Solve for x, y, z and w: (1/3)x − (1/3)y = −1, −(1/3)x + (1/2)y = −w, (1/3)z = w, and x = 30 + z.

Step-by-step solution

1. 1

   Clear the fractions in the first equation. Multiplying by $3$:

   $x - y = -3 \ \Longrightarrow \ y = x + 3$

   Four equations in four unknowns look daunting, but two of them ($\tfrac13 z = w$ and $x = 30 + z$) are already explicit substitutions, so the real work is small.

2. 2

   Express everything in terms of z. From $x = 30 + z$ and $y = x + 3$:

   $x = 30 + z, \qquad y = 33 + z, \qquad w = \frac{z}{3}$

   Every unknown is now a function of the single variable $z$ — only the second equation is still unused, and it will pin $z$ down.

3. 3

   Substitute into the remaining equation.

   $-\frac{1}{3}(30 + z) + \frac{1}{2}(33 + z) = -\frac{z}{3}$

   $-10 - \frac{z}{3} + \frac{33}{2} + \frac{z}{2} = -\frac{z}{3}$

4. 4

   Cancel and solve for z. The $-\frac{z}{3}$ appears identically on both sides and cancels, leaving

   $-10 + \frac{33}{2} + \frac{z}{2} = 0 \ \Longrightarrow \ \frac{z}{2} = 10 - \frac{33}{2} = -\frac{13}{2} \ \Longrightarrow \ z = -13$

   That cancellation is what makes the system solvable in one line rather than requiring elimination across all four equations.

5. 5

   Back-substitute for the other three unknowns.

   $x = 30 + (-13) = 17, \qquad y = 33 + (-13) = 20, \qquad w = \frac{-13}{3} = -\frac{13}{3}$

6. 6

   Verify all four equations. $\tfrac13(17) - \tfrac13(20) = -1 \ \checkmark$; $-\tfrac13(17) + \tfrac12(20) = -\tfrac{17}{3} + 10 = \tfrac{13}{3} = -w \ \checkmark$; $\tfrac13(-13) = -\tfrac{13}{3} = w \ \checkmark$; $30 + (-13) = 17 = x \ \checkmark$. All four hold exactly.