The chemist combines [tex]x[/tex] L of solution A, [tex]y[/tex] of solution B, and [tex]z[/tex] of solution C to get a 60 L solution, which means
[tex]x+y+z=60[/tex]
The chemist used twice as much of solution C as solution A, so
[tex]z=2x[/tex]
This mixture contains 29% acid, which means 29% of the total 60 L, or 17.4 L, is acid. For every liter of solution A, there are 0.15 L of acid. Similarly, every liter of solution B contributes 0.05 L of acid, and solution C contributes 0.40 L. This means we have
[tex]0.15x+0.05y+0.40z=17.4[/tex]
So the system you have to solve is
[tex]\begin{cases}x+y+z=60\\0.15x+0.05y+0.4z=17.4\\z=2x\end{cases}[/tex]
Substitute [tex]z=2x[/tex] into the first two equations:
[tex]\begin{cases}x+y+2x=60\\0.15x+0.05y+0.4(2x)=17.4\end{cases}\implies\begin{cases}3x+y=60\\0.95x+0.05y=17.4\end{cases}[/tex]
Write the first equation as
[tex]3x+y=60\implies y=60-3x[/tex]
and substitute this into the other equation:
[tex]0.95x+0.05(60-3x)=17.4\implies0.8x+3=17.4\implies0.8x=14.4\implies x=18[/tex]
We can solve for [tex]z[/tex] at this point:
[tex]z=2x\implies z=36[/tex]
Then solve for [tex]y[/tex]:
[tex]y=60-3x\implies y=6[/tex]
So the chemist used 18 L of solution A, 6 L of solution B, and 36 L of solution C.
