Is there any easy way to solve two equations with three unknowns? Is there a way to solve the below simultaneous equations?
One possible solution is $a_1=20.0948$, $a_2=10.0948$, $a_3=6.3448$.
The variables are actually dual variables of the binding constraints.
The system of equations is actually encompassing the relationship between the dual variables.
$$a_1-a_2=10$$
$$a_3-a_2=3.75$$
One can also derive that:
$$a_1-a_3=6.25$$
 A: This system has infinitely many solutions of the form $a_2=t$, $a_1=t+10$ and $a_3=t+3.75$. The solution set is in fact a line in $\mathbb R^3$ given by $(a_1,a_2,a_3)=(10,0,3.75)+t(1,1,1)$.
A: This is classically solved by looking at the augmented matrix associated to your system
$$
\left[\begin{array}{rrr|r}
1 & -1 & 0 & 10 \\
0 & -1 & 1 & \frac{15}{4}
\end{array}\right]
$$
Using elementary row operations allows us to write our system in reduced row echelon form. In our case we have
$$
\DeclareMathOperator{rref}{rref}\rref
\left[\begin{array}{rrr|r}
1 & -1 & 0 & 10 \\
0 & -1 & 1 & \frac{15}{4}
\end{array}\right]
=
\left[\begin{array}{rrrr}
1 & 0 & -1 & \frac{25}{4} \\
0 & 1 & -1 & -\frac{15}{4}
\end{array}\right]
$$
This allows us to read off the solutions "cleanly". The reduced matrix  corresponds to the equation
$$
\begin{bmatrix}
a_1\\ a_2\\ a_3
\end{bmatrix}
=\begin{bmatrix}
25/4+a_3\\ -15/4+a_3\\a_3
\end{bmatrix}
=
\begin{bmatrix}
25/4\\ -15/4\\ 0
\end{bmatrix}
+
a_3
\begin{bmatrix}
1\\ 1\\ 1
\end{bmatrix}\tag{1}
$$
That is, plugging any choice of $a_3$ into (1) gives a solution. Thus the original system has infinitely many solutions.
