A dehydrated patient needs a 3.82% saline IV. Unfortunately, the hospital only has bags of 4% and 3% saline solutions. How many liters of each of these solutions should be mixed together to yield 5 liters of the desired concentration?
So far I've come up with this:
x= 4% solution; y= 3% solution
x + y = 5
4x + 3(5-x) = 3.82
4x + 15 -3x = 3.82 subtracting 15 from both sides
1x = -11.18
x= -11.18
To find the correct values for x and y, you made a mistake in your calculations. Let's correct that:
Let:
x = liters of 4% saline solution
y = liters of 3% saline solution
We have two equations from the information given:
Equation 1: x + y = 5 (total volume of the solution)
Equation 2: 4x + 3y = 3.82 (desired concentration of saline)
Let's solve this system of equations using substitution.
From Equation 1, we get x = 5 - y.
Substituting this value of x into Equation 2:
4(5 - y) + 3y = 3.82
20 - 4y + 3y = 3.82
20 - y = 3.82
-y = 3.82 - 20
-y = -16.18
To isolate y, multiply by -1 on both sides:
y = 16.18
Now substitute the value of y back into Equation 1 to get the value of x:
x + 16.18 = 5
x = 5 - 16.18
x = -11.18
It seems that there might be an error in the problem statement. Negative volumes are not possible, so we need to reassess the problem.