A \( 20 \% \) acid solution is mixed with a \( 70 \% \) acid solution to get 50 liters of a \( 40 \% \) solution. Which of the following expressions is equal to \( 4(50) \) ? \( 2 x+7 y \) \( x+y \) \( 0.2 x+0.7 y \)

To solve this problem, let's denote \( x \) as the amount of the \( 20\% \) acid solution and \( y \) as the amount of the \( 70\% \) acid solution. The first equation we can set up based on the total volume is: \[ x + y = 50 \] Next, for the acid concentration, we can set up the second equation as follows: \[ 0.2x + 0.7y = 0.4(50) \] Now, since \( 0.4(50) = 20 \), we rewrite the second equation: \[ 0.2x + 0.7y = 20 \] Now we have a system of equations: 1. \( x + y = 50 \) 2. \( 0.2x + 0.7y = 20 \) To find which expression equals \( 4(50) \), which is \( 200 \), let's evaluate the expressions: 1. \( 2x + 7y \): We can express \( y \) in terms of \( x \) using the first equation, which gives us \( y = 50 - x \). Then, substituting this into the expression gives \( 2x + 7(50 - x) = 2x + 350 - 7x = 350 - 5x \). 2. \( x + y \): This simplifies directly to \( 50 \), which does not match. 3. \( 0.2x + 0.7y \): As mentioned, this equals \( 20 \). Clearly, none of these expressions directly equate to \( 200 \). However, through substitution and manipulation of expressions, you can evaluate combinations to create equivalent forms if additional information were provided. This scenario illustrates how you can set up a system of equations to solve concentration problems, which often appear in chemistry and real-world applications such as mixing solutions or even cooking!