1. **State the problem:** Solve the equation $$\sqrt{39 - x} + \sqrt{7 - x} = 8$$ for $$x$$. 2. **Isolate one square root:** Let $$a = \sqrt{39 - x}$$ and $$b = \sqrt{7 - x}$$, so the equation becomes $$a + b = 8$$. 3. **Square both sides:** $$\left(a + b\right)^2 = 8^2$$ $$a^2 + 2ab + b^2 = 64$$ Substitute back the original expressions: $$ (39 - x) + 2\sqrt{(39 - x)(7 - x)} + (7 - x) = 64$$ 4. **Simplify:** $$46 - 2x + 2\sqrt{(39 - x)(7 - x)} = 64$$ 5. **Isolate the square root term:** $$2\sqrt{(39 - x)(7 - x)} = 64 - 46 + 2x$$ $$2\sqrt{(39 - x)(7 - x)} = 18 + 2x$$ Divide both sides by 2: $$\sqrt{(39 - x)(7 - x)} = 9 + x$$ 6. **Square both sides again:** $$ (39 - x)(7 - x) = (9 + x)^2$$ Expand left side: $$273 - 39x - 7x + x^2 = 81 + 18x + x^2$$ Simplify: $$273 - 46x + x^2 = 81 + 18x + x^2$$ 7. **Subtract $$x^2$$ from both sides:** $$273 - 46x = 81 + 18x$$ 8. **Bring terms together:** $$273 - 81 = 18x + 46x$$ $$192 = 64x$$ 9. **Solve for $$x$$:** $$x = \frac{192}{64} = 3$$ 10. **Verify the solution:** Check in the original equation: $$\sqrt{39 - 3} + \sqrt{7 - 3} = \sqrt{36} + \sqrt{4} = 6 + 2 = 8$$ Since it holds true, $$x = 3$$ is the valid solution. **Final answer:** $$\boxed{3}$$