let:
x=ml of 30% alcohol solution
y=ml of 80% alcohol solution.
100 ml of 50% alcohol solution has 50 ml of alcohol and 50 ml of water.
50 % of 100 ml=(50/100)100 ml=50 ml
ml of alcohol in a 30% alcohol solution=0.3 (number of ml of 30% alcohol solution). In this case would be: 03x
ml of alcohol in a 80% alcohol solution=0.3 (number of ml of 30% alcohol solution). In this case would be: 0.8y
We can suggest this system of equations:
x+y=100
0.3x+0.8y=50
We can solve this system of equations by substitution method:
x+y=100 ⇒ y=100-x
0.3x+0.8(100-x)=50
0.3x+80-0.5x=50
0.3x-0.5x=50-80
-0.2x=-30
x=-30/(-0.2)
x=60
y=100-x
y=100-60
y=40
Answer: She should mix 60 ml of a 30% alcohol soluiton with 40 ml of 80% alcohol solution.
