Let <em>x</em> and <em>y</em> be the amounts (in mL) of the 4% and 15% solutions, respectively, that the scientist needs to use.
He wants to end up with a 44 mL solution, so
<em>x</em> + <em>y</em> = 44 mL
Each milliliter of 4% solution contains 0.04 mL of acid, while each mL of 15% contains 0.15 mL of acid. The resulting solution should have a concentration of 12%, so that each mL of it contains 0.12 mL of acid. Then the solution will contain
0.04<em>x</em> + 0.15<em>y</em> = 0.12 × (44 mL) = 5.28 mL
of acid.
Solve for <em>x</em> and <em>y</em>. In the first equation, we have <em>y</em> = 44 mL - <em>x</em>, and substituting into the second equation gives
0.04<em>x</em> + 0.15 (44 mL - <em>x</em>) = 5.28 mL
0.04<em>x</em> + 6.6 mL - 0.15<em>x</em> = 5.28 mL
1.32 mL = 0.19<em>x</em>
<em>x</em> ≈ 6.95 mL
==> <em>y</em> ≈ 37.05 mL