really complex word problem! HELP?

Question

A chemist has one solution containing a 16% concentration of acid and a second solution containing a 26% concentration of acid. How many mL of each should be mixed in order to obtain 30mL of a solution containing an 18% concentration of acid?

Answer 1

let x be 16% acid in mL
let y be 26% acid in mL


x(0.16) + y(0.26) = 30(0.18) #Eq.1

x + y = 30
x = 30 - y #Eq.2

sub Eq2 into Eq1

(30 - y) (0.16) + y(0.26) = 30(0.18)
4.8 - 0.16y + 0.26y = 5.4
0.1y = 0.6
y = 6

x + y = 30
x + 6 = 30
x = 24

Therefore u need 24 mL of 16% and 6 mL of 26% to make 30mL of 18%

Answer 2

Suppose you use x ml of 16% concentration of acid and
(30 -x) ml of 26% concentration of acid to obtain
30mL of a solution containing an 18% concentration of acid
so

x * 16/100 + (30 - x) * 26/100 = 30 * 18/100
or 16x + 30 *26 - 26x = 30 * 18
or 10x = 30(26 - 18)
or x = 24
so 24 ml of 16% concentration of acid
and 6ml of 26% concentration of acid

Answer 3

We know that the volumes of both acids, a and b add up to 30
We know that the a amount times its percent mixed with the b amount times its percent will equal them together times that percent
So a+b = 30
and .16a +.26b = .18(a+b)
Simplifying the second equation: 4b=a
Substitute in the first 5a = 30 so a = 24ml of 16% and b = 6ml of 26%.