How many gallons of a 50 antifreeze solution must be mixed with 90 gallons of 20% antifreeze to get a mixture that is 40% antifreeze

Try this:

1) note that weight of pure antifreeze before mixing and after mixing is the same. So, if 'x' is weight of pure antifreeze in 50% solution, it is possible to make up equation before mixing: 0.5x+0.2*90.

2) there are 0.2*90=18 gal. of pure antifreeze in the 20% solution. If 'x' gal. is the weight of pure antifreeze in 50% sol. and 18 gal. is the weight of pure antifreeze in 20% sol., it is possible to make up an equation after mixing: 0.4 (x+18).

3) using the both parts: 0.5x+0.2*90=0.4 (x+18) ⇒ x=54 gal. of pure weight.

4) to find 50% solution of 54 gal. pure weight just 54:0.5=108 gal.

Answer: 108 gal.