Problem 28 - Understanding Elementary Algebra with Geometry

Question

Solve each of the following verbal problems algebraically. You may use either a one or a two-variable approach. A \(60 \%\) acid solution is to be mixed with an \(80 \%\) acid solution to produce 20 liters of a \(65 \%\) acid solution. How many liters of each solution are needed?

Step-by-Step Solution

Answer

15 liters of 60% acid solution and 5 liters of 80% acid solution.

Step 1: Define the Variables

Let the amount of the 60% acid solution be represented by x liters. Similarly, let the amount of the 80% acid solution be represented by y liters.

Step 2: Set Up the System of Equations

The total volume of the final mixture is 20 liters. Therefore, the equation can be written as:\[ x + y = 20 \]

Step 3: Set Up the Concentration Equation

The total amount of acid in the mixture is equal to the sum of acid from each solution. This gives the equation:\[ 0.60x + 0.80y = 0.65 \times 20 \]

Step 4: Simplify the Concentration Equation

Simplify the right side of the concentration equation:\[ 0.60x + 0.80y = 13 \]

Step 5: Substitution Method

Solve the first equation for y:\[ y = 20 - x \]Substitute this expression into the second equation:\[ 0.60x + 0.80(20 - x) = 13 \]

Step 6: Solve for x

Distribute the 0.80 and solve for x:\[ 0.60x + 16 - 0.80x = 13 \]Combine like terms:\[ -0.20x + 16 = 13 \]Subtract 16 from both sides:\[ -0.20x = -3 \]Divide by -0.20:\[ x = 15 \]

Step 7: Solve for y

Use the equation \( y = 20 - x \) to find y:\[ y = 20 - 15 \]\[ y = 5 \]

Step 8: Conclusion

The solution is 15 liters of the 60% acid solution and 5 liters of the 80% acid solution.

Key Concepts

Algebraic Equations

Algebraic equations are mathematical statements that use variables, constants, and arithmetic operations to express relationships between quantities. In mixture problems, they help us find unknown quantities based on given conditions. For instance, in this exercise, we defined the variables as:

• \(x\) represents the liters of the 60% acid solution
• \(y\) represents the liters of the 80% acid solution

These variables were then used to set up equations based on the problem’s conditions. Understanding how to translate word problems into algebraic equations is crucial for solving mixture problems.

Systems of Equations

A system of equations consists of two or more equations with the same set of variables. To solve for these variables, we need to determine the values that satisfy all equations simultaneously. In this exercise, we created a system of equations to find the amounts of each acid solution:

• The first equation, \(x + y = 20\), represents the total volume of the mixture
• The second equation, \(0.60x + 0.80y = 13\), represents the total acid content

By solving this system, we can find out the amounts of each acid solution needed to achieve the desired concentration.

Concentration Calculations

Concentration calculations are used to determine the amount of substance in a given volume of a mixture. In this exercise, concentration was expressed in percentages. We used the given concentrations (60% and 80%) and the desired concentration (65%) to set up the equations. The key steps included:

• Recognizing that \(0.60x\) represents the acid from the 60% solution
• Recognizing that \(0.80y\) represents the acid from the 80% solution
• Combining these terms to form the concentration equation \(0.60x + 0.80y = 13\)

Understanding how to convert percentages into decimal form and set up equations is essential for solving problems involving mixtures.

Substitution Method

The substitution method is a technique to solve systems of equations by expressing one variable in terms of another and substituting this expression into another equation. In this exercise, after setting up our system of equations, we solved the first equation for \(y\): \(y = 20 - x\). We then substituted this expression into the concentration equation:
\(0.60x + 0.80(20 - x) = 13\).
This substitution allowed us to focus on one variable, simplify the equation, and solve for \(x\). We then used the found value of \(x\) to determine \(y\). The step-by-step process of isolation and substitution helps systematically solve for unknowns in algebraic equations.