A bleach is \( 5.25 \% \) sodium hypochlorite and swimming pool shock treatment is \( 10.5 \% \) sodium hypochlorite. How much of each item must be mixed to get 6 cups of a mixture that is \( 7.3 \% \) sodium hypochlorito? To get 6 cups of a mixture that is \( 7.3 \% \) sodium hypochlorite, approximately \( \square \) cups of a bleach and approximately \( \square \) cups of a swimming pool shock treatment must be mixed. (Round to two decimal places as needed.)

To solve this problem, we can use a system of equations. Let \( x \) be the cups of bleach and \( y \) be the cups of swimming pool shock treatment. We have the following conditions based on the question: 1. The total volume: \[ x + y = 6 \] 2. The concentration of sodium hypochlorite: \[ 0.0525x + 0.105y = 0.073(6) \] Now, calculate the right side of the concentration equation: \[ 0.073 \times 6 = 0.438 \] So, we can rewrite our second equation as: \[ 0.0525x + 0.105y = 0.438 \] Now, we can solve these equations simultaneously. From the first equation, we express \( y \): \[ y = 6 - x \] Substitute \( y \) in the second equation: \[ 0.0525x + 0.105(6 - x) = 0.438 \] \[ 0.0525x + 0.63 - 0.105x = 0.438 \] \[ -0.0525x + 0.63 = 0.438 \] \[ -0.0525x = 0.438 - 0.63 \] \[ -0.0525x = -0.192 \] \[ x = \frac{-0.192}{-0.0525} \approx 3.66 \] Now substitute \( x \) back into the equation for \( y \): \[ y = 6 - x = 6 - 3.66 \approx 2.34 \] Finally, the approximate amounts to mix are: To get 6 cups of a mixture that is \( 7.3\% \) sodium hypochlorite, approximately \( 3.66 \) cups of bleach and approximately \( 2.34 \) cups of swimming pool shock treatment must be mixed.