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Mixing Acid Solutions to a Certain Concentration

A chemist requires 100 litres of a solution containing 25% hydrochloric acid (HCl) to be made up from three solutions:
1. A 10% HCl solution
2. A 20% HCl solution
3. A 40% HCl solution
Let  
\[x_1 , \: x_2, \: x_3\]
  be the amounts of solutions 1, 2 and 3 respectively.
We require 100 litres of the mixture so  
\[x_1+x_2+x_3=100\]

In the mixture there are 25 litres of HCl, of which  
\[0.1x_1, \: 0.2x_2, \: 0.4x_3\]
  are from solutions 1, 2, 3 respectively, so  
\[0.1x_1+0.2x_2+0.4x_3=25\]
.
We have the simultaneous equations
\[x_1+x_2+x_3=100\]

\[0.1x_1+0.2x_2+0.4x_3=25\]

We can write this as
\[ \left( \begin{array}{ccc} 1 & 1 & 1 \\ 0.1 & 0.2 & 0.43 \end{array} \right) \begin{pmatrix}x_1 \\ x_2 \end{pmatrix} = \begin{pmatrix}100 \\ 25 \end{pmatrix}\]
.
The rank of the coefficient matrix is 2 but the number of unknowns is 3, so the solution to this system is not unique. Let  
\[x_3=x\]
  then we can write the equations as
\[x_1+x_2=100-x\]

\[0.1x_1+0.2x_2=25-0.4x\]

The solution to this system is  
\[x_1=2x-50, \: x_2=-3x+150\]
.
Because  
\[0 \le x_1, \: x_2, \: x_3=x \le 100\]
  we must have  
\[25 \le x=x_3 \le 50, \: 0 \le x_1 \le 50, \: 0 \le x_2 \le 75\]
.

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