Logan wants to mix a \( 19 \% \) acid solution with a \( 41 \% \) acid solution to get 14 L of a \( 37 \% \) acid solution. How many liters of the \( 19 \% \) solution and how many liters of the \( 41 \% \) solution should be mixed? Part: \( 0 / 2 \)

To solve for how many liters of the \( 19\% \) and \( 41\% \) solutions Logan needs to mix to obtain \( 14 \) liters of a \( 37\% \) solution, let’s denote \( x \) as the volume of the \( 19\% \) solution and \( y \) as the volume of the \( 41\% \) solution. We have two equations based on the problem statement: 1. The total volume equation: \[ x + y = 14 \] 2. The total amount of acid equation: \[ 0.19x + 0.41y = 0.37(14) \] Simplifying the second equation: \[ 0.19x + 0.41y = 5.18 \] Now, we can use the first equation to express \( y \) in terms of \( x \): \[ y = 14 - x \] Substituting \( y \) in the second equation gives: \[ 0.19x + 0.41(14 - x) = 5.18 \] Distributing the \( 0.41 \): \[ 0.19x + 5.74 - 0.41x = 5.18 \] Combining like terms: \[ -0.22x + 5.74 = 5.18 \] Subtracting \( 5.74 \) from both sides: \[ -0.22x = 5.18 - 5.74 \] \[ -0.22x = -0.56 \] Dividing by \( -0.22 \): \[ x = \frac{-0.56}{-0.22} \approx 2.545 \text{ L (of the } 19\% \text{ solution)} \] Now, substituting \( x \) back to find \( y \): \[ y = 14 - 2.545 \approx 11.455 \text{ L (of the } 41\% \text{ solution)} \] In summary, Logan should mix approximately \( 2.55 \, \text{L} \) of the \( 19\% \) acid solution and \( 11.45 \, \text{L} \) of the \( 41\% \) acid solution.