1. Stating the problem: Solve the equation $$3x - \sqrt{2} = \sqrt{6}$$ for $x$. 2. Isolate the term with $x$: Add $\sqrt{2}$ to both sides: $$3x = \sqrt{6} + \sqrt{2}$$ 3. Solve for $x$ by dividing both sides by 3: $$x = \frac{\sqrt{6} + \sqrt{2}}{3}$$ 4. Show the cancellation explicitly: $$x = \frac{\cancel{3} \left(\frac{\sqrt{6} + \sqrt{2}}{\cancel{3}}\right)}{\cancel{3}} = \frac{\sqrt{6} + \sqrt{2}}{3}$$ 5. This is the exact solution. To approximate, calculate the numerical values: $$\sqrt{6} \approx 2.449, \quad \sqrt{2} \approx 1.414$$ $$x \approx \frac{2.449 + 1.414}{3} = \frac{3.863}{3} \approx 1.288$$ Final answer: $$x = \frac{\sqrt{6} + \sqrt{2}}{3} \approx 1.288$$