Problem 166 - Elementary Algebra

Question

Joe's wallet contains \(\$ 1\) and \(\$ 5\) bills worth \(\$ 47\). The number of \(\$ 1\) bills is five more than the number of \(\$ 5\) bills. How many of each bill does he have?

Step-by-Step Solution

Answer

Joe has 12 \(1 bills and 7 \)5 bills.

Step 1 - Define Variables

Let’s denote the number of \(1 bills as x and the number of \)5 bills as y.

Step 2 - Set Up Equations

From the problem, we have two pieces of information that can be turned into equations:1. The total value of the bills: \[ x + 5y = 47 \] 2. The number of \(1 bills is five more than the number of \)5 bills: \[ x = y + 5 \]

Step 3 - Substitute

Substitute the second equation (\[ x = y + 5 \]) into the first equation (\[ x + 5y = 47 \]) to get: \[ (y + 5) + 5y = 47 \]

Step 4 - Simplify and Solve

Simplify the equation: \[ y + 5 + 5y = 47 \] This simplifies to: \[ 6y + 5 = 47 \] Subtract 5 from both sides: \[ 6y = 42 \] Divide by 6: \[ y = 7 \]

Step 5 - Find x

Now use the value of y (number of \(5 bills) in the second equation to find x (number of \)1 bills): \[ x = y + 5 \] Substitute y = 7: \[ x = 7 + 5 = 12 \]

Step 6 - Verification

Verify the solution by checking the total value: \(1 bills: 12 \times 1 = 12\) \(5 bills: 7 \times 5 = 35\) Total: \[ 12 + 35 = 47 \] This is correct.

Key Concepts

Linear Equations

Linear equations are equations where the highest power of the variable is one.
These equations form a straight line when graphed on a coordinate plane.
In our wallet problem, we use the linear equation \( x + 5y = 47 \) to express the total value of the bills.
Here, \( x \) represents the number of \( \$1 \) bills and \( y \) represents the number of \( \$5 \) bills.
Linear equations are easy to solve because they involve basic arithmetic operations.
They are fundamental in algebra and help us model real-world situations, such as finding the number of bills in Joe's wallet.

Substitution Method

The substitution method involves solving one of the equations for one variable, and then substituting this expression into another equation.
This method helps make the equations simpler and easier to manage.
In our example, we are given two equations:
1. \( x + 5y = 47 \)
2. \( x = y + 5 \)
We solve the second equation for \( x \) and substitute it into the first equation:
\( (y + 5) + 5y = 47 \)
This reduces the problem to a single equation with one variable, \( y \).
The substitution method is a powerful tool for solving systems of equations, especially when dealing with real-world word problems.

Problem Solving

Problem solving in algebra involves converting a word problem into mathematical equations.
This process helps us find unknown values systematically.
For Joe's wallet problem, we define our variables:
- \( x \) is the number of \( \$1 \) bills
- \( y \) is the number of \( \$5 \) bills
Next, we set up our equations based on the information given:
- The total value of the bills: \( x + 5y = 47 \)
- The number of \( \$1 \) bills is five more than the number of \( \$5 \) bills: \( x = y + 5 \)
We then solve these equations using substitution and algebraic manipulation.
Finally, we verify our solution by checking the total value of the bills.
Problem-solving skills are crucial for tackling algebraic word problems, helping bridge the gap between words and mathematical expressions.