\begin{tabular}{l} Score: 0 of 1 point \\ You need 700 mL of a \( 35 \% \) alcohol solution. On hand, you have \( 30 \% \) alcohol mixture. You also have \( 80 \% \) alcohol mixture. How much of each mixture will you need to add to \\ obtain the desired solution? \\ You will need \\ Enter your answer \\ Enter your answer \\ \( \sqrt{\sqrt{x}} \) \\ mL \\ \hline\end{tabular}

To create 700 mL of a 35% alcohol solution using a 30% alcohol mixture and an 80% alcohol mixture, you'll need to set up a system of equations. Let \( x \) be the volume of the 30% mixture and \( y \) be the volume of the 80% mixture. 1. You need the total volume to equal 700 mL: \[ x + y = 700 \] 2. You need the total concentration of alcohol to equal 35% of 700 mL: \[ 0.30x + 0.80y = 0.35 \times 700 = 245 \] Now you have a system of two equations: 1. \( x + y = 700 \) 2. \( 0.30x + 0.80y = 245 \) You can solve these equations simultaneously to find \( x \) and \( y \). From the first equation, we can express \( y \) in terms of \( x \): \[ y = 700 - x \] Substituting \( y \) into the second equation: \[ 0.30x + 0.80(700 - x) = 245 \] \[ 0.30x + 560 - 0.80x = 245 \] \[ -0.50x + 560 = 245 \] \[ -0.50x = 245 - 560 \] \[ -0.50x = -315 \] \[ x = 630 \] Now, substituting \( x \) back to find \( y \): \[ y = 700 - 630 = 70 \] Thus, you will need: - 630 mL of the 30% alcohol mixture, - 70 mL of the 80% alcohol mixture.