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Coin problems are a common type of algebra problem that revolve around determining the quantity of one or more types of coins, given certain constraints. In this kind of problem, you'll generally deal with two main pieces of information: the total number of coins and the total monetary value of these coins.

For this specific problem, we're looking at dimes and quarters, which are U.S. coins. Dimes are worth 10 cents (\\(0.10) and quarters are worth 25 cents (\\)0.25). The problem states that Jill has a total of $9.20 in dimes and quarters, and there are 68 coins altogether.

To solve coin problems like these, we use systems of equations to mathematically model the relationships and constraints given in the problem. This helps us solve for the unknowns, typically the number of each type of coin.

A system of equations consists of two or more equations with the same set of unknowns. In coin problems, systems of equations are essential for organizing information and finding solutions.

For Jill's coin problem, we create a system of equations based on the information provided:

• Equation 1 represents the total number of coins: \( d + q = 68 \) where \( d \) is the number of dimes and \( q \) is the number of quarters.
• Equation 2 represents the total value of the coins: \( 0.10d + 0.25q = 9.20 \) reflecting the monetary contribution of each kind of coin.

By solving this system, we determine how many of each type of coin Jill has.

Systems of equations can be solved using various methods, but we'll focus on the method of variable elimination next.

The method of variable elimination, also known as the substitution or elimination method, is a powerful technique to solve systems of linear equations. The goal is to eliminate one of the variables by combining the equations, thus making it easier to solve for the other variable.

For Jill's coin problem:

• First, simplify Equation 2 by converting currency from dollars to cents to remove decimals, making it more manageable: \( 10d + 25q = 920 \).
• Second, solve Equation 1 for one variable, say \( d = 68 - q \).
• Third, substitute \( d \) in the transformed Equation 2: \( 10(68 - q) + 25q = 920 \).

This substitution leads to an equation in terms of a single variable (\( q \)). Solving this equation gives us the value of \( q \), which we can use to find \( d \) by substitution.

This method efficiently simplifies the problem, making it easier to solve.

Once we've utilized variable elimination, we continue to find the algebraic solution by solving the simplified equations. Here's how it works for Jill's problem:

• Begin by solving the equation after substitution: let's simplify and solve \( 680 - 10q + 25q = 920 \).
• Combine like terms to get \( 15q = 240 \).
• Divide by 15 to find \( q = 16 \).

Now, substitute back to determine \( d \). Since \( d = 68 - q \), substitute \( q = 16 \) to find \( d = 52 \).

The algebraic approach ensures that all relationships and constraints are respected and gives us an accurate count of each coin. Jill's 52 dimes and 16 quarters match the problem's conditions, confirming our solution is correct.