Solve n=frac{sqrt{k-1}}{3}+m

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

Solve for k

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

$$3n-3m\geq 0$$

Steps for Solving Radical Equations

Swap sides so that all variable terms are on the left hand side.

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

Subtract m from both sides.

$$\frac{\sqrt{k-1}}{3}=n-m$$

Multiply both sides of the equation by 3.

$$\sqrt{k-1}=3n-3m$$

Square both sides of the equation.

$$k-1=9\left(n-m\right)^{2}$$

Add 1 to both sides of the equation.

$$k-1-\left(-1\right)=9\left(n-m\right)^{2}-\left(-1\right)$$

Subtracting -1 from itself leaves 0.

$$k=9\left(n-m\right)^{2}-\left(-1\right)$$

Subtract -1 from 9\left(n-m\right)^{2}.

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

Solve for m

$$m=n-\frac{\sqrt{k-1}}{3}$$

$$k\geq 1$$