# 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:

$$a_1-a_3=6.25$$

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You have two equations and three unknowns. Does that sound solvable to you? – Arthur Jan 28 at 23:17
yes, by keeping one variable constant :) – Dirk Jan 28 at 23:18
I posted one of the solution to this equation. One can recognize a pattern in the solution. I wonder about the pattern in the solution. – Dirk Jan 28 at 23:21
You can think of a linear equation with three variables as a plane. If you solve two equations simultaneously, you are finding the intersection of two planes, which is (usually) a line. So you will have a 1D set of solutions, not just 1, – mlainz Jan 28 at 23:24
A set of solutions is true; but some how the dual variables are related with each other according to a implicit scaling factor. – Dirk Jan 28 at 23:28

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 $(a_1,a_2,a_3)=(10,0,3.75)+t(1,1,1)$.
It's vector notation. It means that the solution is a line that you get by starting in $(10,0,3.75)$ an continuing in the direction $(1,1,1)$ – mlainz Jan 28 at 23:34
You can give any value to $t$ and get a solution. For example: t=2, (10,0,3.75)+(2,2,2)=(12,2,5.75). – mlainz Jan 28 at 23:38
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.