Problem 18 - Elementary Algebra

Question

Set up an equation and solve each of the following problems. (Objectives 1-3) We have a \(20 \%\) alcohol solution and a \(50 \%\) solution. How many pints must be used from each to obtain 8 pints of a \(30 \%\) solution?

Step-by-Step Solution

Answer

Use 5.33 pints of 20% solution and 2.67 pints of 50% solution.

Step 1: Define Variables

Let's denote the amount of the 20% solution needed as \( x \) pints and the amount of the 50% solution needed as \( y \) pints.

Step 2: Set Up Equation for Total Volume

The total volume of the solution is 8 pints. Therefore, we have the equation: \( x + y = 8 \).

Step 3: Set Up Equation for Alcohol Content

The total alcohol from both solutions must equal the alcohol in the 8 pints of 30% solution. For this, we have the equation: \( 0.2x + 0.5y = 0.3 \times 8 \).

Step 4: Simplify the Second Equation

Calculate the right-hand side of the second equation: \( 0.3 \times 8 = 2.4 \). So the equation becomes: \( 0.2x + 0.5y = 2.4 \).

Step 5: Solve the System of Equations using Substitution

From \( x + y = 8 \), we express \( y \) in terms of \( x \): \( y = 8 - x \). Substitute \( y = 8 - x \) into \( 0.2x + 0.5y = 2.4 \):\( 0.2x + 0.5(8 - x) = 2.4 \).

Step 6: Solve for \( x \)

Distribute and simplify: \( 0.2x + 4 - 0.5x = 2.4 \). Combine like terms: \( -0.3x + 4 = 2.4 \). Subtract 4 from both sides: \( -0.3x = -1.6 \). Divide by -0.3: \( x = \frac{-1.6}{-0.3} = \frac{16}{3} = 5.33 \) pints.

Step 7: Solve for \( y \)

Substitute \( x = 5.33 \) back into \( y = 8 - x \):\( y = 8 - 5.33 = 2.67 \) pints.

Key Concepts

Linear Equations

Linear equations are mathematical statements where two expressions are set equal, involving constants and variables raised to the first power. They look like simple sentences in math that describe relationships.

In the given problem, we set up two linear equations to relate the amounts of solutions. For example, if we want the total volume to be 8 pints, we wrote the equation:

• \( x + y = 8 \), where \( x \) and \( y \) are the pints for 20% and 50% solutions respectively.

To understand this, think about solving simple puzzles. You have certain pieces (variables like \( x \) and \( y \)) and rules (equations) that you need to fit together to find the solution.

Linear equations form the backbone of algebra and are a powerful tool in breaking down word problems into manageable parts.

System of Equations

When you have more than one equation working with the same set of variables, you have what's called a system of equations. Solving these systems allows you to discover the values of the variables that satisfy all the equations at the same time.

In our exercise, you came across a system of equations:

• \( x + y = 8 \) (for the total volume)
• \( 0.2x + 0.5y = 2.4 \) (for the alcohol content)

By solving these together, you find the exact amount of each solution needed.

One method to solve these is substitution, as shown in the exercise. You start by expressing one variable in terms of the other and replacing it in the second equation. This reduces the problem to a single-variable equation that can be solved straightforwardly. Using substitution helps simplify and solve the system efficiently.

Solution Concentration

Solution concentration tells us how much of a substance is mixed into a solution. It's often expressed as a percentage. In our problem, different concentrations tell us the strength of each alcohol solution.

The aim was to mix a 20% and a 50% alcohol solution to create an 8-pint solution with a 30% concentration. Here’s why concentration matters:

• A solution's concentration denotes the ratio of solute (what’s being dissolved, like alcohol) to solvent.
• By finding the balance of different concentrations, we tailor solutions to desired specifications (30% in this case).

The equation \( 0.2x + 0.5y = 2.4 \) helps achieve this right concentration by ensuring the total amount of alcohol equals 2.4 pints in an 8-pint mixture. Understanding solution concentrations is crucial in many fields such as chemistry and pharmacology, where precise mixing can be vital.