Solve the system x + y = z, 9.5 − 25.33333x − 55z = 0, and 3.5 − 3y − 55z = 0.

Step-by-step solution

1. 1

   Put the last two equations in standard form. Moving the constants to the right:

   $25.33333x + 55z = 9.5, \qquad 3y + 55z = 3.5$

   Both now have the unknowns on the left, which makes the elimination bookkeeping much less error-prone than working with the mixed signs as given.

2. 2

   Eliminate z using the first equation. Substituting $z = x + y$ into each:

   $25.33333x + 55(x+y) = 9.5 \ \Longrightarrow \ 80.33333x + 55y = 9.5$

   $3y + 55(x+y) = 3.5 \ \Longrightarrow \ 55x + 58y = 3.5$

   Using $z = x + y$ as a substitution rather than as a third row is what keeps this a $2\times 2$ problem.

3. 3

   Solve the 2×2 system. From the second equation, $x = \dfrac{3.5 - 58y}{55}$. Substituting into the first and multiplying through by $55$:

   $80.33333(3.5 - 58y) + 3025y = 522.5$

   $281.16666 - 4659.3331y + 3025y = 522.5 \ \Longrightarrow \ -1634.3331y = 241.33334$

4. 4

   Find y, then x, then z.

   $y = \frac{241.33334}{-1634.3331} \approx -0.147665$

   $55x = 3.5 - 58(-0.147665) = 12.06459 \ \Longrightarrow \ x \approx 0.219356$

   $z = x + y \approx 0.219356 - 0.147665 = 0.071691$

   Carry six significant figures throughout: the coefficient $1634.33$ is large, so early rounding of $y$ shows up in the fourth digit of $x$.

5. 5

   Recognise the repeating decimal for an exact answer. The coefficient $25.33333$ is almost certainly $\tfrac{76}{3} = 25.\overline{3}$. Redoing the elimination with exact fractions gives

   $x = \frac{2151}{9806}, \qquad y = -\frac{724}{4903}, \qquad z = \frac{703}{9806}$

   which evaluate to $0.2193555$, $-0.1476647$, $0.0716908$ — matching the decimal solution to seven digits.

6. 6

   Check all three equations. With $x = 0.219356$, $y = -0.147665$, $z = 0.071691$: $x + y = 0.071691 = z \ \checkmark$; $9.5 - 25.33333(0.219356) - 55(0.071691) = 9.5 - 5.5570 - 3.9430 \approx 0 \ \checkmark$; $3.5 - 3(-0.147665) - 55(0.071691) = 3.5 + 0.4430 - 3.9430 \approx 0 \ \checkmark$.