Determine the amounts of 20% and 50% salt solutions that should be mixed to obtain 300 gallons of 41% salt solution answer

 Let x be the number of gallons of 20% solution to be mixed. 
Then the amount of 50% solution to be mixed must be 300-x. 
.20 (x) + .50 (300-x) = .41 (300) 
.2x + 150 - .5x = 123 
-.3x = - 27 
x = 90 gallons of 20% 
300-x = 210 gallons of 50%

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Idon’t know the answer to this