This is about three variable systems but i can’t figure out how to solve it.

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.