3a+b-c=21
4a+5b-c=0
4a-16b+20c+0
You asked:
Given the following equations: \[\begin{aligned}3 a + b - c &= 21\\ 4 a + 5 b - c &= 0\end{aligned}\] Evaluate \(4 a - 16 b + 20 c + 0\).
MathBot Answer:
The system of linear equations has infinitely many solutions.
The substitutions are \( a = \frac{4 c}{11} + \frac{105}{11} \), \( b = - \frac{c}{11} - \frac{84}{11} \).
\(c\) is a free variable.
\[4 a - 16 b + 20 c + 0 = \frac{252 c}{11} + \frac{1764}{11}\]