Solve the system 100a + 10b + c = 5.9, 400a + 20b + c = 8.4, 900a + 30b + c = 10.4 for a, b and c.

Step-by-step solution

1. 1

   Notice the structure before grinding. The coefficients are $x^{2}$, $x$, $1$ at $x=10,20,30$, so this system is exactly "fit $y=ax^{2}+bx+c$ through $(10,5.9)$, $(20,8.4)$, $(30,10.4)$". Because $c$ appears with coefficient $1$ in all three equations, subtracting consecutive pairs eliminates it immediately.

2. 2

   Subtract equation 1 from equation 2.

   $(400-100)a+(20-10)b=8.4-5.9\;\Longrightarrow\;300a+10b=2.5\;\Longrightarrow\;30a+b=0.25.$

3. 3

   Subtract equation 2 from equation 3.

   $(900-400)a+(30-20)b=10.4-8.4\;\Longrightarrow\;500a+10b=2\;\Longrightarrow\;50a+b=0.2.$

   The two reduced equations differ only in the $a$-coefficient, so one more subtraction finishes $a$.

4. 4

   Solve for $a$, then $b$, then $c$.

   $(50a+b)-(30a+b)=0.2-0.25\;\Longrightarrow\;20a=-0.05\;\Longrightarrow\;a=-0.0025.$

   Substituting into $30a+b=0.25$: $-0.075+b=0.25$, so $b=0.325$. Substituting both into the first original equation: $-0.25+3.25+c=5.9$, so $c=2.9$.

5. 5

   Verify all three equations. With $a=-0.0025$, $b=0.325$, $c=2.9$: $100(-0.0025)+10(0.325)+2.9=-0.25+3.25+2.9=5.9$ ✓; $400(-0.0025)+20(0.325)+2.9=-1+6.5+2.9=8.4$ ✓; $900(-0.0025)+30(0.325)+2.9=-2.25+9.75+2.9=10.4$ ✓. Solving the same system with a linear-algebra routine returns $(-0.0025,\,0.325,\,2.9)$ as well. The negative $a$ means the fitted parabola opens downward, consistent with the data rising by $2.5$ then only $2.0$.