Solve 3m+1/3m-1-3m-1/3m+1=2

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

Solve for m

$$m=\frac{1-\sqrt{2}}{3}\approx -0.138071187$$

$$m=\frac{\sqrt{2}+1}{3}\approx 0.804737854$$

Steps Using the Quadratic Formula

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

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

Multiply 3m+1 and 3m+1 to get \left(3m+1\right)^{2}.

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

Multiply 3m-1 and 3m-1 to get \left(3m-1\right)^{2}.

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

Use binomial theorem \left(a+b\right)^{2}=a^{2}+2ab+b^{2} to expand \left(3m+1\right)^{2}.

$$9m^{2}+6m+1-\left(3m-1\right)^{2}=2\left(3m-1\right)\left(3m+1\right)$$

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(3m-1\right)^{2}.

$$9m^{2}+6m+1-\left(9m^{2}-6m+1\right)=2\left(3m-1\right)\left(3m+1\right)$$

To find the opposite of 9m^{2}-6m+1, find the opposite of each term.

$$9m^{2}+6m+1-9m^{2}+6m-1=2\left(3m-1\right)\left(3m+1\right)$$

Combine 9m^{2} and -9m^{2} to get 0.

$$6m+1+6m-1=2\left(3m-1\right)\left(3m+1\right)$$

Combine 6m and 6m to get 12m.

$$12m+1-1=2\left(3m-1\right)\left(3m+1\right)$$

Subtract 1 from 1 to get 0.

$$12m=2\left(3m-1\right)\left(3m+1\right)$$

Use the distributive property to multiply 2 by 3m-1.

$$12m=\left(6m-2\right)\left(3m+1\right)$$

Use the distributive property to multiply 6m-2 by 3m+1 and combine like terms.

$$12m=18m^{2}-2$$

Subtract 18m^{2} from both sides.

$$12m-18m^{2}=-2$$

Add 2 to both sides.

$$12m-18m^{2}+2=0$$

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.

$$-18m^{2}+12m+2=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute -18 for a, 12 for b, and 2 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$m=\frac{-12±\sqrt{12^{2}-4\left(-18\right)\times 2}}{2\left(-18\right)}$$

Square 12.

$$m=\frac{-12±\sqrt{144-4\left(-18\right)\times 2}}{2\left(-18\right)}$$

Multiply -4 times -18.

$$m=\frac{-12±\sqrt{144+72\times 2}}{2\left(-18\right)}$$

Multiply 72 times 2.

$$m=\frac{-12±\sqrt{144+144}}{2\left(-18\right)}$$

Add 144 to 144.

$$m=\frac{-12±\sqrt{288}}{2\left(-18\right)}$$

Take the square root of 288.

$$m=\frac{-12±12\sqrt{2}}{2\left(-18\right)}$$

Multiply 2 times -18.

$$m=\frac{-12±12\sqrt{2}}{-36}$$

Now solve the equation m=\frac{-12±12\sqrt{2}}{-36} when ± is plus. Add -12 to 12\sqrt{2}.

$$m=\frac{12\sqrt{2}-12}{-36}$$

Divide -12+12\sqrt{2} by -36.

$$m=\frac{1-\sqrt{2}}{3}$$

Now solve the equation m=\frac{-12±12\sqrt{2}}{-36} when ± is minus. Subtract 12\sqrt{2} from -12.

$$m=\frac{-12\sqrt{2}-12}{-36}$$

Divide -12-12\sqrt{2} by -36.

$$m=\frac{\sqrt{2}+1}{3}$$

The equation is now solved.

$$m=\frac{1-\sqrt{2}}{3} m=\frac{\sqrt{2}+1}{3}$$

Steps for Completing the Square

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

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

Multiply 3m+1 and 3m+1 to get \left(3m+1\right)^{2}.

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

Multiply 3m-1 and 3m-1 to get \left(3m-1\right)^{2}.

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

Use binomial theorem \left(a+b\right)^{2}=a^{2}+2ab+b^{2} to expand \left(3m+1\right)^{2}.

$$9m^{2}+6m+1-\left(3m-1\right)^{2}=2\left(3m-1\right)\left(3m+1\right)$$

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(3m-1\right)^{2}.

$$9m^{2}+6m+1-\left(9m^{2}-6m+1\right)=2\left(3m-1\right)\left(3m+1\right)$$

To find the opposite of 9m^{2}-6m+1, find the opposite of each term.

$$9m^{2}+6m+1-9m^{2}+6m-1=2\left(3m-1\right)\left(3m+1\right)$$

Combine 9m^{2} and -9m^{2} to get 0.

$$6m+1+6m-1=2\left(3m-1\right)\left(3m+1\right)$$

Combine 6m and 6m to get 12m.

$$12m+1-1=2\left(3m-1\right)\left(3m+1\right)$$

Subtract 1 from 1 to get 0.

$$12m=2\left(3m-1\right)\left(3m+1\right)$$

Use the distributive property to multiply 2 by 3m-1.

$$12m=\left(6m-2\right)\left(3m+1\right)$$

Use the distributive property to multiply 6m-2 by 3m+1 and combine like terms.

$$12m=18m^{2}-2$$

Subtract 18m^{2} from both sides.

$$12m-18m^{2}=-2$$

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.

$$-18m^{2}+12m=-2$$

Divide both sides by -18.

$$\frac{-18m^{2}+12m}{-18}=-\frac{2}{-18}$$

Dividing by -18 undoes the multiplication by -18.

$$m^{2}+\frac{12}{-18}m=-\frac{2}{-18}$$

Reduce the fraction \frac{12}{-18} to lowest terms by extracting and canceling out 6.

$$m^{2}-\frac{2}{3}m=-\frac{2}{-18}$$

Reduce the fraction \frac{-2}{-18} to lowest terms by extracting and canceling out 2.

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

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.

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

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

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

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

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

Factor m^{2}-\frac{2}{3}m+\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}.

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

Take the square root of both sides of the equation.

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

Simplify.

$$m-\frac{1}{3}=\frac{\sqrt{2}}{3} m-\frac{1}{3}=-\frac{\sqrt{2}}{3}$$

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

$$m=\frac{\sqrt{2}+1}{3} m=\frac{1-\sqrt{2}}{3}$$