1. **State the problem:** Solve the simultaneous equations: $$x + y + z = 16$$ $$\frac{5}{2}x + y + 10z = 40$$ $$2x + \frac{1}{2}y + 4z = 21$$ 2. **Write the system clearly:** $$\begin{cases} x + y + z = 16 \\ \frac{5}{2}x + y + 10z = 40 \\ 2x + \frac{1}{2}y + 4z = 21 \end{cases}$$ 3. **Use substitution or elimination.** Let's eliminate $y$ by subtracting equations. From equation 1: $y = 16 - x - z$ Substitute $y$ into equations 2 and 3: Equation 2: $$\frac{5}{2}x + (16 - x - z) + 10z = 40$$ Simplify: $$\frac{5}{2}x - x + 16 - z + 10z = 40$$ $$\left(\frac{5}{2}x - x\right) + 16 + 9z = 40$$ $$\frac{3}{2}x + 9z = 24$$ Equation 3: $$2x + \frac{1}{2}(16 - x - z) + 4z = 21$$ Simplify: $$2x + 8 - \frac{1}{2}x - \frac{1}{2}z + 4z = 21$$ $$\left(2x - \frac{1}{2}x\right) + 8 + \left(-\frac{1}{2}z + 4z\right) = 21$$ $$\frac{3}{2}x + \frac{7}{2}z + 8 = 21$$ $$\frac{3}{2}x + \frac{7}{2}z = 13$$ 4. **Now solve the two equations:** $$\begin{cases} \frac{3}{2}x + 9z = 24 \\ \frac{3}{2}x + \frac{7}{2}z = 13 \end{cases}$$ Subtract second from first: $$\left(\frac{3}{2}x + 9z\right) - \left(\frac{3}{2}x + \frac{7}{2}z\right) = 24 - 13$$ $$9z - \frac{7}{2}z = 11$$ $$\frac{18}{2}z - \frac{7}{2}z = 11$$ $$\frac{11}{2}z = 11$$ Multiply both sides by 2: $$\cancel{\frac{11}{2}}z \times 2 = 11 \times 2$$ $$11z = 22$$ Divide both sides by 11: $$\cancel{11}z = \frac{22}{\cancel{11}}$$ $$z = 2$$ 5. **Substitute $z=2$ back into one of the two equations:** $$\frac{3}{2}x + 9(2) = 24$$ $$\frac{3}{2}x + 18 = 24$$ $$\frac{3}{2}x = 6$$ Multiply both sides by $\frac{2}{3}$: $$\cancel{\frac{3}{2}}x \times \frac{2}{3} = 6 \times \frac{2}{3}$$ $$x = 4$$ 6. **Find $y$ using equation 1:** $$x + y + z = 16$$ $$4 + y + 2 = 16$$ $$y + 6 = 16$$ $$y = 10$$ **Final answer:** $$x=4, \quad y=10, \quad z=2$$