Solve 1/3(1/5n-3)=1/2[2-1/n]

$$\frac { 1 } { 3 } ( \frac { 1 } { 5 n } - 3 ) = \frac { 1 } { 2 } [ 2 - \frac { 1 } { n } ]$$

Solve for n

$$n=\frac{17}{60}\approx 0.283333333$$

Steps Using Factoring

Variable n cannot be equal to 0 since division by zero is not defined. Multiply both sides of the equation by 30n, the least common multiple of 3,5n,2,n.

$$\frac{1}{3}\left(\frac{1}{5n}-3\right)\times 30n=\frac{1}{2}\left(2-\frac{1}{n}\right)\times 30n$$

To add or subtract expressions, expand them to make their denominators the same. Multiply 3 times \frac{5n}{5n}.

$$\frac{1}{3}\left(\frac{1}{5n}-\frac{3\times 5n}{5n}\right)\times 30n=\frac{1}{2}\left(2-\frac{1}{n}\right)\times 30n$$

Since \frac{1}{5n} and \frac{3\times 5n}{5n} have the same denominator, subtract them by subtracting their numerators.

$$\frac{1}{3}\times \frac{1-3\times 5n}{5n}\times 30n=\frac{1}{2}\left(2-\frac{1}{n}\right)\times 30n$$

Do the multiplications in 1-3\times 5n.

$$\frac{1}{3}\times \frac{1-15n}{5n}\times 30n=\frac{1}{2}\left(2-\frac{1}{n}\right)\times 30n$$

Multiply \frac{1}{3} and 30 to get 10.

$$10\times \frac{1-15n}{5n}n=\frac{1}{2}\left(2-\frac{1}{n}\right)\times 30n$$

Express 10\times \frac{1-15n}{5n} as a single fraction.

$$\frac{10\left(1-15n\right)}{5n}n=\frac{1}{2}\left(2-\frac{1}{n}\right)\times 30n$$

Cancel out 5 in both numerator and denominator.

$$\frac{2\left(-15n+1\right)}{n}n=\frac{1}{2}\left(2-\frac{1}{n}\right)\times 30n$$

Express \frac{2\left(-15n+1\right)}{n}n as a single fraction.

$$\frac{2\left(-15n+1\right)n}{n}=\frac{1}{2}\left(2-\frac{1}{n}\right)\times 30n$$

To add or subtract expressions, expand them to make their denominators the same. Multiply 2 times \frac{n}{n}.

$$\frac{2\left(-15n+1\right)n}{n}=\frac{1}{2}\left(\frac{2n}{n}-\frac{1}{n}\right)\times 30n$$

Since \frac{2n}{n} and \frac{1}{n} have the same denominator, subtract them by subtracting their numerators.

$$\frac{2\left(-15n+1\right)n}{n}=\frac{1}{2}\times \frac{2n-1}{n}\times 30n$$

Multiply \frac{1}{2} and 30 to get 15.

$$\frac{2\left(-15n+1\right)n}{n}=15\times \frac{2n-1}{n}n$$

Express 15\times \frac{2n-1}{n} as a single fraction.

$$\frac{2\left(-15n+1\right)n}{n}=\frac{15\left(2n-1\right)}{n}n$$

Express \frac{15\left(2n-1\right)}{n}n as a single fraction.

$$\frac{2\left(-15n+1\right)n}{n}=\frac{15\left(2n-1\right)n}{n}$$

Use the distributive property to multiply 2 by -15n+1.

$$\frac{\left(-30n+2\right)n}{n}=\frac{15\left(2n-1\right)n}{n}$$

Use the distributive property to multiply -30n+2 by n.

$$\frac{-30n^{2}+2n}{n}=\frac{15\left(2n-1\right)n}{n}$$

Use the distributive property to multiply 15 by 2n-1.

$$\frac{-30n^{2}+2n}{n}=\frac{\left(30n-15\right)n}{n}$$

Use the distributive property to multiply 30n-15 by n.

$$\frac{-30n^{2}+2n}{n}=\frac{30n^{2}-15n}{n}$$

Subtract \frac{30n^{2}-15n}{n} from both sides.

$$\frac{-30n^{2}+2n}{n}-\frac{30n^{2}-15n}{n}=0$$

Since \frac{-30n^{2}+2n}{n} and \frac{30n^{2}-15n}{n} have the same denominator, subtract them by subtracting their numerators.

$$\frac{-30n^{2}+2n-\left(30n^{2}-15n\right)}{n}=0$$

Do the multiplications in -30n^{2}+2n-\left(30n^{2}-15n\right).

$$\frac{-30n^{2}+2n-30n^{2}+15n}{n}=0$$

Combine like terms in -30n^{2}+2n-30n^{2}+15n.

$$\frac{-60n^{2}+17n}{n}=0$$

Variable n cannot be equal to 0 since division by zero is not defined. Multiply both sides of the equation by n.

$$-60n^{2}+17n=0$$

Factor out n.

$$n\left(-60n+17\right)=0$$

To find equation solutions, solve n=0 and -60n+17=0.

$$n=0 n=\frac{17}{60}$$

Variable n cannot be equal to 0.

$$n=\frac{17}{60}$$

Steps Using the Quadratic Formula

Variable n cannot be equal to 0 since division by zero is not defined. Multiply both sides of the equation by 30n, the least common multiple of 3,5n,2,n.

$$\frac{1}{3}\left(\frac{1}{5n}-3\right)\times 30n=\frac{1}{2}\left(2-\frac{1}{n}\right)\times 30n$$

To add or subtract expressions, expand them to make their denominators the same. Multiply 3 times \frac{5n}{5n}.

$$\frac{1}{3}\left(\frac{1}{5n}-\frac{3\times 5n}{5n}\right)\times 30n=\frac{1}{2}\left(2-\frac{1}{n}\right)\times 30n$$

Since \frac{1}{5n} and \frac{3\times 5n}{5n} have the same denominator, subtract them by subtracting their numerators.

$$\frac{1}{3}\times \frac{1-3\times 5n}{5n}\times 30n=\frac{1}{2}\left(2-\frac{1}{n}\right)\times 30n$$

Do the multiplications in 1-3\times 5n.

$$\frac{1}{3}\times \frac{1-15n}{5n}\times 30n=\frac{1}{2}\left(2-\frac{1}{n}\right)\times 30n$$

Multiply \frac{1}{3} and 30 to get 10.

$$10\times \frac{1-15n}{5n}n=\frac{1}{2}\left(2-\frac{1}{n}\right)\times 30n$$

Express 10\times \frac{1-15n}{5n} as a single fraction.

$$\frac{10\left(1-15n\right)}{5n}n=\frac{1}{2}\left(2-\frac{1}{n}\right)\times 30n$$

Cancel out 5 in both numerator and denominator.

$$\frac{2\left(-15n+1\right)}{n}n=\frac{1}{2}\left(2-\frac{1}{n}\right)\times 30n$$

Express \frac{2\left(-15n+1\right)}{n}n as a single fraction.

$$\frac{2\left(-15n+1\right)n}{n}=\frac{1}{2}\left(2-\frac{1}{n}\right)\times 30n$$

To add or subtract expressions, expand them to make their denominators the same. Multiply 2 times \frac{n}{n}.

$$\frac{2\left(-15n+1\right)n}{n}=\frac{1}{2}\left(\frac{2n}{n}-\frac{1}{n}\right)\times 30n$$

Since \frac{2n}{n} and \frac{1}{n} have the same denominator, subtract them by subtracting their numerators.

$$\frac{2\left(-15n+1\right)n}{n}=\frac{1}{2}\times \frac{2n-1}{n}\times 30n$$

Multiply \frac{1}{2} and 30 to get 15.

$$\frac{2\left(-15n+1\right)n}{n}=15\times \frac{2n-1}{n}n$$

Express 15\times \frac{2n-1}{n} as a single fraction.

$$\frac{2\left(-15n+1\right)n}{n}=\frac{15\left(2n-1\right)}{n}n$$

Express \frac{15\left(2n-1\right)}{n}n as a single fraction.

$$\frac{2\left(-15n+1\right)n}{n}=\frac{15\left(2n-1\right)n}{n}$$

Use the distributive property to multiply 2 by -15n+1.

$$\frac{\left(-30n+2\right)n}{n}=\frac{15\left(2n-1\right)n}{n}$$

Use the distributive property to multiply -30n+2 by n.

$$\frac{-30n^{2}+2n}{n}=\frac{15\left(2n-1\right)n}{n}$$

Use the distributive property to multiply 15 by 2n-1.

$$\frac{-30n^{2}+2n}{n}=\frac{\left(30n-15\right)n}{n}$$

Use the distributive property to multiply 30n-15 by n.

$$\frac{-30n^{2}+2n}{n}=\frac{30n^{2}-15n}{n}$$

Subtract \frac{30n^{2}-15n}{n} from both sides.

$$\frac{-30n^{2}+2n}{n}-\frac{30n^{2}-15n}{n}=0$$

Since \frac{-30n^{2}+2n}{n} and \frac{30n^{2}-15n}{n} have the same denominator, subtract them by subtracting their numerators.

$$\frac{-30n^{2}+2n-\left(30n^{2}-15n\right)}{n}=0$$

Do the multiplications in -30n^{2}+2n-\left(30n^{2}-15n\right).

$$\frac{-30n^{2}+2n-30n^{2}+15n}{n}=0$$

Combine like terms in -30n^{2}+2n-30n^{2}+15n.

$$\frac{-60n^{2}+17n}{n}=0$$

Variable n cannot be equal to 0 since division by zero is not defined. Multiply both sides of the equation by n.

$$-60n^{2}+17n=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute -60 for a, 17 for b, and 0 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$n=\frac{-17±\sqrt{17^{2}}}{2\left(-60\right)}$$

Take the square root of 17^{2}.

$$n=\frac{-17±17}{2\left(-60\right)}$$

Multiply 2 times -60.

$$n=\frac{-17±17}{-120}$$

Now solve the equation n=\frac{-17±17}{-120} when ± is plus. Add -17 to 17.

$$n=\frac{0}{-120}$$

Divide 0 by -120.

$$n=0$$

Now solve the equation n=\frac{-17±17}{-120} when ± is minus. Subtract 17 from -17.

$$n=-\frac{34}{-120}$$

Reduce the fraction \frac{-34}{-120} to lowest terms by extracting and canceling out 2.

$$n=\frac{17}{60}$$

The equation is now solved.

$$n=0 n=\frac{17}{60}$$

Variable n cannot be equal to 0.

$$n=\frac{17}{60}$$

Steps for Completing the Square

Variable n cannot be equal to 0 since division by zero is not defined. Multiply both sides of the equation by 30n, the least common multiple of 3,5n,2,n.

$$\frac{1}{3}\left(\frac{1}{5n}-3\right)\times 30n=\frac{1}{2}\left(2-\frac{1}{n}\right)\times 30n$$

To add or subtract expressions, expand them to make their denominators the same. Multiply 3 times \frac{5n}{5n}.

$$\frac{1}{3}\left(\frac{1}{5n}-\frac{3\times 5n}{5n}\right)\times 30n=\frac{1}{2}\left(2-\frac{1}{n}\right)\times 30n$$

Since \frac{1}{5n} and \frac{3\times 5n}{5n} have the same denominator, subtract them by subtracting their numerators.

$$\frac{1}{3}\times \frac{1-3\times 5n}{5n}\times 30n=\frac{1}{2}\left(2-\frac{1}{n}\right)\times 30n$$

Do the multiplications in 1-3\times 5n.

$$\frac{1}{3}\times \frac{1-15n}{5n}\times 30n=\frac{1}{2}\left(2-\frac{1}{n}\right)\times 30n$$

Multiply \frac{1}{3} and 30 to get 10.

$$10\times \frac{1-15n}{5n}n=\frac{1}{2}\left(2-\frac{1}{n}\right)\times 30n$$

Express 10\times \frac{1-15n}{5n} as a single fraction.

$$\frac{10\left(1-15n\right)}{5n}n=\frac{1}{2}\left(2-\frac{1}{n}\right)\times 30n$$

Cancel out 5 in both numerator and denominator.

$$\frac{2\left(-15n+1\right)}{n}n=\frac{1}{2}\left(2-\frac{1}{n}\right)\times 30n$$

Express \frac{2\left(-15n+1\right)}{n}n as a single fraction.

$$\frac{2\left(-15n+1\right)n}{n}=\frac{1}{2}\left(2-\frac{1}{n}\right)\times 30n$$

To add or subtract expressions, expand them to make their denominators the same. Multiply 2 times \frac{n}{n}.

$$\frac{2\left(-15n+1\right)n}{n}=\frac{1}{2}\left(\frac{2n}{n}-\frac{1}{n}\right)\times 30n$$

Since \frac{2n}{n} and \frac{1}{n} have the same denominator, subtract them by subtracting their numerators.

$$\frac{2\left(-15n+1\right)n}{n}=\frac{1}{2}\times \frac{2n-1}{n}\times 30n$$

Multiply \frac{1}{2} and 30 to get 15.

$$\frac{2\left(-15n+1\right)n}{n}=15\times \frac{2n-1}{n}n$$

Express 15\times \frac{2n-1}{n} as a single fraction.

$$\frac{2\left(-15n+1\right)n}{n}=\frac{15\left(2n-1\right)}{n}n$$

Express \frac{15\left(2n-1\right)}{n}n as a single fraction.

$$\frac{2\left(-15n+1\right)n}{n}=\frac{15\left(2n-1\right)n}{n}$$

Use the distributive property to multiply 2 by -15n+1.

$$\frac{\left(-30n+2\right)n}{n}=\frac{15\left(2n-1\right)n}{n}$$

Use the distributive property to multiply -30n+2 by n.

$$\frac{-30n^{2}+2n}{n}=\frac{15\left(2n-1\right)n}{n}$$

Use the distributive property to multiply 15 by 2n-1.

$$\frac{-30n^{2}+2n}{n}=\frac{\left(30n-15\right)n}{n}$$

Use the distributive property to multiply 30n-15 by n.

$$\frac{-30n^{2}+2n}{n}=\frac{30n^{2}-15n}{n}$$

Subtract \frac{30n^{2}-15n}{n} from both sides.

$$\frac{-30n^{2}+2n}{n}-\frac{30n^{2}-15n}{n}=0$$

Since \frac{-30n^{2}+2n}{n} and \frac{30n^{2}-15n}{n} have the same denominator, subtract them by subtracting their numerators.

$$\frac{-30n^{2}+2n-\left(30n^{2}-15n\right)}{n}=0$$

Do the multiplications in -30n^{2}+2n-\left(30n^{2}-15n\right).

$$\frac{-30n^{2}+2n-30n^{2}+15n}{n}=0$$

Combine like terms in -30n^{2}+2n-30n^{2}+15n.

$$\frac{-60n^{2}+17n}{n}=0$$

Variable n cannot be equal to 0 since division by zero is not defined. Multiply both sides of the equation by n.

$$-60n^{2}+17n=0$$

Divide both sides by -60.

$$\frac{-60n^{2}+17n}{-60}=\frac{0}{-60}$$

Dividing by -60 undoes the multiplication by -60.

$$n^{2}+\frac{17}{-60}n=\frac{0}{-60}$$

Divide 17 by -60.

$$n^{2}-\frac{17}{60}n=\frac{0}{-60}$$

Divide 0 by -60.

$$n^{2}-\frac{17}{60}n=0$$

Divide -\frac{17}{60}, the coefficient of the x term, by 2 to get -\frac{17}{120}. Then add the square of -\frac{17}{120} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$n^{2}-\frac{17}{60}n+\left(-\frac{17}{120}\right)^{2}=\left(-\frac{17}{120}\right)^{2}$$

Square -\frac{17}{120} by squaring both the numerator and the denominator of the fraction.

$$n^{2}-\frac{17}{60}n+\frac{289}{14400}=\frac{289}{14400}$$

Factor n^{2}-\frac{17}{60}n+\frac{289}{14400}. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(n-\frac{17}{120}\right)^{2}=\frac{289}{14400}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(n-\frac{17}{120}\right)^{2}}=\sqrt{\frac{289}{14400}}$$

Simplify.

$$n-\frac{17}{120}=\frac{17}{120} n-\frac{17}{120}=-\frac{17}{120}$$

Add \frac{17}{120} to both sides of the equation.

$$n=\frac{17}{60} n=0$$

Variable n cannot be equal to 0.

$$n=\frac{17}{60}$$