Solve the system a + b + c = 100, 600a + c = 1200, 600b + 599c = 0.

Step-by-step solution

1. 1

   Choose the easiest variable to eliminate. The second equation contains only $a$ and $c$, so it gives $c$ in terms of $a$ with no fractions:

   $c=1200-600a$

   Starting from the sparsest equation keeps the arithmetic small.

2. 2

   Substitute into the third equation to get $b$ in terms of $a$.

   $600b+599(1200-600a)=0\ \Longrightarrow\ 600b+718800-359400a=0$

   $600b=359400a-718800\ \Longrightarrow\ b=599a-1198$

   Both $359400$ and $718800$ are divisible by $600$, so $b$ stays an exact linear expression.

3. 3

   Substitute both into the first equation.

   $a+(599a-1198)+(1200-600a)=100$

4. 4

   Watch every $a$ disappear. Collecting the $a$ terms: $1+599-600=0$. Collecting the constants: $-1198+1200=2$. So the equation reduces to

   $2=100$

   which is false. There is no value of $a$ that repairs it, because $a$ is no longer present.

5. 5

   Interpret the contradiction correctly. A vanished variable with a false leftover means the system is inconsistent: no solution at all. (Had the leftover been a true statement such as $0=0$, the opposite conclusion would apply — infinitely many solutions.) Geometrically, the three planes have no common point.

6. 6

   Sanity-check with a concrete value. Take $a=2$: then $c=1200-1200=0$ and $b=1198-1198=0$, so the second and third equations hold, but $a+b+c=2\neq 100$. Take $a=1$: $c=600$, $b=-599$, and $a+b+c=2$ again. The sum is stuck at $2$ for every $a$, which is exactly what the algebra predicted.