Question

Pure acid is to be added to a 10% acid solution to obtain 99L of a 20% acid solution. What amounts of each should be used? (answer 1: L of pure acid solution, answer 2: L of 10% acid solution)

Answer 1

Let

x = amount of 100% acid solution (pure acid, no water)
y = amount of 10% acid solution (acid + water)

both amounts are in liters

Answer 2

we have a solution that is 10% acid. We have y liters of this solution, so we have 0.10y liters of pure acid with this 10% solution (the rest is water)

x liters of pure acid are added to this to get x + 0.10y liters total of pure acid

Answer 3

99L of 20% acid ---> 0.20*99 = 19.8

so 19.8 liters of the final solution are pure acid

Answer 4

this means

x + 0.10y = 19.8

is one equation

Answer 5

the two amounts x & y must add to 99 because this is the total amount of liters in the final solution of water and acid

x+y = 99

Answer 6

solve x+y = 99 for y to get y = 99-x

then plug that into the first equation

x + 0.10y = 19.8

x + 0.10(99-x) = 19.8

I'll let you finish up