Exercise 15 Page 411

Let's start by defining variables that represent the amounts of each of type of solution to be used. & x = number of liters of 30 % insecticide solution & y = number of liters of 50 % insecticide solution Since the chemist wants to mix the solutions, it implies that both x and y add to be 200. x + y=200Now, let's make a table where we can see how much insecticide is contained in each solution per liter used.

Now, since we want to mix Solutions 1 and 2 in such a way that we will get the Final Mix, the blue and green amounts have to add to be the red amount. 0.3* x + 0.5* y = 0.42* 200 Therefore, we have a system of linear equations that models the described situation. x+y=200 & (I) 0.3x+0.5y=0.42* 200 & (II) In order to solve this system, we can use the Substitution Method. First, let's solve the first equation for y.

Now that we've isolated the y, we can solve the system by substitution.

Now, we can substitute this value of x into the first equation to find the value of y.

This implies that the chemist has to mix 80 L of the 30 % insecticide solution with 120 L of the 50 % insecticide solution to obtain 200 L of a 42 % insecticide solution.