Question

Question #53b61

Answer 1

A. `(a, b, c) = (-5, 0, 3)`

Explanation:

`{(4a+7b = -20),(8a+3c = -31),(6b+3c=9):}`

Subtracting twice the first equation from the second equation, we have

`8a+3c - 2(4a+7b) = -31 - 2(-20)`

`=> 8a+3c - 8a - 14b = -31 + 40`

`=> -14b + 3c = 9`

Subtracting this from the third equation, we get

`6b + 3c - (-14b + 3c) = 9 - 9`

`=> 6b + 3c + 14b - 3c = 0`

`=> 20b = 0`

`=> b = 0`

Substituting this into the third equation, we get

`6(0)+3c = 9`

`=> 3c = 9`

`=> c = 3`

Substituting `b=0` into the first equation, we get

`4a+7(0) = -20`

`=> 4a = -20`

`=> a = -5`

Thus, we find the solution

`{(a = -5),(b = 0),(c = 3):}`

Checking, this solution indeed fulfills the given conditions. Furthermore, the steps we took determine exactly that solution. No other solutions could suffice.