Solve for a, b and c: 10a + 10b + c = 2.9, 400a + 20b + c = 4.0, and 900a + 30b + c = 4.9.

Step-by-step solution

1. 1

   Pick the variable that is cheapest to eliminate. In all three equations $c$ appears with coefficient exactly $1$, so subtracting any two equations removes $c$ with no scaling at all. Eliminating $a$ instead would require multiplying by awkward factors like $40$ and $90$.

2. 2

   Subtract equation 1 from equation 2.

   $(400a + 20b + c) - (10a + 10b + c) = 4.0 - 2.9$

   $390a + 10b = 1.1 \ \Longrightarrow \ 39a + b = 0.11 \qquad (1)$

3. 3

   Subtract equation 2 from equation 3.

   $(900a + 30b + c) - (400a + 20b + c) = 4.9 - 4.0$

   $500a + 10b = 0.9 \ \Longrightarrow \ 50a + b = 0.09 \qquad (2)$

   Now there are two equations in two unknowns, both with $b$-coefficient $1$.

4. 4

   Subtract (1) from (2) to find a.

   $(50a + b) - (39a + b) = 0.09 - 0.11 \ \Longrightarrow \ 11a = -0.02$

   $a = -\frac{0.02}{11} = -\frac{1}{550} \approx -0.0018182$

   Switch to exact fractions here: the decimals do not terminate, and carrying rounded values into the next two substitutions would visibly corrupt $b$ and $c$.

5. 5

   Back-substitute for b. From (1) with $a = -\frac{1}{550}$:

   $b = 0.11 + \frac{39}{550} = \frac{11}{100} + \frac{39}{550} = \frac{121}{1100} + \frac{78}{1100} = \frac{199}{1100} \approx 0.1809091$

6. 6

   Back-substitute for c and verify all three equations. From the first equation:

   $c = 2.9 - 10a - 10b = 2.9 + \frac{1}{55} - \frac{199}{110} = \frac{61}{55} \approx 1.1090909$

   Checking: equation 2 gives $400(-\tfrac{1}{550}) + 20(\tfrac{199}{1100}) + \tfrac{61}{55} = -\tfrac{8}{11} + \tfrac{199}{55} + \tfrac{61}{55} = 4$, and equation 3 gives $4.9$. Exact Gaussian elimination reproduces the same three fractions.