MATH MIXTURE PROBLEM

We are dealing with two alcohol solutions here:
Solution A : 10% alcohol
Solution B : 50% alcohol

The question is how much of Solution A is to be added to 2 liters of Solution B to create a new solution (say C) that contains 15% alcohol. Let the unknown quantity of Solution A that is required be x.

Then we can say this:
Alcohol content in x liters of A + Alcohol content in 2 liters of B = Alcohol content in x+2 liters of C
(10% of x) + (50% of 2) = 15% of (x+2)
.1x + 1 = .15x + .3
.05x = .7
x = 14

So the answer is 14 liters.
You can verify this. 10% of 14 liter + 50% of 2 liter = 1.4 + 1 = 2.4 liters.
Combined solution is 14+2 = 16 liters. 15% of 16 = 2.4 liters.

Let the solution that needs to be added is x

Lets make sure we have same volume of alcohol on both sides, i.e. LHS = RHS

10% of x + 50% of 2 ltr= 15% of (x+2)
=> x/10 + 1 = 15(x+2)/100
=> (x+10)/10 = (3x + 6)/20
=> (x+10) *2 = 3x + 6
=> 2x+20 = 3x + 6
=> x = 14 ltr(answer)