Solve (2(1+2k/n))=n

$$( 2 ( 1 + \frac { 2 k } { n } ) ) = n$$

Solve for k

$$k=-\frac{n\left(2-n\right)}{4}$$

$$n\neq 0$$

Steps for Solving Linear Equation

Multiply both sides of the equation by n.

$$2\left(1+\frac{2k}{n}\right)n=nn$$

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

$$2\left(1+\frac{2k}{n}\right)n=n^{2}$$

To add or subtract expressions, expand them to make their denominators the same. Multiply 1 times \frac{n}{n}.

$$2\left(\frac{n}{n}+\frac{2k}{n}\right)n=n^{2}$$

Since \frac{n}{n} and \frac{2k}{n} have the same denominator, add them by adding their numerators.

$$2\times \frac{n+2k}{n}n=n^{2}$$

Express 2\times \frac{n+2k}{n} as a single fraction.

$$\frac{2\left(n+2k\right)}{n}n=n^{2}$$

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

$$\frac{2\left(n+2k\right)n}{n}=n^{2}$$

Cancel out n in both numerator and denominator.

$$2\left(n+2k\right)=n^{2}$$

Use the distributive property to multiply 2 by n+2k.

$$2n+4k=n^{2}$$

Subtract 2n from both sides.

$$4k=n^{2}-2n$$

Divide both sides by 4.

$$\frac{4k}{4}=\frac{n\left(n-2\right)}{4}$$

Dividing by 4 undoes the multiplication by 4.

$$k=\frac{n\left(n-2\right)}{4}$$

Solve for n

$$\left\{\begin{matrix}n=\sqrt{4k+1}+1\text{, }&k\geq -\frac{1}{4}\\n=-\sqrt{4k+1}+1\text{, }&k\neq 0\text{ and }k\geq -\frac{1}{4}\end{matrix}\right.$$