Solve the system: (1/3)x + (1/3)y = −1; −1/3 + y − (1/2)z = −w; (5/6)z − (1/2)y = w; x = 30 + z.

Step-by-step solution

1. 1

   Clear the first equation and fold in the fourth. Multiplying by $3$:

   $x+y=-3.$

   Substituting $x=30+z$ gives $30+z+y=-3$, so

   $y+z=-33\quad\Longrightarrow\quad y=-33-z.$

   Reading the second equation literally matters here: its leading term is the constant $-\tfrac13$, so $x$ appears in only two of the four equations.

2. 2

   Get two expressions for $w$. Multiplying the second equation by $-1$,

   $w=\frac13-y+\frac12z,$

   while the third equation gives directly

   $w=\frac56z-\frac12y.$

3. 3

   Equate them and clear denominators. Setting the two right-hand sides equal and multiplying by $6$:

   $2-6y+3z=5z-3y\;\Longrightarrow\;2-3y-2z=0\;\Longrightarrow\;3y+2z=2.$

4. 4

   Substitute $y=-33-z$ and solve for $z$.

   $3(-33-z)+2z=2\;\Longrightarrow\;-99-z=2\;\Longrightarrow\;z=-101.$

   Then $y=-33-(-101)=68$ and $x=30+z=-71$.

5. 5

   Recover $w$ and verify all four equations.

   $w=\frac56(-101)-\frac12(68)=-\frac{505}{6}-34=-\frac{709}{6}.$

   Exact checks: $\tfrac13(-71)+\tfrac13(68)=-1$ ✓; $-\tfrac13+68-\tfrac12(-101)=\tfrac{709}{6}=-w$ ✓; $\tfrac56(-101)-\tfrac12(68)=-\tfrac{709}{6}=w$ ✓; $30+(-101)=-71=x$ ✓.