1. **State the problem**: Solve the system of equations using substitution: $$2I_1 - 3I_2 + I_3 = 4$$ $$3I_1 + 2I_2 - 2I_3 = 2$$ $$4I_1 - I_2 + 3I_3 = 16$$ 2. **Express one variable from the first equation:** Solve for $I_3$ in equation (i): $$I_3 = 4 - 2I_1 + 3I_2$$ 3. **Substitute $I_3$ into equations (ii) and (iii):** In (ii): $$3I_1 + 2I_2 - 2(4 - 2I_1 + 3I_2) = 2$$ Simplify: $$3I_1 + 2I_2 - 8 + 4I_1 - 6I_2 = 2$$ $$7I_1 - 4I_2 = 10$$ In (iii): $$4I_1 - I_2 + 3(4 - 2I_1 + 3I_2) = 16$$ Simplify: $$4I_1 - I_2 + 12 - 6I_1 + 9I_2 = 16$$ $$-2I_1 + 8I_2 = 4$$ 4. **Solve the two-variable system:** $$7I_1 - 4I_2 = 10$$ $$-2I_1 + 8I_2 = 4$$ Multiply second equation by 3.5 to align coefficients of $I_1$: $$7I_1 - 4I_2 = 10$$ $$-7I_1 + 28I_2 = 14$$ Add these: $$(7I_1 - 4I_2) + (-7I_1 + 28I_2) = 10 + 14$$ $$24I_2 = 24$$ Thus, $$I_2 = 1$$ 5. **Find $I_1$ substitute back:** From: $$7I_1 - 4(1) = 10$$ $$7I_1 - 4 = 10$$ $$7I_1 = 14$$ $$I_1 = 2$$ 6. **Find $I_3$ substitute into equation found in step 2:** $$I_3 = 4 - 2(2) + 3(1) = 4 - 4 + 3 = 3$$ 7. **Final values of currents to two decimals:** $$I_1 = 2.00, \, I_2 = 1.00, \, I_3 = 3.00$$