## Practice Exercises on Subtraction of Integers

In subtraction of integers, we have learned two rules:

(1) a – b = a + (-b)
(2) a – (-b) = a + b

We will use these rules in answering the exersises below.

Exercises

1. 2 – 5
2. 18 – ( – 2)
3. 16 – 7
4. -17 – 3
5. -9 – (-3)
6. 0 – (-11)
7. -18 – (-25)
8. -10 – 9
9. 12 – (-9)
10. -6 – 3

1. 2 – 5

Solution 1: 5 is greater than 2. If you subtract two numbers, if the subtrahend is larger than the minuend, the answer will be negative. So, the answer is -3.

Solution 2: From rule 1, a – b = a + (-b), so 2 + 5 = 2 + (-5) = -3

2. 18 – ( – 2)

Solution: From rule 2, a – (-b) = a + b, so 18 + 2 = 20.

3. 16 – 7

4. -17 – 3

Solution: From rule 1, -17 – 3 = -17 + (- 3) = -20. Recall that in adding two negative numbers, we just add the numbers and then the answer will be negative.

5. -9 – (-3)

Solution: From rule 2, -9 – (-3) = -9 + 3 = -6.

6. 0 – (-11)

Solution: From rule 2, 0 – (-11) = 0 + 11 = 11.

7. -18 – (-25)

Solution: From rule 2, a –(-b) = a + b. So, -18 + 25 = 7.

8. -10 – 9

Solution: From rule 1, a – b = a + (-b), so -10 + (- 9) = – 19.

9. 12 – (-9)

Solution: From rule 2, a –(-b) = a + b, so 12 + 9 = 21.

10.- 6 – 3

Solution: From rule 1, a – b = a + (-b) = -6 + -3 = -9.

## Four Effective Techniques in Adding Integers

We have already discussed addition of integers. In this post, I am going to discuss four different techniques in adding integers.

Adding Numbers with the Same Sign

In adding integers with the same sign, we just add them and then copy the sign. For example, in adding 2 + 8, 2 and 8 are positive integers, so we just add them, and the answer will be positive. So, 2 + 8 = 10. On the other hand, if both integers are negative we also do the same: add them, then copy the sign. For example, -9 + -3 = -12 since both of them are negative integers.

The techniques below are for adding integers with different signs. These strategies are important because you can visualize addition even without memorizing the rules.

Techniques in Adding Integers with Different Signs

Technique 1: Using Positive and Negative Chips

You can imagine integers as positive and negative chips. Since +1 + -1 = 0, a pair of positive and negative chips will give a sum of 0. So, 3 means 3 positive chips and -4 means 4 negative chips. Since each pair of positive and negative chip cancels out each other (their sum is 0), then the remaining chips after the pairing will be the answer. So, 3 + (-4) as represented below is -1 since only one negative chip remains.

Another example is -4 + 6 = 2.

Technique 2: Decomposing the Numbers

This strategy uses the fact that a + (-a) = 0. Using this strategy, we can split one of the addends. For example, in 8 + (-5), we split 8 to 3 + 5 so that 5 and -5 will become 0. So,

8 + (-5) = 3 + 5 + (-5)

=3 + (5 + -5)= 3 + 0 = 3.

In -11 + 3, we can split -11 to -8 + -3. So,

-11 + 3 = -8 + (-3 + 3) = -8 + 0 = -8.

Technique 3: Using the Number Line

Integers can also be represented as movement on the number line. A positive integer is a movement to the right of 0 and a negative integer is a movement to the left. Positive 3 and -2 can be represented as shown below.

So, 3 + (-5) can be represented as a movement of 3 units to the right of 0, then a movement of 5 units to the left. As we can see in the next diagarm, it the movement stopped at -2. So, 3 + (-5) = -2.

Also, -2 + 3 can be represented as a movement 2 units to the left of 0 and then a movement of 3 units to the right. The movement stopped at 1, so -2 + 3 = 1.

Technique 4: Grouping Numbers with Similar Signs

In adding more than 2 addends with different signs, group the numbers with the same signs. For example, 9 + (-2) + 4 + (-1), we can do the following:

(1) add the positive integers first: 9 + 4 = 13
(2) add the negative integers (-2 + -1 = -3)
(3): finally, add the two sums: 13 + (-3) = 10.

Try using the techniques above by answering the following and share to us which technique do you like most.

1.) 5 + (-10)
2.) -3 + 8
3.) 12 + – 10 + 7
4.) -10 + -3 + 4

## Solving Word Problems by Working Backwards: Summary

In the previous posts, I have shown to you a series on how to solve word problems by working backward.  Although I recommend that you learn this method, it is also very important to learn algebraic methods and others since some problems are difficult to solve by working backward. Below are the posts in this series.

How to Solve Word Problems by Working Backward Part 1 discusses solving number problems by working backward.  Two examples are given.

How to Solve Word Problems by Working Backward Part 2 discusses solving age problems by working backward. Two sample problems are solved in this post.

How to Solve Word Problems by Working Backward Part 3 discusses more complicated number problems. Two sample problems are discussed in this post.

If you are interested to solve more word problems, please visit the Word Problems page. It contains word problems about number, age, motion, work, discount and many more.

## How to Solve Word Problems by Working Backwards Part 3

In part 1 and part 2 of this series, we have learned how to solve number age problems by working backward. In this post, we are going to learn how to solve backward using inverse operations. Recall that multiplication and division are inverse operations and addition and subtraction are inverse operations.

Example 5

A number is multiplied by 4 and then, 3 is added to the product. The result is 31. What is the number?

Solution

The key phrases in this problem are (1) multiplied by 4 and (2) added to (3) the result is 31. Since we are working backward, we start with 31, and then find the inverse of “added to 3” which is “subtract 3.” So, 31 – 3 = 28.

Next, we find the inverse of “multiplied by 4,” which is “divided by 4.” So, 28/4 = 7.

So, the answer to this problem is 7.

Check: 7(4) + 3 = 31

Example 6

Think of a number. Divide it by 8. Then subtract 4 from the quotient. The result is 5. What is the number?

Solution

The key phrases in this problem are (1) divided by 8 (2) subtract 4 and (3) the result is (3) the result is 5.

We start with the result which is 5 and find the inverse of “subtract 4” which is “add 4.” So, 5 + 4 = 9. Next, we find the inverse of “divide by 8” which is “multiply by 8.” So, 9(8) = 72.

So, the correct answer is 72.

Check: 72/8 – 4 = 9 – 4 = 5.

In the next post, we will discuss more about solving math word problems by working backward.

## How to Solve Problems by Working Backwards Part 2

In the previous post, we have learned how to solve number problems by working backward.  In this post, we discuss age problems. We already had 2 examples in the previous post in this series, so we start with the third example.

Example 3

Arvin is 5 years older than Michael.  The sum of their ages is 37.  What are their ages?

Solution

This is very similar to the problems in the previous post in this series.  Arvin is 5 years older than Michael, so if we subtract 5 from Arvin’s age, their ages will be equal.  But if we subtract 5 from Arvin’s age, we also have to subtract 5 from the sum of their ages. That is,

37 – 5 = 32.

Now, their ages are equal, so we can divide the sum by 2.  That is 32 ÷ 2 = 16.  This means that the younger person is 16 since we subtracted 5 from Arvin’s age. Therefore, Michael is 16 and Arvin is 16 + 5 = 21.

Check

16 + 21 = 37.

Example 4

Mia is 3 years older than Pia.  In 4 years, the sum of their ages is 35. What are their ages?

Solution

There are two of them and we added 4 to both ages, so subtracting 8 from 35 (the sum) will determine the sum of their present age.  That is 35 – 8 = 27 is the sum of their present age.

Next, Mia is 3 years older than Pia, so if we subtract 3 from the sum of their present age, their ages will be equal. So, 27 – 3 = 24.

We can now divide the sum of their ages by 2.  That is 24/2 = 12.

This means that 12 is the age of the younger person because we subtracted 3 from the age of the older person.

So, Pia is 12 and Mia is 15.

Check 12 + 15 = 22.

In the next post, we will discuss more problems that can be solved by working backward.

## How to Solve Word Problems by Working Backwards Part 1

Most of us would always take a pen and solve for x if we see word problems. But did you know that you can solve them by working backward or even mentally? In this post, I am going to teach you some techniques on solving problems by working backward.

Example 1: One number is three more than the other. Their sum is 45. What are the numbers?

Solution

In the given, one number is 3 more than the other. This means that if you subtract 3 from the larger number they will be equal. Note that if we subtract 3 from one of the numbers, then we should also subtract 3 from their sum. Therefore, their sum will be 45 – 3 = 42. Since the numbers are equal, we now divide the sum by 2. That is, 42/2 = 21.

So, the smaller number is 21 and the larger is 21 + 3 = 24.

Check: 21 + 24 = 45

Example 2: One number is 5 less than the other. Their sum is 43. What is the smaller number?

Solution

This is very similar to Example 1. Here, one number is 5 less than the other; so, if we add 5 to the smaller number, they will be equal. If we add 5 to the smaller number, we should also add 5 to their sum. Therefore, their sum will be 43 + 5 = 48. Since the two numbers are equal, we can divide the sum by 2. That is 48/2 = 24. Since we added 5, it means that 24 is the larger number. So, the smaller number is 24 – 5 = 19.

Check: 19 + 24 = 43

In the next post, we are going to discuss more examples.

## Practice Exercises on Subtracting Decimals

We have already learned how to add and subtract numbers with decimals. In this post, we practice subtracting decimals. Recall that in subtracting decimals, the decimal points should be aligned.

Practice Exercises

1.) 2.32 – 1.82
2.) 6.71 – 3.9
3.) 6 – 0.52
4.) 5.03 – 4.25
5.) 0.53 – 0.33
6.) 4 – 1.26
7.) 7.28 – 2.4
8.) 7.08 – 0.29
9.) 3 – 0.305
10.) 40 – 12.5

1.) 0.5
2.) 2.81
3.) 5.48
4.) 0.78
5.) 0.2
6.) 2.74
7.) 4.88
8.) 6.79
9.) 2.695
10.) 27.5

Enjoy learning!

## How to Explore PH Civil Service Exam Reviewer

The PH Civil Service Exam Reviewer website (civilservicereview.com) has already provided free reviewers for the past two years and so far many testimonials have already reached me that this website has been very useful in this review. Below are some of the useful links within the website that are often ignored or missed especially when browsing using mobile phones.

1.) Math – The math page contains basic concepts in numerical ability such as operations on fractions, decimals,  solving equations, as well as finding areas of geometric figures. This also includes topics like LCM, GCF, and addition of positive and negative integers. If you are not very good in math, I suggest that you start with this page.

2.) Word Problems – The Word Problems page contains detailed tutorials about word problems. This includes number problems, age problems, motion problems, work problems, mixture problems, coin problems, investment problems, ratio problems, digit problems, and more.

3.) English – The English page contains reviewers on vocabulary, grammar and correct usage, and paragaraph organization. It also contains basic grammar tutorials particularly the tenses of the verbs.

4.) Practice Tests – This contains practice tests and exercises on both English and Math topics discussed in the site.

5.) Post List – contains the complete of posts of this website. This includes list of Civil Service Exam Passers.

6.) Videos – Useful videos from Sipnayan for learning mathematics.

Another useful site is the Sipnayan Youtube Channel which contains more than 300 Tagalog Math Tutorial videos. 🙂

Lastly, as a bonus, if you already passed the Civil Service Exam, then you may want to explore our partner website, the Philippine Government Jobs, if you want to work for the Philippine Government.

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