Problem 69 - College Algebra with Corequisite Support

Question

For the following exercises, use a system of linear equations with two variables and two equations to solve. If a scientist mixed \(10 \%\) saline solution with \(60 \%\) saline solution to get 25 gallons of \(40 \%\) saline solution, how many gallons of \(10 \%\) and \(60 \%\) solutions were mixed?

Step-by-Step Solution

Answer

10 gallons of 10% saline and 15 gallons of 60% saline were mixed.

Step 1: Define the Variables

Let \( x \) represent the gallons of \( 10\% \) saline solution and \( y \) represent the gallons of \( 60\% \) saline solution.

Step 2: Set up the First Equation

The total volume of the solution must be 25 gallons. Thus, the equation is: \( x + y = 25 \).

Step 3: Set up the Second Equation

The total amount of salt from both solutions should equal the amount in 25 gallons of \( 40\% \) saline. Hence, the equation is: \( 0.10x + 0.60y = 0.40 \times 25 \).

Step 4: Simplify the Second Equation

First, calculate the right-hand side: \( 0.40 \times 25 = 10 \). Now the second equation becomes: \( 0.10x + 0.60y = 10 \).

Step 5: Solve the System of Equations

Solve the system of equations:1. \( x + y = 25 \)2. \( 0.10x + 0.60y = 10 \)Multiply the first equation by 0.10 to help eliminate \( x \):\( 0.10x + 0.10y = 2.5 \)Now subtract this from the second equation:\( (0.10x + 0.60y) - (0.10x + 0.10y) = 10 - 2.5 \)\( 0.50y = 7.5 \)Solve for \( y \): \( y = 15 \).

Step 6: Find the Value of x

Substitute \( y = 15 \) back into the first equation:\( x + 15 = 25 \)\( x = 25 - 15 \)\( x = 10 \).

Key Concepts

Linear Equations with Two Variables

Linear equations with two variables are foundational to solving many types of mathematical problems. These equations typically take the form \( ax + by = c \), where \( x \) and \( y \) are variables and \( a \), \( b \), and \( c \) are constants. In simpler terms, they represent straight lines when graphed on a coordinate plane.
In the exercise we are reviewing, the linear equations were formed based on the given problem of mixing solutions. The first equation derived was \( x + y = 25 \), representing the total gallons of saline solution. The simplicity of these equations is that they provide a clear way to find variables efficiently by just needing basic mathematics such as addition, subtraction, and multiplication.

Mixture Problems

Mixture problems often involve combining different substances that result in a new mixture with a desired composition. The challenge is to determine how much of each substance is needed. This could be anything from paint to different concentrations of solutions.
In our particular problem, a scientist is mixing two saline solutions with different concentrations to achieve a specific target mixture. This is a classic example of a mixture problem because it involves determining the right proportions needed to reach the desired concentration, which requires setting up equations that represent the total amount and the desired concentrations in the mixture.

Saline Solution Concentration

Understanding saline solution concentration is key in chemistry and related fields. A saline solution's concentration is the amount of salt dissolved in a particular amount of water, usually expressed as a percentage. In the given problem, the concentrations were \(10\%\), \(60\%\), and \(40\%\).
This concept is essential as it helps in setting up accurate equations for calculations. Converting the percentage concentration to a numerical term helps in utilizing it within an equation to represent the total amount of salt effectively. This ensures that when the solutions are combined, they yield the correct concentration.

Solving Systems of Equations

Solving systems of equations is a critical skill that allows us to find the values of variables that satisfy more than one equation at once. In cases like these, where we deal with two equations and two unknown variables, we often have several methods: substitution, elimination, and graphical representation.
The step-by-step solution involves the elimination method to simplify the solving process. By aligning the equations in such a way that one variable is eliminated, they become easier to solve, leading us to find that \( y = 15 \) gallons of \( 60\% \) saline and \( x = 10 \) gallons of \( 10\% \) saline are needed. This breaking down into simpler steps makes the process manageable and highlights the solution methodology.