With amounts measured in gallons, let
x = amount of 65% antifreeze
y = amount of 90% antifreeze
1 gal of the 65% brand contains 0.65 gal of pure antifreeze; x gal would contain 0.65x gal. Similarly, y gal of the 90% brand contains 0.90y gal of pure antifreeze.
To obtain 120 gal of 80% antifreeze solution (which contains 0.80•120 = 96 gal of pure antifreeze), we must have
x + y = 120 … … … … … [total volume of antifreeze solution]
0.65x + 0.90y = 96 … [total volume of pure antifreeze]
Solve the first equation for y :
y = 120 - x
Substitute this into the second equation and solve for x :
0.65x + 0.90 (120 - x) = 96
0.65x + 108 - 0.90x = 96
0.25x = 12
x = 48
Solve for y :
y = 120 - 48
y = 72
Basically you substitute 16 in for x.
f(16) = 3/4(16) - 5
f(16) = 12 - 5
f(16) = 7
In this equation x is equal to 5/26.
