A dehydrated patient needs a 7.9% saline IV. Unfortunately, the hospital only has bags of 3% and 10% saline solutions. How many liters of each of these solutions should be mixed together to yield 3 liters of the desired concentration?
To solve this problem, we can use the concept of the equation of dilution. Let's assume we mix x liters of the 3% saline solution and y liters of the 10% saline solution to obtain 3 liters of the desired 7.9% saline solution.
The equation of dilution can be expressed as follows:
(x * 0.03) + (y * 0.10) = 3 * 0.079
Here, x * 0.03 represents the amount of saline in the 3% solution, and y * 0.10 represents the amount of saline in the 10% solution. On the right side of the equation, (3 * 0.079) represents the amount of saline in the final 7.9% solution.
Simplifying the equation:
0.03x + 0.10y = 0.237
Now, we have an equation with two variables. To solve for x and y, we need another equation.
Given that the total volume of the final solution is 3 liters:
x + y = 3
Now, we have a system of two equations with two variables:
0.03x + 0.10y = 0.237 (Equation 1)
x + y = 3 (Equation 2)
We can solve this system of equations to find the values of x and y, representing the amounts of the 3% and 10% saline solutions, respectively.
One way to solve the system is to use a method called "substitution."
First, rearrange Equation 2 to get:
x = 3 - y
Now, substitute this value of x into Equation 1:
0.03(3 - y) + 0.10y = 0.237
Simplify and solve for y:
0.09 - 0.03y + 0.10y = 0.237
0.07y = 0.237 - 0.09
0.07y = 0.147
y = 0.147 / 0.07
y ≈ 2.1
Now, substitute the value of y back into Equation 2 to solve for x:
x = 3 - 2.1
x ≈ 0.9
Therefore, to obtain 3 liters of the desired 7.9% saline solution, you need to mix approximately 0.9 liters of the 3% saline solution and 2.1 liters of the 10% saline solution.