1. **Problem:** Solve the system of linear equations: $$\begin{cases} x + 2y + 2z = 13 \\ 2x + 3y + z = 11 \\ 4x + 0.5z = 2 \end{cases}$$ 2. **Step 1: Express $z$ from the third equation** $$4x + 0.5z = 2 \implies 0.5z = 2 - 4x \implies z = \frac{2 - 4x}{0.5} = 4 - 8x$$ 3. **Step 2: Substitute $z = 4 - 8x$ into the first two equations** - First equation: $$x + 2y + 2(4 - 8x) = 13$$ $$x + 2y + 8 - 16x = 13$$ $$\cancel{x} - 16x + 2y + 8 = 13$$ $$-15x + 2y + 8 = 13$$ $$-15x + 2y = 5$$ - Second equation: $$2x + 3y + (4 - 8x) = 11$$ $$2x + 3y + 4 - 8x = 11$$ $$-6x + 3y + 4 = 11$$ $$-6x + 3y = 7$$ 4. **Step 3: Solve the two-variable system:** $$\begin{cases} -15x + 2y = 5 \\ -6x + 3y = 7 \end{cases}$$ Multiply the first equation by 3 and the second by 2 to eliminate $y$: $$\begin{cases} -45x + 6y = 15 \\ -12x + 6y = 14 \end{cases}$$ Subtract second from first: $$(-45x + 6y) - (-12x + 6y) = 15 - 14$$ $$-45x + 6y + 12x - 6y = 1$$ $$-33x = 1 \implies x = -\frac{1}{33}$$ 5. **Step 4: Substitute $x$ back to find $y$** Using $-6x + 3y = 7$: $$-6\left(-\frac{1}{33}\right) + 3y = 7$$ $$\frac{6}{33} + 3y = 7$$ $$3y = 7 - \frac{6}{33} = \frac{231}{33} - \frac{6}{33} = \frac{225}{33}$$ $$y = \frac{225}{33 \times 3} = \frac{225}{99} = \frac{25}{11}$$ 6. **Step 5: Find $z$ using $z = 4 - 8x$** $$z = 4 - 8\left(-\frac{1}{33}\right) = 4 + \frac{8}{33} = \frac{132}{33} + \frac{8}{33} = \frac{140}{33}$$ 7. **Final answer:** $$\boxed{\left(x, y, z\right) = \left(-\frac{1}{33}, \frac{25}{11}, \frac{140}{33}\right)}$$