Problem 54 - Algebra 2

Question

For a party, Lenora bought several pounds of cashews and several pounds of almonds. The cashews cost \(\$ 8\) per pound, and the almonds cost \(\$ 6\) per pound. Lenora bought a total of 7 pounds and paid a total of \(\$ 48 .\) How many pounds of cashews did she buy? F. 2 pounds G. 3 pounds H. 4 pounds J. 5 pounds

Step-by-Step Solution

Answer

Lenora bought 3 pounds of cashews.

Step 1: Define Variables

Let \( x \) represent the pounds of cashews Lenora bought, and \( y \) represent the pounds of almonds she bought. We need to find the value of \( x \).

Step 2: Set Up Equations

We have two pieces of information:1. The total weight equation: \( x + y = 7 \).2. The total cost equation: \( 8x + 6y = 48 \).

Step 3: Solve the Weight Equation for y

From the first equation, solve for \( y \):\[ y = 7 - x \]

Step 4: Substitute y in the Cost Equation

Substitute \( y = 7 - x \) into the cost equation:\[ 8x + 6(7 - x) = 48 \]

Step 5: Simplify and Solve for x

Expand and simplify the equation:\[ 8x + 42 - 6x = 48 \]Combine like terms and solve for \( x \):\[ 2x + 42 = 48 \]\[ 2x = 6 \]\[ x = 3 \]

Step 6: Verification

Verify by substituting \( x = 3 \) back into the first equation to find \( y \):\[ y = 7 - 3 = 4 \]Check the total cost:\[ 8(3) + 6(4) = 24 + 24 = 48 \]The values satisfy both equations.

Key Concepts

Systems of Equations

When tackling problems involving systems of equations, you're dealing with more than one equation at the same time. This is necessary when you have multiple unknowns to find. In our example with Lenora, there are two key pieces of information concerning the total weight and the total cost.

• Total Weight Equation: Represents the collective weight of the items she bought, here this is given as \( x + y = 7 \), where \( x \) is the pounds of cashews, and \( y \) the pounds of almonds.
• Total Cost Equation: Accounts for the money spent on the items based on their individual costs. This is expressed as \( 8x + 6y = 48 \), based on the cost per pound of cashews and almonds.

These equations together form what is called a system of equations. The goal is to incorporate both equations to solve for the unknowns \( x \) and \( y \). By aligning more than one equation, you can potentially determine precise values for different variables which otherwise would remain unresolved with only one equation.

Linear Equations

Linear equations are the backbone of many algebraic problems, including our system of equations. A linear equation is an equation that makes a straight line when graphed on a coordinate plane.

• Each of the two equations in Lenora's problem is a linear equation. They can be written as equations of a straight line.

The equation \( x + y = 7 \) is a straightforward line equation where the sum of the variables equals a constant value. Similarly, \( 8x + 6y = 48 \) can also be graphed as a line.
Linear equations are characterized by their uniform order of variables (no exponents or fanciful terms) and simplicity, making them easier to work with than their nonlinear counterparts. In our scenario, their simplicity allows us to apply the substitution method to find specific solutions efficiently.

Substitution Method

The substitution method is a common approach to solve systems of equations. It involves isolating one variable in one equation and substituting its value into the other equation.
In our problem, we begin by solving the first equation for one of the variables. Here’s a step-by-step rundown:

• Isolate Variable: From \( x + y = 7 \), we solve for \( y \) which results in \( y = 7 - x \).
• Substitute: Replace \( y \) with \( 7 - x \) in the second equation: \( 8x + 6(7 - x) = 48 \).
• Simplify & Solve: This results in a single equation with \( x \) only, \( 8x + 42 - 6x = 48 \). This is simplified to find \( x \).

This method reduces the complexity of our system by removing one of the variables, making the calculation straightforward. Once you have \( x \), you can quickly determine \( y \) with minimal calculation.

Cost Analysis

Cost analysis is an essential step in many real-world problems, especially those involving budgets or purchases. In algebra, it involves understanding how different costs accumulate based on given variables.

• In Lenora's scenario, each pound of cashews and almonds had a distinct price (\\(8 and \\)6 respectively).
• Cost calculations help us determine how many pounds of each type were bought given a total expenditure.

The cost equation \( 8x + 6y = 48 \) represents this analysis where the total money spent is tied directly to the number of pounds of cashews and almonds.
To verify if a solution is correct after solving the equations, always plug back the values into the cost equation to ensure that the total cost remains consistent with the problem statement.