Solve the system: (1/3)x − (1/3)y = −1 − m; −(1/3)x + y − (1/2)z = −4w; (5/6)z − (1/2)y = 4w + m; x = 30 + z; y = 6w.

Step-by-step solution

1. 1

   Use the two explicit substitutions to cut the unknown count. The last two equations give $x=30+z$ and $y=6w$ outright, so only $z$, $w$ and $m$ remain. Substituting into the first equation:

   $\frac13(30+z)-\frac13(6w)=-1-m\;\Longrightarrow\;10+\frac{z}{3}-2w=-1-m,$

   so

   $m=-11-\frac{z}{3}+2w.$

2. 2

   Reduce the second equation to a link between $w$ and $z$.

   $-\frac13(30+z)+6w-\frac{z}{2}=-4w\;\Longrightarrow\;-10-\frac{z}{3}-\frac{z}{2}+6w=-4w.$

   Since $\tfrac13+\tfrac12=\tfrac56$, this is $-10-\tfrac56z+10w=0$, hence

   $w=1+\frac{z}{12}.$

   This equation involves no $m$ at all, which is what makes the system solvable without heavy elimination.

3. 3

   Reduce the third equation. With $y=6w$,

   $\frac56z-3w=4w+m\;\Longrightarrow\;m=\frac56z-7w.$

4. 4

   Equate the two expressions for $m$ and solve for $z$.

   $\frac56z-7w=-11-\frac{z}{3}+2w\;\Longrightarrow\;\frac76z+11=9w.$

   Substituting $w=1+\tfrac{z}{12}$, so $9w=9+\tfrac34z$:

   $\frac76z+11=9+\frac34z\;\Longrightarrow\;\left(\frac{14}{12}-\frac{9}{12}\right)z=-2\;\Longrightarrow\;\frac{5}{12}z=-2\;\Longrightarrow\;z=-\frac{24}{5}.$

5. 5

   Back-substitute for the other four unknowns.

   $w=1-\frac{24}{60}=\frac35,\quad y=6w=\frac{18}{5},\quad x=30-\frac{24}{5}=\frac{126}{5},\quad m=\frac56\left(-\frac{24}{5}\right)-7\cdot\frac35=-4-\frac{21}{5}=-\frac{41}{5}.$

6. 6

   Verify all five equations exactly. In decimals, $x=25.2$, $y=3.6$, $z=-4.8$, $w=0.6$, $m=-8.2$. Checking with rational arithmetic: $\tfrac13(25.2)-\tfrac13(3.6)=7.2=-1-(-8.2)$ ✓; $-8.4+3.6+2.4=-2.4=-4(0.6)$ ✓; $-4+(-1.8)=-5.8=4(0.6)+(-8.2)$ ✓; $30-4.8=25.2$ ✓; $6(0.6)=3.6$ ✓. All five hold, so the system is consistent with a unique solution.