Systems of Linear Equations

From the third equation, $z = 4 – y$. From the second, $y = 2x – 3$. [2]

Substitute into the first: $x + (2x – 3) + (4 – (2x – 3)) = 6$. [1]

This simplifies to $x + 2x – 3 + 4 – 2x + 3 = 6$, so $x + 4 = 6$ and $x = 2$. [2]

$y = 2(2) – 3 = 1$ and $z = 4 – 1 = 3$. Check: $2 + 1 + 3 = 6$, $2(2) – 1 = 3$, and $1 + 3 = 4$. [1]

Answer: $x = 2$, $y = 1$, $z = 3$.

Total: 6 marks