Matching Mathematical Expressions A Step By Step Guide

In this article, we will delve into the process of matching mathematical expressions with their corresponding results. This involves performing various arithmetic operations, such as addition, division, multiplication, and subtraction, to arrive at the correct solutions. We will systematically evaluate each expression and match it with its appropriate result, providing a step-by-step explanation for each calculation.

Let's dive into the task of matching the given mathematical expressions with their respective results. We will tackle each expression individually, performing the necessary calculations to arrive at the correct answer. By meticulously working through each step, we will ensure accuracy and a clear understanding of the process.

a. 1125 + 3271 + 4258

In this first expression, we're presented with an addition problem involving three numbers: 1125, 3271, and 4258. To find the correct result, we need to add these numbers together. Let's break down the process step by step to ensure accuracy.

We will begin by adding the first two numbers, 1125 and 3271. Starting with the ones place, 5 + 1 equals 6. Moving to the tens place, 2 + 7 equals 9. In the hundreds place, 1 + 2 equals 3. Finally, in the thousands place, 1 + 3 equals 4. So, 1125 + 3271 equals 4396.

Now, we need to add this sum to the third number, 4258. Again, we start with the ones place: 6 + 8 equals 14. We write down the 4 and carry over the 1 to the tens place. In the tens place, we have 9 + 5 + 1 (the carry-over), which equals 15. We write down the 5 and carry over the 1 to the hundreds place. In the hundreds place, we have 3 + 2 + 1 (the carry-over), which equals 6. Finally, in the thousands place, we have 4 + 4, which equals 8.

Therefore, 4396 + 4258 equals 8654. This means that the correct match for the expression 1125 + 3271 + 4258 is iv. 8654. This multi-step addition showcases the importance of careful alignment and carrying over digits to arrive at the correct sum. Accurate addition is a foundational skill in mathematics, and this example reinforces the need for precision in every step.

b. 7218 ÷ 9

Moving on to the second expression, we encounter a division problem: 7218 divided by 9. Division is the process of splitting a number into equal groups, and in this case, we want to determine how many groups of 9 are contained within 7218. To solve this, we will use long division, a method that systematically breaks down the problem into smaller, more manageable steps.

We start by looking at the first digit of the dividend, 7. Since 7 is less than 9, we consider the first two digits, 72. How many times does 9 go into 72? It goes in exactly 8 times (8 x 9 = 72). So, we write 8 above the 2 in the quotient. Subtracting 72 from 72, we get 0. Next, we bring down the next digit, which is 1. Now we have 1. How many times does 9 go into 1? It goes in 0 times. We write 0 next to the 8 in the quotient. Since 9 doesn't go into 1, we bring down the next digit, which is 8. Now we have 18. How many times does 9 go into 18? It goes in exactly 2 times (2 x 9 = 18). We write 2 next to the 0 in the quotient. Subtracting 18 from 18, we get 0.

So, 7218 ÷ 9 equals 802. Therefore, the correct match for this expression is v. 802. This division problem highlights the importance of understanding the steps involved in long division and the ability to accurately perform each step. Mastering division is crucial for solving a wide range of mathematical problems and understanding numerical relationships.

c. 321 x 172

Our third expression involves multiplication: 321 multiplied by 172. Multiplication is the process of repeated addition, and in this case, we are finding the total when 321 is added to itself 172 times. To solve this, we will use the standard multiplication algorithm, which involves multiplying each digit of the second number by the first number and then adding the results.

First, we multiply 321 by 2 (the ones digit of 172). 2 x 1 = 2, 2 x 2 = 4, and 2 x 3 = 6. So, 321 x 2 = 642. Next, we multiply 321 by 7 (the tens digit of 172). Since we are multiplying by the tens digit, we add a 0 as a placeholder in the ones place. 7 x 1 = 7, 7 x 2 = 14 (write down 4 and carry over 1), and 7 x 3 = 21 + 1 (carry-over) = 22. So, 321 x 70 = 22470. Finally, we multiply 321 by 1 (the hundreds digit of 172). Since we are multiplying by the hundreds digit, we add two 0s as placeholders. 1 x 321 = 321. So, 321 x 100 = 32100.

Now, we add the results: 642 + 22470 + 32100. Starting with the ones place, 2 + 0 + 0 = 2. In the tens place, 4 + 7 + 0 = 11 (write down 1 and carry over 1). In the hundreds place, 6 + 4 + 1 + 1 (carry-over) = 12 (write down 2 and carry over 1). In the thousands place, 2 + 2 + 1 (carry-over) = 5. In the ten-thousands place, we have 3. So, the sum is 55212.

Therefore, 321 x 172 equals 55212. This matches ii. 55212. This multiplication problem demonstrates the importance of understanding place value and the systematic approach required for multi-digit multiplication. Proficient multiplication skills are essential for tackling more complex mathematical problems and real-world calculations.

d. 48 x 135

Our fourth expression is another multiplication problem: 48 multiplied by 135. Similar to the previous example, we will use the standard multiplication algorithm to find the product of these two numbers. This process involves multiplying each digit of the second number by the first number and then adding the results together.

We start by multiplying 48 by 5 (the ones digit of 135). 5 x 8 = 40 (write down 0 and carry over 4). 5 x 4 = 20 + 4 (carry-over) = 24. So, 48 x 5 = 240. Next, we multiply 48 by 3 (the tens digit of 135). Since we are multiplying by the tens digit, we add a 0 as a placeholder in the ones place. 3 x 8 = 24 (write down 4 and carry over 2). 3 x 4 = 12 + 2 (carry-over) = 14. So, 48 x 30 = 1440. Finally, we multiply 48 by 1 (the hundreds digit of 135). Since we are multiplying by the hundreds digit, we add two 0s as placeholders. 1 x 48 = 48. So, 48 x 100 = 4800.

Now, we add the results: 240 + 1440 + 4800. Starting with the ones place, 0 + 0 + 0 = 0. In the tens place, 4 + 4 + 0 = 8. In the hundreds place, 2 + 4 + 8 = 14 (write down 4 and carry over 1). In the thousands place, 1 + 4 + 1 (carry-over) = 6. So, the sum is 6480.

Therefore, 48 x 135 equals 6480. This matches i. 6480. This problem reinforces the importance of a systematic approach to multiplication, ensuring that each digit is multiplied correctly and that place values are properly accounted for. Consistent practice with multi-digit multiplication helps build fluency and accuracy.

e. 52190 - 35798

Lastly, we have a subtraction problem: 52190 minus 35798. Subtraction is the process of finding the difference between two numbers, and in this case, we want to determine the amount that remains when 35798 is taken away from 52190. We will use the standard subtraction algorithm, which involves subtracting corresponding digits, starting from the ones place.

Starting with the ones place, we have 0 - 8. Since 0 is less than 8, we need to borrow from the tens place. The 9 in the tens place becomes 8, and the 0 in the ones place becomes 10. Now we have 10 - 8 = 2. Moving to the tens place, we have 8 - 9. Again, 8 is less than 9, so we borrow from the hundreds place. The 1 in the hundreds place becomes 0, and the 8 in the tens place becomes 18. Now we have 18 - 9 = 9. In the hundreds place, we have 0 - 7. Since 0 is less than 7, we borrow from the thousands place. The 2 in the thousands place becomes 1, and the 0 in the hundreds place becomes 10. Now we have 10 - 7 = 3. In the thousands place, we have 1 - 5. Since 1 is less than 5, we borrow from the ten-thousands place. The 5 in the ten-thousands place becomes 4, and the 1 in the thousands place becomes 11. Now we have 11 - 5 = 6. Finally, in the ten-thousands place, we have 4 - 3 = 1.

So, 52190 - 35798 equals 16392. Therefore, the correct match for this expression is iii. 16392. This subtraction problem illustrates the importance of borrowing and paying close attention to place values to ensure accurate calculations. Proficiency in subtraction is essential for solving a wide variety of mathematical problems, especially those involving differences and comparisons.

In conclusion, we have successfully matched each mathematical expression with its correct result by performing the necessary arithmetic operations. This exercise underscores the importance of mastering basic mathematical skills such as addition, subtraction, multiplication, and division. Each type of problem requires a specific approach, whether it's the careful alignment in addition, the systematic steps in long division, the multi-digit multiplication algorithm, or the borrowing technique in subtraction. By understanding and applying these methods, we can confidently solve a wide range of mathematical problems. Remember, consistent practice and a thorough understanding of the fundamentals are key to success in mathematics. The ability to accurately perform these operations is not only crucial for academic success but also for everyday problem-solving in various real-life situations. From calculating expenses to understanding measurements, these skills form the backbone of mathematical literacy.