95-2A+B=0,\:B-2C+D=0,\:B-2C+D=0,\:270-41D+23C=0
You asked:
Solve the system of equations:\[\begin{aligned}95 - 2 A + B &= 0\\B - 2 C + D &= 0\\B - 2 C + D &= 0\\270 - 41 D + 23 C &= 0\end{aligned}\]
MathBot Answer:
The system of equations has infinitely many solutions.
$$\begin{aligned}A &= \frac{59 D}{46} + \frac{1645}{46}\\B &= \frac{59 D}{23} - \frac{540}{23}\\C &= \frac{41 D}{23} - \frac{270}{23}\end{aligned}$$
\(D\) is a free variable.