HELP ME SOON PLEASE

Question

A chemical company makes two brands of antifreeze. The first brand is 20%

 pure antifreeze, and the second brand is 45%

 pure antifreeze. In order to obtain 80

 gallons of a mixture that contains 35%

 pure antifreeze, how many gallons of each brand of antifreeze must be used?

Answer 1

V1= Volume of the first antifreeze

V2= Volume of the second antifreeze

V1+V2=80 gallon

The equation for the balance of pure antifreeze is:

0.2V1+0.45V2=(0.35)(80)

If we put V2=80-V1 in this equation and solve it for V1, then V1= 32 gallon and then V2=80-32=48 gallon

Answer 2

20x + 45y = 35(80) x=80-y

20(80-y) + 45y = 2800

1600-20y + 45y = 2800

25y = 2800-1600 = 1200

y = 1200/25 = 48 gallons of 45%

x =80-48 = 32 gallons of 20%

32(20) + 48(45) = 2800 = 80(35)