Problem 49 - Intermediate Algebra

Question

For Problems \(49-60\), solve each problem by setting up and solving an appropriate system of equations. (Objective 3 ) A \(10 \%\)-salt solution is to be mixed with a \(20 \%\)-salt solution to produce 20 gallons of a \(17.5 \%\)-salt solution. How many gallons of the \(10 \%\) solution and how many gallons of the \(20 \%\) solution will be needed?

Step-by-Step Solution

Answer

5 gallons of 10% solution and 15 gallons of 20% solution.

Step 1: Define Variables

Let \( x \) be the number of gallons of the 10% salt solution, and \( y \) be the number of gallons of the 20% salt solution.

Step 2: Set Up Equations Based on Total Volume

The total volume of the mixtures must be 20 gallons, therefore: \[ x + y = 20 \]

Step 3: Set Up Equations Based on Salt Concentration

The mixture's salt content is formed by adding the salt content of each solution. Therefore: \[ 0.10x + 0.20y = 0.175 \times 20 \]

Step 4: Simplify Salt Concentration Equation

Simplify the second equation by calculating the total salt content of the final mixture: \[ 0.10x + 0.20y = 3.5 \]

Step 5: Solve the System of Equations

Use the two equations \( x + y = 20 \) and \( 0.10x + 0.20y = 3.5 \) to find \( x \) and \( y \). Rearrange the first equation for \( y \): \[ y = 20 - x \] Substitute into the second equation: \[ 0.10x + 0.20(20 - x) = 3.5 \] Simplify and solve: \[ 0.10x + 4 - 0.20x = 3.5 \] \[ -0.10x + 4 = 3.5 \] \[ -0.10x = -0.5 \] \[ x = 5 \] Substitute \( x = 5 \) into \( y = 20 - x \), so \[ y = 20 - 5 = 15 \]

Step 6: Conclusion

The solution requires 5 gallons of the 10% solution and 15 gallons of the 20% solution to produce 20 gallons of a 17.5% solution.

Key Concepts

Linear Equations

Linear equations form the backbone for solving many mathematical problems, including systems of equations. In this context, a linear equation is an equation of the first order.
Linear equations are characterized by constant coefficients and the variables have the highest power of one. A simple example of a linear equation is \( ax + b = 0 \).
The main goal when working with linear equations is to find the values of the variables. These equations can be part of a system, which means they work together to find a solution. In the given exercise, two linear equations were used to find how much of each salt solution was necessary.

• The first equation, \( x + y = 20 \), describes the total volume of the solutions needed.
• The second, \( 0.10x + 0.20y = 3.5 \), relates to the necessary concentration of salt within the mixture.

By using substitution or other methods such as elimination or graphing, these equations help determine unknown quantities efficiently.

Solution Concentration

When mixing solutions to achieve a specific concentration, it is important to calculate the amounts of each component correctly. Solution concentration refers to the amount of solute in a given quantity of solvent or solution.
In percentage solution concentration, indicated by a percentage symbol (like 10% or 20% in the problem), it refers to the mass of solute in 100 units of solution. This depiction makes it easy to understand and precisely communicate how much of one substance is present in a solution.
In our exercise, the desired result was a 17.5% salt concentration. The challenge was to mix two available solutions with different concentrations to achieve this. By setting up a system of equations, the problem could be resolved mathematically, ensuring the desired concentration is maintained in the final product.

Problem Solving

Problem-solving is a crucial skill in mathematics, allowing for practical application of concepts like linear equations and solution concentration. It involves understanding the problem, planning a strategy, and systematically working towards a solution.

Let's break down the approach used in the exercise:

• Understanding the Problem: Recognize what's given and what's needed. Here, the goal was to mix solutions to achieve a specific concentration in a specific volume.
• Planning: Identify the equations that embody the problem's conditions, including total mixtures and concentration balance.
• Execution: Use algebraic methods to solve these equations. Substitution or simultaneous methods are crucial for finding satisfactory solutions.
• Conclusion: Verification ensures the solution works, that is, adding 5 gallons of 10% solution and 15 gallons of the 20% solution provides a total of 20 gallons at 17.5% concentration.

This structured approach not only solves the current problem but equips students with a framework for tackling similar mixed-problem challenges in the future.