1. **State the problem:** Solve the system of linear equations: $$\begin{cases} x_1 - 4x_2 - 2x_3 = 0 \\ 3x_1 + 5x_2 + 5x_3 = 4 \\ 2x_1 - 5x_2 + 2x_3 = 15 \end{cases}$$ 2. **Method:** We will use the method of elimination or substitution to find $x_1$, $x_2$, and $x_3$. 3. **From the first equation:** $$x_1 = 4x_2 + 2x_3$$ 4. **Substitute $x_1$ into the second and third equations:** Second equation: $$3(4x_2 + 2x_3) + 5x_2 + 5x_3 = 4$$ Simplify: $$12x_2 + 6x_3 + 5x_2 + 5x_3 = 4$$ $$17x_2 + 11x_3 = 4$$ Third equation: $$2(4x_2 + 2x_3) - 5x_2 + 2x_3 = 15$$ Simplify: $$8x_2 + 4x_3 - 5x_2 + 2x_3 = 15$$ $$3x_2 + 6x_3 = 15$$ 5. **Now solve the system:** $$\begin{cases} 17x_2 + 11x_3 = 4 \\ 3x_2 + 6x_3 = 15 \end{cases}$$ 6. **Multiply the second equation by $\cancel{17}$ and the first by $\cancel{3}$ to eliminate $x_2$: ** $$\cancel{17} \times (3x_2 + 6x_3) = 17 \times 15$$ $$\cancel{3} \times (17x_2 + 11x_3) = 3 \times 4$$ Which gives: $$51x_2 + 102x_3 = 255$$ $$51x_2 + 33x_3 = 12$$ 7. **Subtract the second from the first:** $$ (51x_2 + 102x_3) - (51x_2 + 33x_3) = 255 - 12$$ $$ 69x_3 = 243$$ 8. **Solve for $x_3$: ** $$x_3 = \frac{243}{69} = \frac{81}{23}$$ 9. **Substitute $x_3$ back into the second equation:** $$3x_2 + 6 \times \frac{81}{23} = 15$$ $$3x_2 + \frac{486}{23} = 15$$ 10. **Isolate $x_2$: ** $$3x_2 = 15 - \frac{486}{23} = \frac{345}{23} - \frac{486}{23} = -\frac{141}{23}$$ $$x_2 = \frac{-\frac{141}{23}}{3} = -\frac{141}{69} = -\frac{47}{23}$$ 11. **Substitute $x_2$ and $x_3$ into $x_1 = 4x_2 + 2x_3$: ** $$x_1 = 4 \times \left(-\frac{47}{23}\right) + 2 \times \frac{81}{23} = -\frac{188}{23} + \frac{162}{23} = -\frac{26}{23}$$ 12. **Final solution:** $$\boxed{\left(x_1, x_2, x_3\right) = \left(-\frac{26}{23}, -\frac{47}{23}, \frac{81}{23}\right)}$$