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Solve Stamp Problems: A System Of Equations Guide

Dec 11, 2025 by Andrew McMorgan 50 views

Solve Stamp Problems: A System of Equations Guide

Hey guys, let's dive into a classic word problem that can be solved using a system of equations. Krista was faced with a challenge: figure out how many stamps of each price were bought, given the total number of stamps and the total cost. This is a super common type of math problem, and understanding how to set up and solve the equations is key to mastering it. We're talking about a scenario where you have a total quantity and a total value, and you need to find the breakdown of individual components. This skill is not just useful for stamp collecting homework; it pops up in all sorts of real-world situations, from budgeting to calculating ingredient amounts in recipes.

Setting Up the System of Equations

Krista was given the following information: 45 stamps were purchased for a total of $18.75. Some stamps cost $0.40 each, and others cost $0.55 each. To tackle this, she correctly identified the need for a **system of equations**. A system of equations is basically a set of two or more equations that share the same variables. In this case, we have two unknowns: the number of 40-cent stamps and the number of 55-cent stamps. Let's assign variables: Let '$ x $' represent the number of 40-cent stamps, and let '$ y

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represent the number of 55-cent stamps.

Now, we can translate the word problem into two distinct equations. The first equation will deal with the total number of stamps. Since Krista bought 45 stamps in total, the sum of the 40-cent stamps and the 55-cent stamps must equal 45. So, our first equation is:

$x + y = 45$

This equation is straightforward: the number of stamps of one type plus the number of stamps of the other type equals the total number of stamps. It's like saying, if you have a basket of apples and oranges, and you know the total number of fruits, you can write an equation relating the count of each.

The second equation will deal with the total cost of the stamps. The problem states that the total cost was $18.75. The cost of the 40-cent stamps is the price per stamp ( $0.40) multiplied by the number of those stamps ($ x$). Similarly, the cost of the 55-cent stamps is $0.55$ multiplied by the number of those stamps ( $y$ ). Adding these two costs together must give us the total amount spent. Therefore, our second equation is:

$0.40x + 0.55y = 18.75$

This equation represents the monetary value. The value derived from the cheaper stamps plus the value derived from the more expensive stamps equals the total amount of money spent. It's crucial here to ensure consistency in units. Since the prices are given in dollars (e.g., $0.40), the total cost should also be in dollars ($18.75). Alternatively, you could convert everything to cents (40 cents, 55 cents, and 1875 cents), which sometimes makes the numbers easier to work with, but the principle remains the same.

So, Krista's system of equations, which accurately represents the problem, is:

$x + y = 45$
$0.40x + 0.55y = 18.75$

This setup is the foundation for solving the problem. Without a correctly formed system of equations, any attempt to solve it will likely lead to incorrect answers. It’s all about translating the narrative of the problem into the precise language of mathematics. Remember, every word problem that involves finding two unknown quantities, where you have two distinct relationships between them, is a prime candidate for a system of equations. Pay close attention to the total quantity and the total value (or rate, or sum, etc.) described in the problem – these are your clues for forming the two equations.

Solving the System of Equations

Now that we have Krista's system of equations, let's solve it. There are several methods to solve a system of linear equations, but the most common ones are substitution and elimination. We’ll walk through both, so you can choose the one you find easiest, or the one your teacher prefers. Both methods will lead you to the same correct answer, so don't stress too much about which one to use.

Method 1: Substitution

The substitution method involves solving one of the equations for one variable, and then substituting that expression into the other equation. It's like swapping one piece for another to simplify the puzzle. Let's use the first equation, $x + y = 45$ , because it's simpler.

We can easily solve for $y$ in terms of $x$ :

$y = 45 - x$

Now, we take this expression for $y$ and substitute it into the second equation ( $0.40x + 0.55y = 18.75$ ). Wherever we see ' $y$ ' in the second equation, we replace it with ' $(45 - x)$ ':

$0.40x + 0.55(45 - x) = 18.75$

This single equation now only has one variable, ' $x$ ', which we can solve. First, distribute the $0.55$ into the parentheses:

$0.40x + (0.55 imes 45) - (0.55 imes x) = 18.75$

Calculate $0.55 imes 45$ :

$0.55 imes 45 = 24.75$

So the equation becomes:

$0.40x + 24.75 - 0.55x = 18.75$

Now, combine the ' $x$ ' terms (remember $0.40 - 0.55$ ):

$(0.40 - 0.55)x + 24.75 = 18.75$

$-0.15x + 24.75 = 18.75$

Next, isolate the ' $x$ ' term by subtracting $24.75$ from both sides of the equation:

$-0.15x = 18.75 - 24.75$

$-0.15x = -6$

Finally, solve for ' $x$ ' by dividing both sides by $-0.15$ :

$x = rac{-6}{-0.15}$

$x = 40$

So, there are 40 stamps that cost $0.40.

Now that we know ' $x$ ', we can find ' $y$ ' using the equation we derived earlier: $y = 45 - x$ .

$y = 45 - 40$

$y = 5$

So, there are 5 stamps that cost $0.55.

To recap using substitution: we isolated one variable, substituted its expression into the other equation, solved for the remaining variable, and then plugged that value back in to find the first variable. Pretty slick, right?

Method 2: Elimination

The elimination method, as the name suggests, involves eliminating one of the variables by adding or subtracting the equations. This is super handy when the coefficients of the variables are the same or can be easily made the same. Let's look at Krista's equations again:

$x + y = 45$
$0.40x + 0.55y = 18.75$

To use elimination, we want the coefficients of either ' $x$ ' or ' $y$ ' in both equations to be opposites (so they cancel out when added) or identical (so they cancel out when subtracted). A common strategy is to multiply one or both equations by a constant.

Let's try to eliminate ' $x$ '. The coefficient of ' $x$ ' in the first equation is 1, and in the second equation is 0.40. We can multiply the first equation by $-0.40$ so that the ' $x$ ' coefficients become opposites ($ -0.40x$ and $+0.40x$ ).

Multiply equation (1) by $-0.40$ :

$-0.40(x + y) = -0.40(45)$

This gives us:

$-0.40x - 0.40y = -18$

Now we have a new system:

1a. $-0.40x - 0.40y = -18$ 2. $0.40x + 0.55y = 18.75$

Notice that the ' $x$ ' terms have opposite coefficients. If we add equation (1a) and equation (2) together, the ' $x$ ' terms will cancel out:

$(-0.40x - 0.40y) + (0.40x + 0.55y) = -18 + 18.75$

Combine like terms:

$(-0.40x + 0.40x) + (-0.40y + 0.55y) = 0.75$

$0x + 0.15y = 0.75$

$0.15y = 0.75$

Now, solve for ' $y$ ' by dividing both sides by $0.15$ :

$y = rac{0.75}{0.15}$

$y = 5$

So, there are 5 stamps that cost $0.55.

Once we have the value of ' $y$ ', we can substitute it back into either of the original equations to find ' $x$ '. Let's use the simpler first equation: $x + y = 45$ .

$x + 5 = 45$

Subtract 5 from both sides:

$x = 45 - 5$

$x = 40$

So, there are 40 stamps that cost $0.40.

Both methods, substitution and elimination, yield the same result: 40 stamps at $0.40 each and 5 stamps at $0.55 each. The elimination method can sometimes feel a bit cleaner, especially when you can easily make coefficients match or be opposites. The key is to manipulate the equations strategically so that one variable disappears when you combine them.

Checking Your Answer

It's always a good idea, guys, to check your answer to make sure it makes sense and satisfies all the conditions of the original problem. This step can save you from silly mistakes!

We found that $x=40$ (number of 40-cent stamps) and $y=5$ (number of 55-cent stamps).

Let's check the total number of stamps:

$x + y = 40 + 5 = 45$

This matches the problem statement (45 stamps purchased). Check!

Now let's check the total cost:

Cost of 40-cent stamps: $40 imes $0.40 = $16.00$

Cost of 55-cent stamps: $5 imes $0.55 = $2.75$

Total cost: $16.00 + $2.75 = $18.75$

This also matches the problem statement ($18.75 total cost). Double check!

Since both conditions are met, our solution is correct. We have successfully solved Krista's homework problem using a system of equations. This process of setting up equations and then solving them is a fundamental skill in algebra and mathematics, applicable to countless scenarios beyond just stamps. Keep practicing, and you'll become a pro at translating word problems into solvable equations!

Why Systems of Equations Matter

Understanding systems of equations is more than just passing a math test; it's about developing critical thinking and problem-solving skills that are essential in many fields. Whether you're an aspiring engineer, a budding entrepreneur, a scientist, or even just someone trying to manage their personal finances, the ability to model real-world situations with mathematical equations and solve for unknown variables is invaluable. Think about it:

Budgeting: If you have a total amount of money to spend and know the cost of different items, you can use systems of equations to figure out how many of each item you can buy. For instance, if you have $100 to spend on shirts ($20 each) and pants ($30 each), how many of each can you get?
Mixture Problems: In chemistry or cooking, you might need to mix substances with different concentrations or properties. Systems of equations help determine the quantities of each component needed to achieve a desired outcome.
Distance, Rate, and Time: When objects are moving, their speeds, distances, and times are often related. Systems of equations are perfect for solving problems involving multiple moving objects.
Business and Economics: Businesses use systems of equations for forecasting, analyzing market trends, and optimizing production. For example, determining the break-even point for a product involves setting costs equal to revenues, often represented by systems of equations.

Krista's stamp problem is a simplified example, but it perfectly illustrates the power of these mathematical tools. By breaking down a complex situation into smaller, manageable parts (the individual equations) and then combining them logically (solving the system), we can arrive at clear and accurate solutions. So, the next time you encounter a word problem, remember to look for those relationships, define your variables, and set up your system. You've got this, guys!

Keywords: system of equations, substitution method, elimination method, word problem, algebra, mathematics, solving equations, linear equations, problem-solving, quantitative reasoning.

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