Question 146027
Let x=amount of 70% solution needed
Then 400-x=amount of 20% solution needed

Now we know that the amount of pure iodine in the 70% solution (0.70x) plus the amount of pure iodine in the 20% solution (0.20(400-x)) has to equal the amount of pure iodine in the final mixture (400*0.575).  So our equation to solve:

0.70x+0.20(400-x)=400*0.575  get rid of parens (distributive law)

0.70x+80-0.20x=230  subtract 80 from each side
0.70x+80-80-0.20x=230-80  collect like terms
0.50x=150  divide each side by 0.5

x=300 liters-------------------------amount of 70% solution needed

400-x=400-300=100 liters----------------amount of 20% solution needed

CK
300*0.7+100*0.2=400*0.575
210+20=230
230=230

We can also work this problem using two unknowns:

Let x=amount of 70% solution needed
And let y=amount of 20% solution needed

Now we are told that:
x+y=400-------------------------------eq1
0.7x+0.2y=400*0.575---------------------------eq2

From eq1, substitute y=400-x into eq2 and we have the same equation as before


Hope this helps---ptaylor