Systems of Equations

Add the first two equations: $3x + y = 4$, so $y = 4 – 3x$. [1]

Substitute that expression for $y$ into the third equation: $x + (4 – 3x) + z = 6$, so $z = 2 + 2x$. [2]

Substitute into the first equation: $x + 2(4 – 3x) – (2 + 2x) = 3$. This simplifies to $x + 8 – 6x – 2 – 2x = 3$, so $-7x + 6 = 3$ and $x = dfrac{3}{7}$. [2]

$y = 4 – 3cdotdfrac{3}{7} = dfrac{19}{7}$ and $z = 2 + 2cdotdfrac{3}{7} = dfrac{20}{7}$. [1]

Answer: $x = dfrac{3}{7}$, $y = dfrac{19}{7}$, $z = dfrac{20}{7}$.

Total: 6 marks