Problem 6 - Algebra 1: Concepts and Skills

Question

In Exercises 6–8, use the following information. You have 2.65 dollar in your pocket. You have a total of 16 coins, with only quarters and dimes. Let q equal the number of quarters and d equal the number of dimes. $$ Complete: ?+?=16 $$

Step-by-Step Solution

Answer

The student has 7 quarters and 9 dimes.

Step 1: Formulate the equations

We have two equations here; the first one is given which is \(q + d = 16\) and it represents the total number of coins. The second equation can be derived from the total value of the coins which is given as 2.65 dollars. Quarters (q) are worth 0.25 dollars each and dimes (d) are worth 0.10 dollars each. This brings us to our second equation, \(0.25q + 0.10d = 2.65\).

Step 2: Solve the equations

These two equations can be solved easily by either substitution or elimination method. It's convenient to multiply the second equation by 100 to get rid of decimals. This gives us \(25q + 10d = 265\). We now multiply the first equation by 10 to facilitate the elimination process, \(10q + 10d = 160\). Subtracting these two new equations gives \(15q = 105\). Solving for \(q\) gives \(q = 7\). Substituting \(q = 7\) back into the first original equation, we find that \(d = 16 - q = 16 - 7 = 9\). So, there are 7 quarters and 9 dimes.

Step 3: Validate the solution

As a last step, we should verify our result by checking that our \(q\) and \(d\) values satisfy both the original equations. For the first equation, \(7 + 9 = 16\), which is the total number of coins, and for the second equation, \(0.25*7 + 0.10*9 = 2.65\), which is the total value of the coins. Both are correct, so our solution is valid.

Key Concepts

System of Equations

A system of equations consists of two or more algebraic equations that share variables. It's like a puzzle where you need to find values that make all the equations true simultaneously. In our coin problem, we have two equations related to the number and value of quarters and dimes:

• \( q + d = 16 \) ensures the total number of coins is correct;
• \( 0.25q + 0.10d = 2.65 \) verifies the total monetary amount is as expected.

These equations are interrelated: the same quarters \( q \) and dimes \( d \) are accounted for in both. Solving the system means finding a pair \( (q, d) \) that satisfies both conditions. This requires employing strategies such as substitution or elimination to simplify and solve the equations.

Substitution Method

The substitution method involves solving one of the equations for one variable and substituting this expression into the other equation. This method is straightforward when one of the equations is simple enough to express one variable in terms of another. In our case, we can start with the equation:

• \( q + d = 16 \) can be rearranged to solve for \( d \) as \( d = 16 - q \).

Once we have \( d \) in terms of \( q \), substitute \( d = 16 - q \) in the second equation:

• \( 0.25q + 0.10(16 - q) = 2.65 \).

Simplify and solve the resulting equation considering only one variable. This technique often yields a swift solution when efficient substitution is possible.

Elimination Method

The elimination method aims to eliminate one variable by adding or subtracting equations, making it easier to solve for the remaining variable. In the coin problem, we modified the equations to facilitate elimination:

• The second equation was multiplied by 100: \( 25q + 10d = 265 \) to clear decimals;
• The first equation by 10: \( 10q + 10d = 160 \).

By subtracting the second equation from the first, we removed the variable \( d \):

• \( (25q + 10d) - (10q + 10d) = 265 - 160 \), resulting in \( 15q = 105 \).

Solve \( 15q = 105 \) for \( q \) to get \( q = 7 \). Then, substitute back to find \( d \). This strategy is very useful for systems where straightforward arithmetic can eliminate a variable quickly.