How much of a 12% saline solution should be mixed with a 4% saline solution to obtain 40 milliliters of a 6% saline solution?. Use a system of equations and elimination method to solve. can someone please see if I am on the right track.

Question

12T + 4F=6x40
-4 -4F 4
8T=2x40
8T=80
8T/8=80/8
T=10 ml

Answer 1

First of all, you always define your variables.
I deduced from your work that you:
let the amount of 12% solution used be T ml
let the amount of 4% solution used be F ml

clearly T + F = 40 is our first equation
( don't see that anywhere)

then .12T + .04F = .06(40)
times 100
12T + 4F = 240
or, dividing by 4 ....
3T + F = 60 , again, where is that equation?

(you had 12T + 4T = 6(40)
which is correct but I don't know how you are using it.
the line -4 -4f 4 below it means nothing to me)

so we have 2 equations:
3T + F = 60
T + F = 40
since we are to use elimination, subtract them
2T = 20
T = 10

back in T+F=40
10+F =40
F = 30

They should use 10 ml of the 12%, and 30 ml of the 4% solution.

You did get T = 10, but did not finish the question.
Your instructions were to use elimination, thus you needed two equations.