# Duk Son wants to mix 20% saline solution with a 12% saline solution to make 40 ounces of a solution that is 15% saline. how many ounces of the 12% saline solution will he need?

Nov 7, 2015

He will need $15$ ounces of 20% solution and $25$ ounces of 12% solution

#### Explanation:

If we denote:

$x$- amount of 20% solution (in ounces)
$y$- amount of 12% solution (in ounces)

$x + y = 40$, because the total amount of both solutions is $40$ ounces.

$0.2 x + 0.12 y = 0.15 \cdot 40$, this comes from the calculation of amount of saline in all 3 solutions (the two we mx and the result).

Now we can write the system of equations to solve:

$\left\{\begin{matrix}x + y = 40 \\ 0.2 x + 0.12 y = 0.15 \cdot 40\end{matrix}\right.$

I started calculations with multiplying second equation by $100$ to make the coefficients integer:

$\left\{\begin{matrix}x + y = 40 \\ 20 x + 12 y = 600\end{matrix}\right.$

Now we can multiply first equation by $\left(- 12\right)$ to make $y$ coefficients opposite numbers:

$\left\{\begin{matrix}- 12 x - 12 y = - 480 \\ 20 x + 12 y = 600\end{matrix}\right.$

After adding both equations we get:

$8 x = 120$ which leads to $x = 15$

Now we can calculate $y$ from the firs equation:

$15 + y = 40$

$y = 25$

So the solution of the system of equations is: $\left\{\begin{matrix}x = 15 \\ y = 25\end{matrix}\right.$

The answer is: We need $15$ ounces of 20% solution and $25$ ounces of 12% solution.