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.
108 gal.

200 miles.45 x 3 (charge per day times days)135$205$ - 135$ = 70$0.35x = 70divide both sides by 0.35200