1. **Problem statement:** Solve the system of linear equations: $$\begin{cases} 2x_1 + 5x_2 + 4x_3 + x_4 = 20 \\ x_1 + 3x_2 + 2x_3 + x_4 = 11 \\ 2x_1 + 10x_2 + 9x_3 + 7x_4 = 40 \\ 3x_1 + 8x_2 + 9x_3 + 2x_4 = 37 \end{cases}$$ 2. **Method:** Use substitution or elimination to find $x_1, x_2, x_3, x_4$. 3. From the second equation: $$x_1 = 11 - 3x_2 - 2x_3 - x_4$$ 4. Substitute $x_1$ into the first, third, and fourth equations: First: $$2(11 - 3x_2 - 2x_3 - x_4) + 5x_2 + 4x_3 + x_4 = 20$$ Simplify: $$22 - 6x_2 - 4x_3 - 2x_4 + 5x_2 + 4x_3 + x_4 = 20$$ $$22 - x_2 - x_4 = 20$$ $$- x_2 - x_4 = -2$$ $$x_2 + x_4 = 2$$ Third: $$2(11 - 3x_2 - 2x_3 - x_4) + 10x_2 + 9x_3 + 7x_4 = 40$$ $$22 - 6x_2 - 4x_3 - 2x_4 + 10x_2 + 9x_3 + 7x_4 = 40$$ $$22 + 4x_2 + 5x_3 + 5x_4 = 40$$ $$4x_2 + 5x_3 + 5x_4 = 18$$ Fourth: $$3(11 - 3x_2 - 2x_3 - x_4) + 8x_2 + 9x_3 + 2x_4 = 37$$ $$33 - 9x_2 - 6x_3 - 3x_4 + 8x_2 + 9x_3 + 2x_4 = 37$$ $$33 - x_2 + 3x_3 - x_4 = 37$$ $$- x_2 + 3x_3 - x_4 = 4$$ 5. Now we have the system: $$\begin{cases} x_2 + x_4 = 2 \\ 4x_2 + 5x_3 + 5x_4 = 18 \\ -x_2 + 3x_3 - x_4 = 4 \end{cases}$$ 6. From the first equation: $$x_4 = 2 - x_2$$ 7. Substitute $x_4$ into the second and third equations: Second: $$4x_2 + 5x_3 + 5(2 - x_2) = 18$$ $$4x_2 + 5x_3 + 10 - 5x_2 = 18$$ $$-x_2 + 5x_3 = 8$$ Third: $$-x_2 + 3x_3 - (2 - x_2) = 4$$ $$-x_2 + 3x_3 - 2 + x_2 = 4$$ $$3x_3 - 2 = 4$$ $$3x_3 = 6$$ $$x_3 = 2$$ 8. Substitute $x_3 = 2$ into $-x_2 + 5x_3 = 8$: $$-x_2 + 5(2) = 8$$ $$-x_2 + 10 = 8$$ $$-x_2 = -2$$ $$x_2 = 2$$ 9. Find $x_4$: $$x_4 = 2 - x_2 = 2 - 2 = 0$$ 10. Find $x_1$ from step 3: $$x_1 = 11 - 3(2) - 2(2) - 0 = 11 - 6 - 4 = 1$$ **Final solution:** $$(x_1, x_2, x_3, x_4) = (1, 2, 2, 0)$$ This corresponds to choice B.