Question 234400
Let {{{a}}} = ml of 5% solution to be added 
Let {{{b}}} = ml of 10% solution to be added
given:
Total ml of final solution = {{{50}}} ml
Ml of acetic acid in final solution, {{{.09*50 = 4.5}}} ml
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In words:
( acetic acid in 5% solution) + (acetic acid in 10% solution)/ (total ml of solution) = 9%
(1) {{{(.05a + .1b)/50 = .09}}}
(2) {{{a + b = 50}}}
This is 2 equations and 2 unknowns, so it's solvable
Multiply both sides of (1) by {{{50}}}
(1) {{{.05a + .1b = 4.5}}}
(1) {{{5a + 10b = 450}}}
Now multiply both sides of (2) by {{{5}}} and subtract from (1)
(1) {{{5a + 10b = 450}}}
(2) {{{-5a - 5b = -250}}}
{{{5b = 200}}}
{{{b = 40}}}
and, since
{{{a + b = 50}}}
{{{a = 50 - 40}}}
{{{a = 10}}}
10 ml of 5% solution and 40 ml of 10% solution are needed
check answer:
(1) {{{.05a + .1b)/50 = .09}}}
(1) {{{.05*10 + .1*40)/50 = .09}}}
{{{(.5 + 4)/50 = .09}}}
{{{4.5/50 = .09}}}
{{{.09 = .09}}}
OK