algerbra homework help needed

Question

how many gallons of 40% antifreeze should be mixed with 10 gallons of 67% antifreeze to obtain a 55% antifreeze mixture?

Answer 1

0.4 X + 6.7 = 0.55(x+10)

0.4X + 6.7 = 0.55X + 5.5

6.7 - 5.5 + 0.4X = 0.55X

1.2 + 0.4X = 0.55x

1.2 = 0.15x

120 = 15x

x = 8

8 gallons of the 40% = 3.2 gallons of anitfreeze

10 gallons of 67% = 6.7 gallons of antifreeze

for a total of...

18 gallons contain 9.9 gallons of antifreeze ---> 9/9 / 18 = 0.55 = 55%