Answer
$40$ mL of $10\%$ acid, $50$ mL of $20\%$ acid, $10$ mL of $40\%$ acid
Work Step by Step
Step 1. Establish the system of equations, assuming $x$ amount of $10\%$ acid, $y$ amount of $20\%$ acid, and $z$ amount of $40\%$ acid are needed:
$\begin{cases} x+y+z=100 \\ 0.1x+0.2y+0.4z=0.18\times100 \\ x+0y-4z=0 \end{cases}$
Step 2. Establish the augmented matrix of the system and use the Gauss Eliminations method:
$\begin{vmatrix} 1 & 1 & 1 & 100 \\0.1 & 0.2 & 0.4 & 18\\1 & 0 & -4 & 0 \end{vmatrix} \begin{array}( .\\10R_2-R_1\to R_2\\R_1-R_3\to R_3\\ \end{array}$
Step 3. Do the operations given on the right side of the matrix.
$\begin{vmatrix} 1 & 1 & 1 & 100 \\0 & 1 & 3 & 80\\0 & 1 & 5 & 100 \end{vmatrix} \begin{array}( .\\.\\R_3-R_2\to R_3\\ \end{array}$
Step 4. Do the operations given on the right side of the matrix.
$\begin{vmatrix} 1 & 1 & 1 & 100 \\0 & 1 & 3 & 80\\0 & 0 & 2 & 20 \end{vmatrix} \begin{array}( .\\.\\.\\ \end{array}$
Step 4. Row 3 gives $2z=20$ and $z=10$mL. Row 2 gives $y+3z=80$ and $y=50$mL, and row 1 gives $x+y+z=100$ and $x=40$mL.
Step 5. The final results are: $40$ mL of $10\%$ acid, $50$ mL of $20\%$ acid, and $10$ mL of $40\%$ acid are needed
