How to Add and Subtract Rational Numbers

How to Add and Subtract Rational Numbers

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Rational numbers, forming a core classification of numbers, include values that can be denoted as a distinct ratio of two integers. In this piece, we're going to delve into the mechanics of adding and subtracting these rational numbers.

An Easy-to-Follow Tutorial on Adding and Subtracting Rational Numbers

Let's dive into the step-by-step process of adding and subtracting rational numbers:

Phase 1: Understanding Rational Numbers
Kickstarting our numerical journey, the first stage calls for a clear comprehension of rational numbers. Essentially, a rational number can be defined as a value expressible as a fraction, with both the numerator and the denominator being integers, and the denominator being non-zero.

Phase 2: Appreciating the Concept of Common Denominators
Building upon the foundational knowledge, the next step is grasping an important mathematical tenet: for adding or subtracting fractions, the denominators need to be identical, or 'common'. If the denominators differ, the fractions can't be directly combined or subtracted.

Phase 3: Creating Common Denominators
If the denominators are unlike, it's necessary to tweak the fractions to have common denominators. This can be done by finding the least common multiple (LCM) of the denominators and subsequently adjusting the fractions.

Phase 4: Modifying the Numerators
When changing fractions to have common denominators, it's important to similarly adjust the numerators. If you multiply or divide the denominator of a fraction, the numerator must be treated likewise. This maintains the fraction's value, despite its altered appearance.

Phase 5: Executing the Addition or Subtraction
With the fractions now sharing common denominators, the addition or subtraction can be carried out. In case of addition, merge the numerators and place the result over the common denominator. For subtraction, deduct the second numerator from the first and place the result over the common denominator.

Exercises for Add or Subtract Rational Numbers

1) \({7 \over 4} \ + \ {5 \over 6} = \)

2) \({8 \over 10} \ + \ {2 \over 3} = \)

3) \({9 \over 8} \ + \ {8 \over 2} = \)

4) \({5 \over 3} \ - \ {3 \over 5} = \)

5) \({6 \over 3} \ - \ {8 \over 8} = \)

6) \({5 \over 1} \ - \ {5 \over 4} = \)

7) \({3 \over 2} \ - \ {7 \over 8} = \)

8) \({9 \over 3} \ - \ {3 \over 5} = \)

9) \({7 \over 11} \ + \ {16 \over 17} = \)

10) \({8 \over 9} \ + \ {15 \over 10} = \)

1)\({7 \over 4} \ + \ {5 \over 6} = \)\( \ \color{red}{{7 \times 6 \ + \ 5 \times 4 \over 4\times6} = } \)\({\color{red}{62 \over 24}}\)\(\color{red}{ = {62 \div 2 \over 24 \div 2} = {31 \over 12}}\)

GCF(62,24) = 2

Solution
For “unlike” fractions, the first step is to find the same denominator before you can add or subtract fractions with different denominators.
To find the same denominator, Multiply two denominators, and each numerator by the denominator of the other fraction. \({7 \over 4} \ + \ {5 \over 6} = \)\( \ {7 \times 6 \ + \ 5 \times 4 \over 4\times6} =\)\({62 \over 24}\)
Then, simplify the result. \({62 \over 24} = {31 \over 12}\)
2)\({8 \over 10} \ + \ {2 \over 3} = \)\( \ \color{red}{{8 \times 3 \ + \ 2 \times 10 \over 10 \times 3} = } \) \({\color{red}{44 \over 30}}\)\(\color{red}{ = {44 \div 2 \over 30 \div 2} = {22 \over 15}} \)

GCF(44,30) = 2

Solution
For “unlike” fractions, the first step is to find the same denominator before you can add or subtract fractions with different denominators.
To find the same denominator, Multiply two denominators, and each numerator by the denominator of the other fraction. \({8 \over 10} \ + \ {2 \over 3} = \)\( \ {8 \times 3 \ + \ 2 \times 10 \over 10\times3} =\) = \( {44 \over 30}\)
Then, simplify the result. \({44 \over 30} = {22 \over 15}\)
 
3)\({9 \over 8} \ + \ {8 \over 2} = \)\( \ \color{red}{{9 \times 2 \ + \ 8 \times 8 \over 8\times2} = } \) \({\color{red}{82 \over 16}}\)\(\color{red}{ = {82 \div 2 \over 16 \div 2} = {41 \over 8}} \)

GCF(82,16) = 2

Solution
For “unlike” fractions, the first step is to find the same denominator before you can add or subtract fractions with different denominators.
To find the same denominator, Multiply two denominators, and each numerator by the denominator of the other fraction. \({9 \over 8} \ + \ {8 \over 2} = \)\( \ {9 \times 2 \ + \ 8 \times 8 \over 8\times2} =\)\({82 \over 16}\) 
Then, simplify the result. \({82 \over 16} = {41 \over 8}\)
 
4)\({5 \over 3} \ - \ {3 \over 5} = \)\( \ \color{red}{{5 \times 5 \ - \ 5 \times 3 \over 3\times5} = } \) \({\color{red}{10 \over 15}}\)\(\color{red}{ = {10 \div 5 \over 15 \div 5} = {2 \over 3}} \)

GCF(10,15) =5

Solution
For “unlike” fractions, the first step is to find the same denominator before you can add or subtract fractions with different denominators.
To find the same denominator, Multiply two denominators, and each numerator by the denominator of the other fraction. \({5 \over 3} \ - \ {3 \over 5} = \)\( \ {5 \times 5 \ - \ 5 \times 3 \over 3\times5} = \)\({15 \over 10}\)
Then, simplify the result. \({2 \over 3} = 1\)
 
5)\({6 \over 3} \ - \ {8 \over 8} = \)\( \ \color{red}{{6 \times 8 \ - \ 8 \times 3 \over 3\times8} = } \) \({\color{red}{24 \over 24}}\)\(\color{red}{ = 1}\)
Solution
For “unlike” fractions, the first step is to find the same denominator before you can add or subtract fractions with different denominators.
To find the same denominator, Multiply two denominators, and each numerator by the denominator of the other fraction. \({6 \over 3} \ - \ {8 \over 8} =  \)\( \ {6 \times 8 \ - \ 8 \times 3 \over 3\times8} = \)\({24 \over 24}\)
Then, simplify the result. \({15 \over 15} = 1\)
 
6)\({5 \over 1} \ - \ {5 \over 4} = \)\( \ \color{red}{{5 \times 4 \ - \ 5 \times 1 \over 1\times4} = } \) \({\color{red}{15 \over 4}}\)
Solution
For “unlike” fractions, the first step is to find the same denominator before you can add or subtract fractions with different denominators.
To find the same denominator, Multiply two denominators, and each numerator by the denominator of the other fraction. \({5 \over 1} \ - \ {5 \over 4} =  \)\( \ {5 \times 4 \ - \ 5 \times 1 \over 1\times4} = \)\({15 \over 4}\)
 
7)\({3 \over 2} \ - \ {7 \over 8} = \)\( \ \color{red}{{3 \times 8 \ - \ 7 \times 2 \over 2\times8} = } \) \({\color{red}{10 \over 16}}\)\(\color{red}{ = {10 \div 2 \over 16 \div 2} = {5 \over 8}}\)

GCF(10,16) = 2

Solution
For “unlike” fractions, the first step is to find the same denominator before you can add or subtract fractions with different denominators.
To find the same denominator, Multiply two denominators, and each numerator by the denominator of the other fraction. \({3 \over 2} \ - \ {7 \over 8} =  \)\( \ {3 \times 8 \ - \ 7 \times 2 \over 2\times8} =  \)\({10 \over 16}\)
Then, simplify the result. \({10 \over 16} = { 5 \over 8}\)
 
8)\({9 \over 3} \ - \ {3 \over 5} = \)\( \ \color{red}{{9 \times 5 \ - \ 3 \times 3 \over 3\times5} = } \) \({\color{red}{36 \over 15}}\)\(\color{red}{ = {36 \div 3 \over 15 \div 3} = {12 \over 5}}\)

GCF(36,15) = 3

Solution
For “unlike” fractions, the first step is to find the same denominator before you can add or subtract fractions with different denominators.
To find the same denominator, Multiply two denominators, and each numerator by the denominator of the other fraction.\({9 \over 3} \ - \ {3 \over 5} = \)\( \ {9 \times 5 \ - \ 3 \times 3 \over 3\times5} = \)\({36 \over 15}\)
Then, simplify the result. \({36 \over 15} = { 12 \over 5}\)
 
9)\({7 \over 11} \ + \ {16 \over 17} = \)\( \ \color{red}{{7 \times 17 \ + \ 16 \times 11 \over 11\times17} = } \) \({\color{red}{295 \over 187}}\)
Solution
For “unlike” fractions, the first step is to find the same denominator before you can add or subtract fractions with different denominators.
To find the same denominator, Multiply two denominators, and each numerator by the denominator of the other fraction. \({7 \over 11} \ + \ {16 \over 17} =  \)\( \ {7 \times 17 \ + \ 16 \times 11 \over 11\times17} =\)\({295 \over 187}\)
 
10)\({8 \over 9} \ + \ {15 \over 10} = \)\( \ \color{red}{{8 \times 10 \ + \ 15 \times 9 \over 9\times10} = } \) \({\color{red}{215 \over 90}}\)\(\color{red}{ = {215 \div 5 \over 90 \div 5} = {43 \over 18}} \)

GCF(215,90) = 5

Solution
For “unlike” fractions, the first step is to find the same denominator before you can add or subtract fractions with different denominators.
To find the same denominator, Multiply two denominators, and each numerator by the denominator of the other fraction. \({8 \over 9} \ + \ {15 \over 10} = \)\( \ {8 \times 10 \ + \ 15 \times 9 \over 9\times10} = \)\({215 \over 90}\)

Then, simplify the result. \({215 \over 90} = { 43 \over 18} \)

Add and Subtract Rational Numbers Quiz