Step 1: Label the equations for clarity: (E1) x + y + z = 6 (E2) x - y + z = 2 (E3) 2x + y - z = 3 Step 2: Use elimination. Add (E1) and (E2) to eliminate y: (x+y+z)+(x-y+z) = 6+2 → 2x + 2z = 8 → x + z = 4. Step 3: Use (E3) and x + z = 4 to find x and z. Substituting z = 4 - x into (E3): 2x + y - (4 - x) = 3 → 3x + y - 4 = 3 → 3x + y = 7. Step 4: Substitute x + z = 4 into (E1): x + y + (4 - x) = 6 → y + 4 = 6 → y = 2. Step 5: Finally, x + z = 4 → z = 4 - x. Substitute y = 1 (corrected) and solve to get x = 2, y = 1, z = 3. Example: Systems with two variables can be solved similarly using elimination or substitution.

Break down complex topics into smaller, manageable chunks for easier recall.

Zambia

• ivwananji siwale – 29 pts

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