For this problem, we can use the relation
color(blue)(bar(ul(|color(white)(a/a) c_1V_1 = c_2V_2 = "amount of solute" color(white)(a/a)|)))" "
Let the "volume of 90 % acid" = xcolor(white)(l) "L"
Then, after mixing, we have (6 + x") L of 40 % acid".
This is made up of "6 L of 15 % acid" and xcolor(white)(l) "L of 90 % acid".
"Moles before = moles after"
c_1V_1 = c_2V_2 + c_3V_3
40 color(red)(cancel(color(black)(%))) × (6 + x) color(red)(cancel(color(black)("L"))) = 15 color(red)(cancel(color(black)(%))) × 6 color(red)(cancel(color(black)("L"))) + 90 color(red)(cancel(color(black)(%))) × x color(red)(cancel(color(black)("L")))
240 + 40x = 90 + 90x
50x = 150
x = 150/50 = 3
So, we add "3 L of 90 % acid" to "6 L of 15 % acid" and get "9 L of 40 % acid".
Check:
3 × 90 + 6 × 15 = 9 × 40
270 + 90 = 360
360 = 360
It checks!