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Algebra · real student question

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.

Question

Solve for xx, yy, zz and ww:

13x+13y=−1,−13+y−12z=−w,\frac13x+\frac13y=-1,\qquad -\frac13+y-\frac12z=-w,
56z−12y=w,x=30+z.\frac56z-\frac12y=w,\qquad x=30+z.

(Note the second equation begins with the constant −13-\tfrac13, not with −13x-\tfrac13x.)

Step-by-step solution

  1. Clear the first equation and fold in the fourth. Multiplying by 33:

    x+y=−3.x+y=-3.

    Substituting x=30+zx=30+z gives 30+z+y=−330+z+y=-3, so

    y+z=−33⟹y=−33−z.y+z=-33\quad\Longrightarrow\quad y=-33-z.

    Reading the second equation literally matters here: its leading term is the constant −13-\tfrac13, so xx appears in only two of the four equations.

  2. Get two expressions for ww. Multiplying the second equation by −1-1,

    w=13−y+12z,w=\frac13-y+\frac12z,

    while the third equation gives directly

    w=56z−12y.w=\frac56z-\frac12y.

  3. Equate them and clear denominators. Setting the two right-hand sides equal and multiplying by 66:

    2−6y+3z=5z−3y  ⟹  2−3y−2z=0  ⟹  3y+2z=2.2-6y+3z=5z-3y\;\Longrightarrow\;2-3y-2z=0\;\Longrightarrow\;3y+2z=2.

  4. Substitute y=−33−zy=-33-z and solve for zz.

    3(−33−z)+2z=2  ⟹  −99−z=2  ⟹  z=−101.3(-33-z)+2z=2\;\Longrightarrow\;-99-z=2\;\Longrightarrow\;z=-101.

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

  5. Recover ww and verify all four equations.

    w=56(−101)−12(68)=−5056−34=−7096.w=\frac56(-101)-\frac12(68)=-\frac{505}{6}-34=-\frac{709}{6}.

    Exact checks: 13(−71)+13(68)=−1\tfrac13(-71)+\tfrac13(68)=-1 ✓; −13+68−12(−101)=7096=−w-\tfrac13+68-\tfrac12(-101)=\tfrac{709}{6}=-w ✓; 56(−101)−12(68)=−7096=w\tfrac56(-101)-\tfrac12(68)=-\tfrac{709}{6}=w ✓; 30+(−101)=−71=x30+(-101)=-71=x ✓.

Answer

x=−71,y=68,z=−101,w=−7096x=-71,\quad y=68,\quad z=-101,\quad w=-\frac{709}{6}

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