Problem 69 - College Algebra

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% solution and 15 gallons of 60% solution were mixed.

Step 1: Define Variables

Let's define the variables for our system of equations. Let \(x\) represent the number of gallons of the 10\(\%\) saline solution and \(y\) represent the number of gallons of the 60\(\%\) saline solution.

Step 2: Set Up Equations

We have two conditions: the total volume of the solution and the concentration of the saline. The first equation comes from the total volume: \(x + y = 25\). The second equation comes from the concentration: \(0.10x + 0.60y = 0.40 \times 25\).

Step 3: Solve for One Variable

First, simplify the concentration equation: \(0.10x + 0.60y = 10\). Solve the first equation \(x + y = 25\) for \(x\) by rearranging it: \(x = 25 - y\).

Step 4: Substitute the Variable

Substitute \(x = 25 - y\) into the concentration equation: \(0.10(25 - y) + 0.60y = 10\). This simplifies the equation to \(2.5 - 0.10y + 0.60y = 10\).

Step 5: Simplify and Solve

Combine like terms: \(2.5 + 0.50y = 10\). Subtract 2.5 from both sides: \(0.50y = 7.5\). Divide by 0.50 to solve for \(y\): \(y = 15\).

Step 6: Find the Second Variable

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

Key Concepts

Systems of Equations

A system of equations is a collection of two or more equations with a shared set of variables. In the given exercise, the scientist's goal is to mix different concentrations of saline solutions to create a final mixture with a desired concentration. The task involves figuring out how much of each solution is needed, using the concept of linear equations. In such systems:

• Each equation represents a linear relationship between variables.
• The solution to the system is a set of values for the variables that satisfies all equations simultaneously.

Here, the two equations are derived from two conditions: the total volume of the solution and its concentration. For instance, one equation ensures that the sum of the mixtures equals 25 gallons, while another ensures that the overall saline concentration in the mixture aligns with what is required. Solving these equations simultaneously provides the quantities of each solution needed.

Saline Solution Concentration

Saline solution concentration refers to the amount of salt present in a liquid mixture, expressed as a percentage. In this exercise, you are tasked with adjusting concentrations of two different saline solutions to achieve a target concentration. Here's how it works:

• A 10\(\%\) saline solution means there are 10 parts of salt per 100 parts of the total solution.
• Similarly, a 60\(\%\) solution has 60 parts of salt per 100 parts of solution.

To reach the desired 40\(\%\) concentration across 25 gallons, the equation \(0.10x + 0.60y = 10\) is used. This equation considers the amount of salt contributed by each solution type, totaling to the required amount for 25 gallons of 40\(\%\) solution. This approach is critical when mixing solutions to achieve a specific concentration in any scientific or medical setting.

Algebraic Problem-Solving

Algebraic problem-solving involves using algebraic methods to logically and systematically resolve problems. In this problem, algebra is used to find the volumes of two saline solutions. The step-by-step approach makes it clearer:

• Define your variables: For example, let \(x\) be the gallons of 10\(\%\) solution, and \(y\) be the gallons of 60\(\%\) solution.
• Set up an equation for total volume and another for concentration.

Each equation is simplified and manipulated to express one variable in terms of the other, making substitution possible. Such techniques are a staple in algebra, enabling the solving of complex real-world problems. In this case, after proper substitution and simplifying the equations, the values \( x = 10 \) and \( y = 15 \) provide the required solution. Algebraic problem-solving is essential in situations that require precise calculations, such as recipe formulations or chemical solution preparations.