you have an unlimited supply of 5% saline solution and 8% saline solution. How much of each type would you have to use to create 60 ounces of 7% saline solution? Write and solve a system of equations to model and solve this problem.
amount of 5% solution used --- x
amount of 8% solution used --- 60-x
.05x + .08(60-x) = .07(60)
times 100
5x +8(60-x) = 420
5x + 480 - 8x = 420
-3x = -60
x = 20
20 ounces of the 5% solution and 40 ounces of the 8% solution.
To solve this problem, you can use a system of equations. Let's call the amount of 5% saline solution x ounces and the amount of 8% saline solution y ounces.
The first equation represents the total amount of solution:
x + y = 60
The second equation represents the concentration of saline in the mixture:
(0.05x + 0.08y) / 60 = 0.07
To solve this system of equations, you can use the substitution or elimination method. Let's use the substitution method:
From the first equation, we have x = 60 - y. Substituting this value into the second equation:
(0.05(60 - y) + 0.08y) / 60 = 0.07
Simplifying:
(3 - 0.05y + 0.08y) / 60 = 0.07
(3 + 0.03y) / 60 = 0.07
3 + 0.03y = 60 * 0.07
3 + 0.03y = 4.2
0.03y = 1.2
y = 1.2 / 0.03
y = 40
Substituting the value of y back into the first equation:
x + 40 = 60
x = 60 - 40
x = 20
Therefore, you would need to use 20 ounces of the 5% saline solution and 40 ounces of the 8% saline solution to create 60 ounces of 7% saline solution.