Problem 45 - Precalculus: Mathematical for Calculus

Question

A man has 14 coins in his pocket, all of which are dimes and quarters. If the total value of his change is \(\$ 2.75,\) how many dimes and how many quarters does he have?

Step-by-Step Solution

Answer

5 dimes and 9 quarters.

Step 1: Understanding the problem

We have a total of 14 coins consisting of dimes and quarters. The total value of these coins is $2.75. We want to find out how many dimes and how many quarters are present among these 14 coins.

Step 2: Define Variables

Let \( d \) be the number of dimes and \( q \) be the number of quarters. We know two things: \( d + q = 14 \) (as there are 14 coins total) and \(0.10d + 0.25q = 2.75\) (as the total value is $2.75).

Step 3: Set up the equations

We can set up the following system of equations from the problem:1. \( d + q = 14 \)2. \( 0.10d + 0.25q = 2.75 \)

Step 4: Solve the first equation for one variable

From equation 1, solve for \( d \):\[ d = 14 - q \]

Step 5: Substitute and solve the system

Substitute \( d = 14 - q \) into equation 2:\[ 0.10(14 - q) + 0.25q = 2.75 \]Simplify and solve for \( q \):\[ 1.4 - 0.10q + 0.25q = 2.75 \]\[ 0.15q = 1.35 \]\[ q = \frac{1.35}{0.15} = 9 \]

Step 6: Calculate the remaining variable

Substitute \( q = 9 \) back into the equation \( d = 14 - q \):\[ d = 14 - 9 = 5 \]

Step 7: Conclusion

The man has 5 dimes and 9 quarters.

Key Concepts

Systems of Equations

A system of equations is a set of two or more equations with the same variables. In our coin problem, we have two equations to solve: one equation represents the total number of coins, and the other represents their total value. These two equations together form a system:

• Equation 1: \( d + q = 14 \)
• Equation 2: \( 0.10d + 0.25q = 2.75 \)

Here, \( d \) stands for dimes, and \( q \) stands for quarters. Each equation gives us a piece of information:

• Equation 1 tells us the total count of coins is 14.
• Equation 2 tells us the total monetary value is $2.75.
  Using both equations together helps us find the exact numbers of dimes and quarters. This is the power of a system of equations: it allows us to find specific solutions where the conditions meet the requirements of all equations involved. Solving a system of equations can reveal relationships and give answers that are precise and consistent with all given information.

Linear Equations

Linear equations are equations of the first degree, meaning that the variables are raised only to the power of one and appear linearly throughout the equation. They generally have the form \( ax + by = c \), where \( a \), \( b \), and \( c \) are constants. For our coin problem, both of the equations:

• \( d + q = 14 \)
• \( 0.10d + 0.25q = 2.75 \)

are linear. Here, the coefficients are either 1 (in \( d + q = 14 \)) or the values are decimal (like 0.10 and 0.25) to reflect the financial context with dimes and quarters.
Linear equations are common in problems involving totals or constraints, such as budget calculations or inventory management. They are straightforward to graph and analyze because they form straight lines in a coordinate plane. In many practical applications, like financial planning or engineering, simplifying and solving linear equations helps in making efficient and sound decisions.

Substitution Method

The substitution method is an algebraic method for solving a system of equations, where you solve one of the equations for one variable and substitute that expression into the other equation. This turns a system of two equations into a single equation with one variable, which is easier to solve.
In our example, we start with the equation \( d + q = 14 \). From this, we solve for \( d \):

• \( d = 14 - q \)

This expression for \( d \) can be substituted into the second equation \( 0.10d + 0.25q = 2.75 \):

• \( 0.10(14 - q) + 0.25q = 2.75 \)

Solving this gives the value of \( q \), the number of quarters. Once we know \( q \), we substitute back to find \( d \), the number of dimes.
The substitution method is especially useful when one equation is easily solved for one variable, helping to break down a potentially complex system into simpler steps. This method emphasizes clarity and precision, as each step logically follows from the previous, leading to a satisfying solution.