Pearson Algebra 1 Common Core, 2011
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Pearson Algebra 1 Common Core, 2011 View details
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Exercise 15 Page 411

Define a variable for each of the two given solutions. What is the amount of insecticide per liter in each solution?

30 % Insecticide Solution to Be Used: 80 L
50 % Insecticide Solution to Be Used: 120 L

Practice makes perfect
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=200

Now, let's make a table where we can see how much insecticide is contained in each solution per liter used.

% of Insecticide Total (L) Part Insecticide (L)
Solution 1 30 x 0.3* x
Solution 2 50 y 0.5* y
Final Mix 42 200 0.42* 200
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.
x+y=200
y=200-x
Now that we've isolated the y, we can solve the system by substitution.
y=200-x & (I) 0.3x+0.5y=0.42* 200 & (II)
y=200-x 0.3x+0.5( 200-x)=0.42* 200
â–Ľ
Solve by substitution
y=200-x 0.3x+100-0.5x=84
y=200-x - 0.2x+100=84
y=200-x - 0.2x=- 16
y=200-x - 15x=- 16
y=200-x x=80
Now, we can substitute this value of x into the first equation to find the value of y.
y=200-x x=80
y=200- 80 x=80
y=120 x=80
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.