Problem 22 - College Algebra

Question

A chemistry teacher needs to mix a \(30 \%\) salt solution with a \(70 \%\) salt solution to make 20 qt of a \(40 \%\) salt solution. How many quarts of each solution should the teacher mix to get the desired result?

Step-by-Step Solution

Answer

Mix 15 quarts of 30% solution and 5 quarts of 70% solution.

Step 1: Define Variables

Let \( x \) be the number of quarts of the 30% salt solution, and \( y \) be the number of quarts of the 70% salt solution. We need two equations to solve for these variables.

Step 2: Set Up the Equation for Total Volume

We know that the total volume of the two solutions mixed should be 20 quarts. Therefore, we can write the equation: \[ x + y = 20 \]

Step 3: Set Up the Equation for Salt Concentration

The total amount of salt from both solutions should equal the amount of salt in 20 quarts of a 40% solution. So, we set up the equation based on concentration: \[ 0.3x + 0.7y = 0.4 imes 20 \]Which simplifies to: \[ 0.3x + 0.7y = 8 \]

Step 4: Solve the System of Equations

We now solve the system of equations:1. \( x + y = 20 \)2. \( 0.3x + 0.7y = 8 \)From equation 1, express \( y \) in terms of \( x \):\[ y = 20 - x \]Substitute \( y = 20 - x \) into the second equation:\[ 0.3x + 0.7(20 - x) = 8 \]Simplify and solve for \( x \):\[ 0.3x + 14 - 0.7x = 8 \]\[ -0.4x + 14 = 8 \]\[ -0.4x = -6 \]\[ x = 15 \]

Step 5: Solve for the Second Variable

Now we use the value of \( x \) to find \( y \). Substitute \( x = 15 \) back into \( y = 20 - x \):\[ y = 20 - 15 \]\[ y = 5 \]

Step 6: Verify the Solution

Make sure that the solution satisfies both original equations.1. \( x + y = 20 \): checks as \( 15 + 5 = 20 \).2. \( 0.3x + 0.7y = 8 \): checks as \( 0.3(15) + 0.7(5) = 4.5 + 3.5 = 8 \).Both equations hold true, confirming the solution is correct.

Key Concepts

Systems of Equations

When we encounter a problem like the one with the salt solutions, we often use **systems of equations** to find our answers. This involves setting up multiple equations based on the relationships provided in the problem. Each equation will represent a different condition or constraint.

In our exercise, we have:

• One equation for the total volume of liquid, \( x + y = 20 \), which ensures the total mix equals the desired quantity.
• Another equation for the percentage of salt content, \( 0.3x + 0.7y = 8 \), which ensures the concentration of salt is correct.

By solving these equations together, we find the specific amounts of each solution needed. This method of using multiple equations allows us to break down what might seem complex at first into more manageable pieces, enabling us to solve for multiple variables.

Percentage Concentration

Percentage concentration is a measurement of how much of one substance is mixed with another. It's often expressed as a percentage and tells us how heavily concentrated a solution is. Knowing this is crucial when mixing solutions, especially in chemistry.

For this problem, we deal with three different concentrations: the initial two solutions and the final mixture:

• The first solution has a \(30\%\) concentration, or \(0.3\) as a decimal.
• The second solution has a \(70\%\) concentration, or \(0.7\) as a decimal.
• The desired mixture has a \(40\%\) concentration, or \(0.4\) as a decimal.

To achieve the desired concentration, we use the equation \(0.3x + 0.7y = 0.4 \times 20\). We find the combination of the two solutions that produces exactly \(40\%\) in the 20-quart mixture. When dealing with such problems, converting percentages into decimals allows us to easily perform calculations.

Solving Linear Equations

Once systems of equations have been set up using the given conditions of total volume and salt concentration, the next step is solving these equations. **Linear equations** are typically solved using substitution or elimination methods.

In our case, the substitution method is used:

• First, from \( x + y = 20 \), solve for one variable, such as \( y = 20 - x \).
• Substitute \( y = 20 - x \) into the second equation \(0.3x + 0.7y = 8\).
• After substituting, simplify to find \( x = 15 \).
• Finally, substitute \( x = 15 \) back into \( y = 20 - x \) to find \( y = 5 \).

This method allows us to find precise values for both variables. By verifying the solution with the original equations, we ensure our values satisfy both conditions, confirming their correctness.