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

Step-by-step solution

1. 1

   Clear the fractions in the simplest equation first. Multiplying $\tfrac13x-\tfrac13y=-1$ by $3$ gives

   $x-y=-3\;\Longrightarrow\;x=y-3.$

   Starting with the equation that has the fewest unknowns keeps the substitutions short.

2. 2

   Link the two expressions for $x$. The fourth equation says $x=30+z$, so

   $30+z=y-3\;\Longrightarrow\;y=33+z.$

   Now every unknown is expressible through $z$ except $w$.

3. 3

   Eliminate $w$ by equating the second and third equations. Multiplying the second by $-1$ gives $\tfrac13x-y+\tfrac12z=w$, and the third gives $\tfrac56z-\tfrac12y=w$. Setting them equal and substituting $x=30+z$:

   $10+\frac13z-y+\frac12z=\frac56z-\frac12y.$

   Because $\tfrac13+\tfrac12=\tfrac56$, the $z$-terms are identical on both sides and cancel completely — that is the step that makes the system solvable in one line.

4. 4

   Solve the surviving equation for $y$.

   $10-y=-\frac12y\;\Longrightarrow\;20-2y=-y\;\Longrightarrow\;y=20.$

   Then from $y=33+z$: $z=20-33=-13$, and from $x=30+z$: $x=17$.

5. 5

   Recover $w$ and verify all four equations. From the third equation,

   $w=\frac56(-13)-\frac12(20)=-\frac{65}{6}-10=-\frac{125}{6}.$

   Checking with exact fractions: $\tfrac13(17)-\tfrac13(20)=-1$ ✓; $-\tfrac13(17)+20-\tfrac12(-13)=\tfrac{125}{6}=-w$ ✓; $\tfrac56(-13)-\tfrac12(20)=-\tfrac{125}{6}=w$ ✓; $30+(-13)=17=x$ ✓.