Unit \( 6- \) Systems of Equations and Inequalities Review Guide 18. A chemist has a \( 40 \% \) and a \( 20 \% \) basic solution. How much of each solution should be used to form 300 ml of a \( 30 \% \) acid solution?

To solve this problem, we can set up a system of equations. Let \( x \) be the amount of the 40% solution and \( y \) be the amount of the 20% solution. We know two things: 1. The total volume of the solution must equal 300 ml: \[ x + y = 300 \] 2. The total amount of acid from both solutions must equal the amount of acid in the final 30% solution: \[ 0.4x + 0.2y = 0.3(300) \] Now let's solve these equations step-by-step. From the first equation, we can express \( y \) in terms of \( x \): \[ y = 300 - x \] Substituting this into the second equation: \[ 0.4x + 0.2(300 - x) = 90 \] \[ 0.4x + 60 - 0.2x = 90 \] \[ 0.2x + 60 = 90 \] \[ 0.2x = 30 \] \[ x = 150 \] Now, substituting back to find \( y \): \[ y = 300 - 150 = 150 \] So, the chemist should use 150 ml of the 40% solution and 150 ml of the 20% solution to obtain 300 ml of a 30% acid solution.