Question 6.7.10: MODELING—A Mixture Problem Pat, a pharmacist, needs 500 mill......

First we set up a system of equations. The unknown quantities are the amount of the 5% solution and the amount of the 25% solution that must be used. Let

x = number of mℓ of 5% solution

y = number of mℓ of 25% solution

We know that 500 mℓ of solution are needed. Thus,

x + y = 500

The total amount of phenobarbital in a solution is determined by multiplying the percent of phenobarbital by the number of milliliters of solution. The second equation comes from the fact that

\left ( \begin{matrix} Total  amount  of \\  phenobarbital  in \\ 5 \%  solution \end{matrix} \right ) + \left ( \begin{matrix} total  amount  of\\ phenobarbital  in \\ 25 \% solution \end{matrix} \right ) = \left ( \begin{matrix} total  amount  of \\ phenobarbital \\ in  10 \%  mixture \end{matrix} \right )

0.05x        +        0.25y        =        0.10(500)

or                 0.05x + 0.25y = 50

The system of equations is

x + y = 500

0.05x + 0.25y = 50

Let’s solve this system of equations by using the addition method. There are various ways of eliminating one variable. To obtain integer values in the second equation, we can multiply both sides of the equation by 100. The result will be an x-term of 5x. If we multiply both sides of the first equation by -5, that will result in an x-term of -5x. By following this process, we can eliminate the x-terms from the system.

-5[x + y = 500]      gives      -5x – 5y = -2500

100[0.05x + 0.25y = 50]      gives      5x + 25y = 5000

\frac{20y}{20} = \frac{2500}{20}

y = 125

Now we determine x.

x + y = 500

x + 125 = 500

x = 375

Therefore, 375 mℓ of a 5% phenobarbital solution must be mixed with 125 mℓ of a 25% phenobarbital solution to obtain 500 mℓ of a 10% phenobarbital solution.