Problem 180 - Elementary Algebra

Question

Josie wants to make 10 pounds of trail mix using nuts and raisins, and she wants the total cost of the trail mix to be \(\$ 54 .\) Nuts cost \(\$ 6\) per pound and raisins cost \(\$ 3\) per pound. Solve the system \(\left\\{\begin{array}{l}n+r=10 \\ 6 n+3 r=54\end{array}\right.\) to find \(n,\) the number of pounds of nuts, and \(r,\) the number of pounds of raisins she should use.

Step-by-Step Solution

Answer

8 pounds of nuts and 2 pounds of raisins.

Step 1: Define Variables

Let n be the number of pounds of nuts and r be the number of pounds of raisins.

Step 2: Set Up the Equations

From the problem, we have the following system of linear equations: \[ n + r = 10 \] \[ 6n + 3r = 54 \]

Step 3: Solve for r in the First Equation

Solve the first equation for r: \[ r = 10 - n \]

Step 4: Substitute r in the Second Equation

Substitute \( r = 10 - n \) into the second equation: \[ 6n + 3(10 - n) = 54 \]

Step 5: Simplify the Equation

Expand and simplify the equation: \[ 6n + 30 - 3n = 54 \]Combine like terms: \[ 3n + 30 = 54 \]

Step 6: Solve for n

Isolate n by subtracting 30 from both sides: \[ 3n = 24 \]Divide both sides by 3: \[ n = 8 \]

Step 7: Find r

Substitute \( n = 8 \) back into the equation \( r = 10 - n \): \[ r = 10 - 8 \]\[ r = 2 \]

Step 8: Verify the Solution

Check the values of n and r in both equations to ensure they satisfy the system:For the first equation: \[ n + r = 8 + 2 = 10 \]For the second equation: \[ 6n + 3r = 6(8) + 3(2) = 48 + 6 = 54 \]

Step 9: Final Answer

Josie should use 8 pounds of nuts and 2 pounds of raisins.

Key Concepts

solving systems of equations

Solving systems of equations involves finding values for variables that satisfy multiple linear equations simultaneously. This method is crucial when dealing with real-world problems where multiple constraints apply. For example, Josie wants to make 10 pounds of trail mix by using nuts and raisins. We need to find the specific amount of each ingredient that meets both the weight and cost requirements.
In this problem, we have two equations: one representing the total weight and another representing the total cost. By solving these equations together, we determine the optimal solution that satisfies both conditions.

substitution method

The substitution method is a straightforward way to solve systems of equations. This method involves solving one equation for one variable and then substituting that expression into the other equation. Let's look at our problem with Josie:
First, we solved for one variable in the first equation, representing the total weight:

• The equation was \[ n + r = 10 \]
• We solved for r: \[ r = 10 - n \]

This new expression for r was then substituted into the second equation, representing the total cost:

• The original cost equation was \[ 6n + 3r = 54 \]
• We substituted \[ r = 10 - n \] into the cost equation: \[ 6n + 3(10 - n) = 54 \]

This allowed us to solve for n easily. Once n was found, we plugged it back into the expression for r to get the value of r.

linear equations

Linear equations are the foundation of many algebraic problem-solving situations. They appear in the form \[ ax + by = c \]. In our problem, both given equations were linear:

• \[ n + r = 10 \] represents the total weight, a simple alignment of quantities in pounds.
• \[ 6n + 3r = 54 \] represents the cost equation, combining unit costs and quantities for the total price.

Linear equations are characterized by their graphs, which are straight lines. Solving a system of linear equations gives us the point of intersection of these lines, representing the values of the variables that satisfy both equations. In this case, that point was 8 pounds of nuts and 2 pounds of raisins.

algebraic problem-solving

Algebraic problem-solving is an essential tool in mathematics that helps solve real-world problems through equations. The problem involving Josie's trail mix is a classic example where algebra comes into play.
First, translating the problem into equations: We identified the constraints and used variables to represent unknowns (pounds of nuts and raisins).
Next, simplifying and solving these equations: Using methods like substitution, we solved the system, breaking down complex relationships into manageable steps.
Finally, verifying the solution: Ensuring the solution fits all the original requirements by plugging the values back into the equations.
This structured approach is what makes algebraic problem-solving a powerful tool for handling a variety of real-life situations.