Cost Of Cards: Solving A System Of Equations
Hey guys! Let's dive into a fun math problem that involves figuring out the cost of birthday cards and thank-you notes. This might seem like a simple question, but it's a great way to flex our algebraic muscles and understand how systems of equations work in real-life scenarios. So, buckle up, and let’s get started!
Setting Up the Equations
Alright, so here’s the deal: We've got two packages, A and B, each containing a mix of birthday cards and thank-you notes. Our mission, should we choose to accept it, is to figure out the individual cost of a birthday card and a thank-you note. To do this, we'll translate the given information into a system of linear equations. This might sound intimidating, but trust me, it's like creating a secret code that only we can decipher!
Package A: This package has 3 birthday cards and 2 thank-you notes, costing a total of $9.60. If we let 'x' be the cost of one birthday card and 'y' be the cost of one thank-you note, we can write our first equation as:
3x + 2y = 9.60
This equation basically says, “Three times the cost of a birthday card, plus two times the cost of a thank-you note, equals $9.60.” Simple enough, right?
Package B: Now, let's look at Package B. It contains 8 birthday cards and 6 thank-you notes, with a total cost of $26.60. Using the same logic, we can create our second equation:
8x + 6y = 26.60
This equation is similar to the first one, just with different quantities and a different total cost. So, we now have a system of two equations:
- 3x + 2y = 9.60
- 8x + 6y = 26.60
This system of equations is the key to unlocking the mystery of the card prices. Solving this system will give us the values of 'x' and 'y', which are the costs we're trying to find. There are several ways to solve a system of equations, such as substitution, elimination, or even using matrices. For this example, let’s use the elimination method. This method involves manipulating the equations so that when we add or subtract them, one of the variables cancels out.
To eliminate 'y', we can multiply the first equation by 3. This will give us a '6y' term in both equations, making it easier to cancel them out. Multiplying the entire first equation by 3, we get:
*** 3 * (3x + 2y) = 3 * 9.60*** *** 9x + 6y = 28.80***
Now, we have a new system of equations:
- 9x + 6y = 28.80
- 8x + 6y = 26.60
See how the '6y' terms are the same? This is perfect! Now, we can subtract the second equation from the first to eliminate 'y':
(9x + 6y) - (8x + 6y) = 28.80 - 26.60 9x - 8x = 2.20 x = 2.20
Woohoo! We’ve found the value of 'x', which represents the cost of a birthday card. A birthday card costs $2.20. Now, we need to find the value of 'y', the cost of a thank-you note. To do this, we can substitute the value of 'x' back into one of our original equations. Let's use the first one: 3x + 2y = 9.60. Plugging in x = 2.20, we get:
3 * 2.20 + 2y = 9.60 6.60 + 2y = 9.60
Now, we can solve for 'y':
*** 2y = 9.60 - 6.60*** *** 2y = 3.00*** *** y = 3.00 / 2*** *** y = 1.50***
And there we have it! The value of 'y', the cost of a thank-you note, is $1.50. So, we’ve successfully solved the system of equations and found the cost of each type of card.
Solving the System of Equations
Now that we have our equations set up, it's time to put on our detective hats and solve for 'x' and 'y'. There are a couple of ways we can tackle this, but let's go with the elimination method. It's like a strategic game where we cancel out one variable to find the other.
The Elimination Method: A Step-by-Step Guide
The elimination method is all about making the coefficients (the numbers in front of 'x' and 'y') of one variable the same (or opposites) in both equations. This way, when we add or subtract the equations, that variable disappears, leaving us with a single equation to solve.
Step 1: Choose a Variable to Eliminate
Looking at our equations:
- 3x + 2y = 9.60
- 8x + 6y = 26.60
It seems easier to eliminate 'y' because we can multiply the first equation by 3 to make the 'y' coefficient 6, which matches the second equation. This saves us from multiplying both equations.
Step 2: Multiply Equations to Match Coefficients
We'll multiply the entire first equation by 3:
- 3 * (3x + 2y) = 3 * 9.60
- 9x + 6y = 28.80
Now, our system looks like this:
- 9x + 6y = 28.80
- 8x + 6y = 26.60
Step 3: Eliminate the Chosen Variable
Since the '6y' terms are the same, we'll subtract the second equation from the first:
- (9x + 6y) - (8x + 6y) = 28.80 - 26.60
- 9x - 8x = 2.20
- x = 2.20
Boom! We've found that x = 2.20. That means a birthday card costs $2.20.
Step 4: Substitute to Find the Other Variable
Now that we know 'x', we can plug it back into either of our original equations to find 'y'. Let's use the first one:
- 3x + 2y = 9.60
- 3 * 2.20 + 2y = 9.60
- 6.60 + 2y = 9.60
Step 5: Solve for the Remaining Variable
Let's isolate 'y':
- 2y = 9.60 - 6.60
- 2y = 3.00
- y = 1.50
So, y = 1.50, meaning a thank-you note costs $1.50.
Alternative Method: Substitution
Just for kicks, let's briefly talk about another method: substitution. In this method, you solve one equation for one variable and then substitute that expression into the other equation. It’s like a mathematical relay race where we pass the baton (the expression) to the other equation.
For instance, we could solve the first equation (3x + 2y = 9.60) for 'y':
- 2y = 9.60 - 3x
- y = (9.60 - 3x) / 2
Then, we'd substitute this expression for 'y' into the second equation and solve for 'x'. Once we have 'x', we can plug it back into our expression for 'y' to find its value. Both methods will lead us to the same answer, so it’s cool to have options!
The Solution and Its Meaning
After all that algebraic maneuvering, we’ve arrived at the solution! We found that:
- x = 2.20 (The cost of a birthday card is $2.20)
- y = 1.50 (The cost of a thank-you note is $1.50)
This means that each birthday card costs $2.20, and each thank-you note costs $1.50. Now, if you were planning to buy a bunch of cards, you’d have a better idea of how much it’s going to cost you! This is super practical, right?
Real-World Applications
Understanding systems of equations isn't just about solving math problems; it's a tool that can be applied to many real-world situations. Think about it: businesses use them to optimize costs and profits, scientists use them to model complex systems, and even economists use them to predict market trends. So, mastering these skills now can open up a whole world of possibilities later on.
For example, imagine you’re planning a party and need to figure out how many pizzas and sodas to buy to maximize your budget. You could use a system of equations to represent your budget constraints and the costs of pizzas and sodas. Or, if you're trying to figure out the nutritional content of a meal, you could use equations to balance the calories, protein, and carbohydrates.
The possibilities are endless! So, the next time you encounter a problem that involves multiple variables and constraints, remember the power of systems of equations. You might just surprise yourself with what you can solve.
Conclusion
So, there you have it, folks! We successfully navigated the world of systems of equations and uncovered the secret prices of birthday cards and thank-you notes. By setting up equations, choosing a method to solve them (like elimination or substitution), and carefully working through the steps, we cracked the code. Isn't it awesome how math can help us solve everyday mysteries?
This exercise not only sharpened our algebra skills but also showed us how mathematical concepts can be applied in practical scenarios. Whether it’s figuring out the cost of greeting cards or planning a party, the ability to solve systems of equations is a valuable asset. Keep practicing, keep exploring, and remember: math is not just about numbers; it's about problem-solving and critical thinking. Keep those brain muscles flexed, and who knows what other mysteries you’ll be able to solve! Keep rocking!