## PCSR REVIEW SERIES WEEK 2: Addition and Subtraction of Fractions

PCSR REVIEW SERIES for October 2016
Week 2: Addition and Subtraction of Fractions

In Week 1 of our PCSR Review Series , we have learned how to get the Least Common Multiple (LCM) and Greatest Common Divisor (GCD) of two or more numbers. The LCM is used to add and subtract dissimilar fractions. To add or subtract fractions, you have to get the LCM of their denominator and convert them to similar fractions.

The GCD is used for converting fractions to lowest terms. In converting fractions to lowest terms, you have to get the GCD of the numerator and denominator of a fraction, and divide both the numerator and the denominator by the GCD.

There are two more skills you need to know: to convert improper fractions to mixed fractions. Most of the time, final answers are required to be in mixed fractions. Links to tutorials on how to do it are included below.

Articles

Videos

More…

If you have time and your internet is fast, I suggest that you watch the entire fraction series (21 videos):

Below are the solutions and answers to the Practice Exercises and Problems for the Week 1 Review on LCM and GCD.

Practice Exercises
I. Find the GCD of each of the following.
a.) 6, 10

b.) 18, 42

c.) 12, 48, 60

d.) 56, 72

e.) 225, 75

II. Find the LCM of each of the following.
a.) 3, 4

b.) 2, 5

c.) 3, 6, 8
d.) 3, 4, 5

e.) 6, 12, 15

III. Practice Problems Solutions and Answers

1.) 24 (GCD of 3, 4, and 8)

2.) 15 (GCD of 3 and 15)

3.) Solution: GCD of 3, 7, and 21 is 42. They will be seen in the gym on the same day in 42 days. Since June has 30 days, we need an additional 12 days to complete the 42 days. Therefore, they will be seen on the same day on July 12 (of the same year of course).

4.) Solution: The LCM of the two sequences is 12 and since we are looking for the tenth common number, we multiply 12 by 10. This gives us 120.

5.) (a) Solution: LCM of 24 and 30 is 6. That is 6 groups.
(b) Solution: From (a) we can form 6 groups. There are 30 + 24 = 54 students. So in each group, there are 54/6 = 9 members. Since there are 6 groups, we divide each Grade level by 6. That is, 24/6 = 6 Grade 11 and 30/6 = 5 Grade 12 students in each group.

6.) Reduce 42/56 to lowest terms.
Solution: GCD of 42 and 56 is 14. 42 divided by 14 is 3 and 56 divided by 14 is 4. Therefore, the lowest terms is 3/4.

7.) Answer: 24 (LCM of 6 and 8)

8.) Answer: 6 (GCD of 18 and 12)

9.) Solution: The GCD of 21, 35, and 84 is 7. So, the cube has a side length of 7 cm.
Answer: 7 by 7 by 7

10.) Answer: 30 (GCD of 3, 5, and 6).

## 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

## How to Compare Decimal Numbers

We have already learned how to compare fractions and in this post, we are going to learn how to compare decimals. In comparing decimals it is important to understand place value. In the number 213.489, the following are their place values. For the whole numbers,

2 – hundred
1 – tens
3 – ones.

For the decimal numbers,

4 – tenths
8 – hundredths
9 – thousandths

In whole numbers, clearly, the larger the number of digits the larger the number. For example, 821 > 92 since 821 has three digits and 92 has only two digits. Since whole numbers are always greater than decimal numbers in comparing decimal numbers, look at the whole numbers first. Therefore, we have the following rule.

Rule 1: In comparing decimal numbers, look at the whole number first. The decimal numbers containing larger whole numbers have larger values.

Example 1: 84.23 > 82.345 since 84 is greater than 82.

Example 2: 12.56 < 15.001 since 12 is less than 15.

Example 3: 141.85 > 123.4 because 141 is greater than 123

Rule 2: If the whole numbers are equal, then compare the numbers by looking at the tenths place first. The number with the larger digit in tenths place is larger.

Example 1: 18.34 > 18.21 since 3 > 2.

Example 2: 12.95 > 12.15 since 9 > 1.

Example 3: 0. 9 > 0.873 since 9 > 8.

Notice that in Rule 2 Example 3, even if 0.873 has more digits, it is still less than 0.9 since 9 is greater than 8 and they are in the tenths place.

Rule 3: If the whole numbers and the tenths place are equal, then compare first the hundredths place. The number with the larger digits in the hundredths place is the larger number.

Rules 2 and 3 can be generalized. That means that you have to compare the digits from the tenths place first, then hundredths, then thousandths, and so on.

Please take note however the rules of negative numbers.

• Positive numbers are always greater than negative numbers.
• If both numbers are negative, do the following:

1.) Make them positive
2.) Apply the rules above

Example: Compare -82.45 and -82.31

1.) Make them positive: 82.45 and 82.31
2.) Apply the rules above:

From the rules above, the whole numbers are both 82, so we look at the tenths place. 0.4 > 0.3, so 82.45 > 82.31

3.) Reverse the answer. Since 82.45 > 82.31, – 82.31 > – 82.45.

## Solving Quadratic Word Problems in Algebra

Quadratic Equations are equations of the form ax^2 + bx + c = 0 where $a$, $b$ and $c$ are real numbers and $a \neq 0$. Depending on the form of the equation, you can solve for $x$ by extracting the quare root, factoring, or using the quadratic formula.This type of equation appears in various problems that involves multiplication and usually appears in the Civil Service Exams.

The following series details the method and strategies in solving problems involving quadratic equations.

How to Solve Quadratic Word Problems Part 1 is about solving problems involving consecutive integers. In this problem, the product of consecutive numbers is given and factoring was used to solve the problem.  » Read more

## How to Solve Quadratic Problems Part 2

In the previous post, we have used quadratic equations to solve a word problem involving consecutive numbers. In this post, we discuss more quadratic problems. This is the second problem in the series.

Problem 2

Miel is 12 years older than Nina. The product of their ages is 540.

Solution

Let x = age of Nina
x + 12 = age of Miel

The product of their ages is 540, so we can multiply the expressions above and equate the product to 540. That is,

x(x + 12) = 540.

Multiplying the expressions, we have

$x^2 + 12x = 540$.

Subtracting 540 from both sides, we obtain

$x^2 + 12x - 540 = 0$.

We want to find two numbers whose product is -540 and whose sum is 12. Those numbers are -18 and 30.

This means that the factors are

(x – 18)(x + 30) = 0.

Equating each expression to 0, we have

x – 18 = 0, x = 18
x + 30 = 0, x = – 30.

Since we are talking about age, we take the positive answer x = 18.

This means that Nina is 18 years old. Therefore, Miel is 18 + 12 = 30 years old.

## How to Solve Quadratic Word Problems Part 1

In the previous posts, we have learned how to solve quadratic equations by getting the extracting the square root, by factoring, and by quadratic formula. We continue this series by learning how to solve math word problems using quadratic equations. Most of the time, we need to rewrite the equation to the general form which is $ax^2 + bx + c = 0$.

Problem 1

The product of two consecutive positive even numbers is 48. What are the numbers?

Solution and Explanation

This problem can be solved mentally and by simple guess and check; however, we will solve it algebraically in order to illustrate the method of using quadratic equations.

Let
$x$ = smaller number
$x + 2$ = larger number

From the given above, we can form the following equation.

Smaller number times larger number = 48

$x(x + 2) = 48$

By the distributive property, this results to

$x^2 + 2x = 48$

Now, we need to make this equation in general form $ax^2 + bx + c = 0$ so we can factor easily. To do this, we subtract 48 from both sides resulting to

$x^2 + 2x - 48 = 0$

By factoring, we need two numbers whose sum is 2 and product is -48 where the absolute value of the larger number is greater than that of the smaller. With this restriction in mind, we have the following pairs of factors whose product is -48.

{48, -1}, {24, -2}, {12, -4}, {8,-6}

From these pairs, 8 and -6 has a sum of 2. Therefore, the factors are

$(x + 8)(x - 6) = 0$

Equating to 0, we have

$x + 8 = 0$, $x = -8$
$x - 6 = 0$, $x = 6$

Since we are looking for positive integers, we will take $x = 6$ and $x + 2 = 6 + 2 = 8$.

Therefore, the two consecutive numbers are 6 and 8.

Check: The numbers 6 and 8 are consecutive numbers and their product is 48. Therefore, we are correct.

In the previous post, we have learned how to solve quadratic equations by factoring. In this post, we are going to learn how to solve quadratic equations using the quadratic formula. In doing this, we must identify the values of $a$, $b$, and $c$, in $ax^2 + bx + c = 0$ and substitute their values to the quadratic formula

$x = \dfrac{-b \pm \sqrt{b^2 - 4ac }}{2a}$.

Note that the value of $a$ is the number in the term containing $x^2$, $b$ is the number in the term containing $x$, and $c$ is the value of the constant (without $x$ or $x^2$).

The results in this calculation which are the values of $x$ are the roots of the quadratic equation. Before you calculate using this formula, it is important that you master properties of radical numbers and how to calculate using them.

Example 1: Find the roots of $x^2 - 4x - 4 = 0$

Solution

From the equation, we can identify $a = 1$, $b = -4$, and $c = -4$.

Substituting these values in the quadratic formula, we have

$x = \dfrac{-(-4) \pm \sqrt{(-4)^2 - 4(1)(-4)}}{2(1)}$
$x = \dfrac{4 \pm \sqrt{16 + 16}}{2}$
$x = \dfrac{4 \pm \sqrt{32}}{2}$.

We know, that $\sqrt{32} = \sqrt{(16)(2)} = \sqrt{16} \sqrt{2} = 4 \sqrt{2}$. So, we have

$x = \dfrac{4 \pm 4 \sqrt{2}}{2} = \dfrac{2(2 + \sqrt{2}}{2} = 2 \pm 2 \sqrt{2}$

Therefore, we have two roots

$2 + 2 \sqrt{2}$ or $2 - 2 \sqrt{2}$

Example 2: Find the roots of $2x^2 - 6x = 15$

Solution

Recall, that it easier to identify the values of $a$, $b$, and $c$ if the quadratic equation is in the general form which is $ax^2 + bx + c = 0$. In order to make the right hand side of the equation above equal to 0, subtract 15 from both sides of the equation by 15. This results to

$2x^2 - 6x - 15$.

As we can see, $a = 2$, $b = -6$ and $c =-15$.

Substituting these values to the quadratic formula, we have

$x = \dfrac{-(-6) \pm \sqrt{(-6)^2 - 4(2)(-15)}}{2(2)}$

$x = \dfrac{6 \pm \sqrt{36 + 120}}{4}$

$x = \dfrac{6 \pm \sqrt{156}}{4}$.

But $\sqrt{156} = \sqrt{(4)(39)} = \sqrt{4} \sqrt{39} = 2 \sqrt{39}$.

Therefore,

$x = \dfrac{6 \pm 2 \sqrt{39}}{4}$.

Factoring out 2, we have

$\dfrac{2(3 \pm \sqrt{39}}{4} = \dfrac{3 \pm \sqrt{39}}{2}$

Therefore, we have two roots:

$\dfrac{3 + \sqrt{39}}{2}$ or $\dfrac{3 - \sqrt{39}}{2}$

That’s it. In the next post, we are going to learn how to use quadratic equations on how to solve word problems.

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