Math Problems: Calculating Expressions With A + B + C = 7
Hey guys! Today, we are going to dive into some fun math problems that involve calculating expressions given the condition a + b + c = 7. We will break down each part step by step, making sure everyone understands the process. Let's put on our math hats and get started!
Understanding the Basics
Before we jump into the calculations, it's crucial to understand the basic principles we'll be using. The main idea here is the distributive property and simple arithmetic operations. We'll be factoring out common terms and then substituting the given value of a + b + c. It's like a math puzzle where we have all the pieces; we just need to put them in the right place.
The Distributive Property
The distributive property states that k(a + b + c) = ka + kb + kc. This means we can multiply a single term by a group of terms inside parentheses by multiplying each term individually. This property will be our best friend in solving these problems. It simplifies complex expressions and makes them easier to manage. You'll see this in action in every part of the problem, so make sure you're comfortable with it. If you're not, no worries! We'll go through it slowly and you'll get the hang of it.
Simple Arithmetic Operations
Of course, we can't forget the basics: addition, subtraction, multiplication, and sometimes division. These are the bread and butter of any math problem. We'll be using these operations to simplify expressions once we've applied the distributive property. Think of it as the clean-up crew after the main work is done. They tie up any loose ends and give us our final answer. So, a solid grasp of these operations is key.
Calculating the Expressions
Now, let's get into the heart of the matter and calculate each expression. We'll take it one step at a time, showing all the work so you can follow along easily. Remember, math isn't about memorizing; it's about understanding the process. So, let's focus on why we do each step, not just how.
a) 2a + 2b + 2c
In this first part, we need to calculate 2a + 2b + 2c. The first thing we should notice is that 2 is a common factor in each term. This is where the distributive property comes in handy. We can factor out the 2 from the expression:
2a + 2b + 2c = 2(a + b + c)
Now, we know that a + b + c = 7, so we can substitute this value into our expression:
2(a + b + c) = 2 * 7
Finally, we multiply to get our answer:
2 * 7 = 14
So, 2a + 2b + 2c = 14. See how we broke it down? Factoring out the common term made the problem much simpler. This is a trick we'll use throughout.
b) 10a + 10b + 10c
Next up, we have 10a + 10b + 10c. This looks pretty similar to the last one, doesn't it? That's because we can use the same strategy. Again, we spot a common factor: 10. Let's factor it out:
10a + 10b + 10c = 10(a + b + c)
We know that a + b + c = 7, so we substitute:
10(a + b + c) = 10 * 7
And multiply:
10 * 7 = 70
So, 10a + 10b + 10c = 70. Notice how factoring makes these problems almost automatic? Once you spot the common factor, it's just a matter of substitution and multiplication.
c) a * 13 + b * 13 + c * 13 + 21
This one looks a little more complicated, but don't let it scare you! We can handle it. We have a * 13 + b * 13 + c * 13 + 21. Let's focus on the first part: a * 13 + b * 13 + c * 13. Do you see a common factor here? It's 13!
a * 13 + b * 13 + c * 13 = 13(a + b + c)
Now we substitute a + b + c = 7:
13(a + b + c) = 13 * 7
Multiply:
13 * 7 = 91
But wait! We're not done yet. We still have that + 21 at the end of the original expression. So, we need to add that back in:
91 + 21 = 112
Therefore, a * 13 + b * 13 + c * 13 + 21 = 112. See? Even with the extra step, it's manageable when we break it down.
d) 8a + 8b + 8c - 12
Let's tackle 8a + 8b + 8c - 12. Just like before, we start by looking for a common factor in the first part of the expression: 8a + 8b + 8c. The common factor is 8:
8a + 8b + 8c = 8(a + b + c)
Substitute a + b + c = 7:
8(a + b + c) = 8 * 7
Multiply:
8 * 7 = 56
Now, don't forget about the - 12 at the end of the original expression. We need to subtract that:
56 - 12 = 44
So, 8a + 8b + 8c - 12 = 44. We're on a roll! Each of these problems follows the same basic pattern, and you're getting the hang of it.
e) 400 - (5a + 5b + 5c)
Now we have 400 - (5a + 5b + 5c). The parentheses are a clue that we should focus on the expression inside them first. Look at 5a + 5b + 5c. What's the common factor? You guessed it: 5!
5a + 5b + 5c = 5(a + b + c)
Substitute a + b + c = 7:
5(a + b + c) = 5 * 7
Multiply:
5 * 7 = 35
Now, let's bring back the rest of the expression: 400 - (5a + 5b + 5c). We now know that 5a + 5b + 5c = 35, so we have:
400 - 35 = 365
Therefore, 400 - (5a + 5b + 5c) = 365. It's like a puzzle with layers, but we're peeling them back one at a time.
f) 6a + 6b + 6c + 100
Last one in this set! We're looking at 6a + 6b + 6c + 100. Just like before, let's focus on the terms with variables: 6a + 6b + 6c. The common factor is 6:
6a + 6b + 6c = 6(a + b + c)
Substitute a + b + c = 7:
6(a + b + c) = 6 * 7
Multiply:
6 * 7 = 42
Now, we add back the + 100 from the original expression:
42 + 100 = 142
So, 6a + 6b + 6c + 100 = 142. We did it! We calculated all six expressions using the same basic strategy. You're becoming math expression masters!
Calculating with a = 11 and b + c = 8
Now, let's switch gears a bit. We're given a = 11 and b + c = 8. This time, we don't have a + b + c = 7. We'll need to use these new values to calculate some expressions. This is a great way to practice substitution with different givens.
Understanding the New Givens
Having a = 11 and b + c = 8 changes the game a bit. Instead of having the sum of all three variables, we have the value of a and the sum of b and c. This means we'll need to adapt our strategy slightly. The distributive property is still our friend, but we'll need to be a bit more creative with our substitutions. Think of it as leveling up in our math game!
Building on Previous Skills
Everything we learned in the first part of this problem will still be useful. We'll still be looking for common factors and using the distributive property. The main difference is how we substitute values. Instead of substituting a + b + c, we'll be substituting a and b + c separately. It's like having two pieces of the puzzle instead of one, and we need to figure out how they fit together.
Example Calculations (Illustrative)
Let's walk through a few examples to illustrate how this works. These examples aren't part of the original problem, but they'll help you understand the process.
Example 1: Calculate a + b + c
This one's straightforward. We know a = 11 and b + c = 8. So, we can simply add them together:
a + b + c = 11 + 8 = 19
Example 2: Calculate 2a + 2b + 2c
We can use the distributive property here: 2a + 2b + 2c = 2(a + b + c). We already know a + b + c = 19 from the previous example, so:
2(a + b + c) = 2 * 19 = 38
Example 3: Calculate a + 3b + 3c
This one's a bit trickier. We can factor out the 3 from the b and c terms: a + 3b + 3c = a + 3(b + c). Now we can substitute a = 11 and b + c = 8:
a + 3(b + c) = 11 + 3 * 8 = 11 + 24 = 35
These examples show how we can use the given values and the distributive property to calculate different expressions. The key is to look for opportunities to substitute and simplify.
Conclusion
So, there you have it! We've walked through calculating expressions with a + b + c = 7, and we've also explored how to handle different givens like a = 11 and b + c = 8. The key takeaway here is the power of the distributive property and careful substitution. Math can be like a fun puzzle if you break it down into smaller steps. Keep practicing, and you'll become even more confident in your math skills. Great job, everyone! You've nailed it! Remember, math isn't about being perfect; it's about learning and growing. So keep exploring, keep questioning, and most importantly, keep having fun with numbers!