How do I set up and solve this problem?

100% acid is to be added to 10% acid solution to obtain 90 liters of 48% solution. (a) What amount of 100% of acid should be used? (b) What amount of 10%?

let the amount of 100% solution to be added be x L

then the amount of the originial 10% solution was 90-x L

.1(90-x) + x = .48(90)
times 100
10(90-x) + 100x = 48(90)
900 - 10x + 100x = 4320
90x = 3420
x = 38

38 L of the 100% acid should be added to
52 L of the 10% solution.