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### Section 2.3 : Applications of Linear Equations

8. We need 100 liters of a 25% saline solution and we only have a 14% solution and a 60% solution. How much of each should we mix together to get the 100 liters of the 25% solution?

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We’ll start by letting $$x$$ be the amount of the 14% solution we’ll need. This in turn means that we’ll need $$100 - x$$gallons of the 60% solution.

The basic word equation is then,

$\left( \begin{array}{c}{\mbox{Amount of salt}}\\ {\mbox{in 14% solution}}\end{array} \right) + \left( \begin{array}{c}{\mbox{Amount of salt}}\\ {\mbox{in 60% solution}}\end{array} \right) = \left( \begin{array}{c}{\mbox{Amount of salt}}\\ {\mbox{in 25% solution}}\end{array} \right)$

We know that Amount of Salt in Solution = Percentage of Solution X Volume of Solution. This gives the following word equation.

$\left( {0.14} \right)\left( \begin{array}{c}\,\,{\mbox{Volume of}}\\ {\mbox{14% solution}}\end{array} \right) + \left( {0.6} \right)\left( \begin{array}{c}\,\,{\mbox{Volume of}}\\ {\mbox{60% solution}}\end{array} \right) = \left( {0.25} \right)\left( \begin{array}{c}\,\,{\mbox{Volume of}}\\ {\mbox{25% solution}}\end{array} \right)$ Show Step 2

So, plugging all the known information in gives the following equation that we can solve for $$x$$.

\begin{align*}0.14x + 0.6\left( {100 - x} \right) & = 0.25\left( {100} \right)\\ 0.14x + 60 - 0.6x & = 25\\ - 0.46x & = - 35\\ x & = 76.09\end{align*}

So, we’ll need 76.09 liters of the 14% saline solution and 23.91 liters of the 60% saline solution.