{2} and -6x^{2} to get -3x^{2}.

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

Subtract x from both sides.

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

Combine 3x and -x to get 2x.

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

Factor out x.

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

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

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

Variable x cannot be equal to 0.

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

Steps Using the Quadratic Formula

Variable x cannot be equal to any of the values -\frac{1}{3},0,\frac{1}{3} since division by zero is not defined. Multiply both sides of the equation by x\left(3x-1\right)\left(3x+1\right), the least common multiple of 3x^{2}+x,x-3x^{2},9x^{2}-1.

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

Use the distributive property to multiply 3x-1 by x+1 and combine like terms.

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

Use the distributive property to multiply -1-3x by 2x-1 and combine like terms.

$$3x^{2}+2x-1+x+1-6x^{2}=x$$

Combine 2x and x to get 3x.

$$3x^{2}+3x-1+1-6x^{2}=x$$

Add -1 and 1 to get 0.

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

Combine 3x^{2} and -6x^{2} to get -3x^{2}.

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

Subtract x from both sides.

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

Combine 3x and -x to get 2x.

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

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

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

Take the square root of 2^{2}.

$$x=\frac{-2±2}{2\left(-3\right)}$$

Multiply 2 times -3.

$$x=\frac{-2±2}{-6}$$

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

$$x=\frac{0}{-6}$$

Divide 0 by -6.

$$x=0$$

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

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

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

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

The equation is now solved.

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

Variable x cannot be equal to 0.

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

Steps for Completing the Square

Variable x cannot be equal to any of the values -\frac{1}{3},0,\frac{1}{3} since division by zero is not defined. Multiply both sides of the equation by x\left(3x-1\right)\left(3x+1\right), the least common multiple of 3x^{2}+x,x-3x^{2},9x^{2}-1.

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

Use the distributive property to multiply 3x-1 by x+1 and combine like terms.

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

Use the distributive property to multiply -1-3x by 2x-1 and combine like terms.

$$3x^{2}+2x-1+x+1-6x^{2}=x$$

Combine 2x and x to get 3x.

$$3x^{2}+3x-1+1-6x^{2}=x$$

Add -1 and 1 to get 0.

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

Combine 3x^{2} and -6x^{2} to get -3x^{2}.

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

Subtract x from both sides.

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

Combine 3x and -x to get 2x.

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

Divide both sides by -3.

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

Dividing by -3 undoes the multiplication by -3.

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

Divide 2 by -3.

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

Divide 0 by -3.

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

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.

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

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

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

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}.

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

Take the square root of both sides of the equation.

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

Simplify.

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

Add \frac{1}{3} to both sides of the equation.

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

Variable x cannot be equal to 0.

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