Your 10- gallon mixture currently consists of 70 percetn liquid A and 30 percent liquid B. How  much liquid B do you need to add to the mixture so that if will contain 42% liquid B?           Let x = liquid B      X+10 x .30    I reach this far and I am lost.
in Algebra 1 Answers by Level 1 User (140 points)

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1 Answer

In a 10 gallon mixture, there are 7 gallons of A and 3 gallons of B.

If we add x gallons of B we will have 7+3+x=10+x gallons of mixture.

In those 10+x gallons there will be x+3 gallons of B so:

(x+3)/(10+x) is the proportion of B in the new mixture. We know this proportion is 42% or 0.42.

Therefore: (x+3)/(10+x)=0.42.

x+3=0.42(10+x)=4.2+0.42x,

x-0.42x=4.2-3=1.2,

0.58x=1.2,

x=1.2/0.58=2.069 gallons approximately.

by Top Rated User (711k points)

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