mixture word problem

how much fresh water should be mixed to a 30% salt water solution, to obtain 90 gallons of a 23% salt water solution?

let x = amount of fresh water
 
then, 90-x = amount of 30% solution
 
then, salt content of fresh water = 0% of x = 0x = 0
 
salt content of 30% solution = 30% of 90-x = 0.30(90-x)
 
salt content of mixture = 23% of 90 = 0.23(90)
 
so, 0 + 0.30(90-x) = 20.70
 
            27 - 0.30x = 20.70
 
                  -0.30x = -6.30
 
                         x = 21 gallons of water
 
             

abigail,
 
let x be the number of gallons of fresh water (no salt) and y be the number of gallons of 30% salt water.
 
this means that all the salt in the final solution comes from the original 30% solution.  so, let's find the total amount of salt in the final solution:
 
salt content in final = 90 * .23 = 20.7 gallons
 
now, since all  salt comes from the 30% solution, we can find the number of gallons of 30% this way:
 
.3y = 20.7 (since 30% of the original solution is salt)
 
y = 69 gallons in the original 30% solution.
 
we also know that the sum of the gallons of fresh water and 30% solution is 90.
 
x + y = 90
 
x + 69 = 90
 
x = 21 gallons of fresh water need to be added.