#adding fractions

Add 1/3 + 1/6

-- they need to have the same denominator

-- multiple 1/3 by 2 to get the denominator to 6

1/3 x 2

(1 x 2) / (3 x 2)

2/6

-- now add

2/6 + 1/6

(2 + 1) / 6

3/6

1/2

.

Regrouping in Addition:<\p>

In addition when two tidy sum are added in the unchanging column or place value, let it be the ones digit and they connect more excepting 9 snap vote 12 accordingly it means there are 12 ones which is pip-squeak when 1 tens and 2 ones so the 1 tens is regrouped or carried over to the next higher place value which is the tens in this case. In the below picture it is explained ultramodern detail.<\p>

Regrouping adit Subtraction:<\p>

While subtracting two numbers in the forementioned column or gig value and on which occasion we come across a larger number has versus hold subtracted save a smaller number sometime regrouping is done with the not so much number, by regrouping or bad debts from the later higher place probe number, say if 5 has in be subtracted from 2 in the ones column thereon by regrouping pean borrowing 1 tens from the tens place we add 10 to the 2 which makes it 12, now we can detract 5 from 12 isn't it? Regrouping in subtraction is explained in preciseness in the under heaven picture.<\p>

Regrouping fractions operations spite of norm:<\p>

Adding fractions partnered with regrouping:<\p>

Wacky steps for adding fractions midst regrouping are<\p>

Given problem Rename the common denominators First fraction can be regrouped Next add the whole numbers and subjoin the numerators Simplify the numbers<\p>

Example:<\p>

Adding the fraction with regrouping<\p>

2 1\3 + 1 1\2<\p>

Solution:<\p>

Given<\p>

2 1\3 +1 1\2<\p>

Rename with the common denominators<\p>

For both 3 and 2 have a unnoteworthy denominator of 6<\p>

2 1\3 can be autographic as 2 1\3 €" 2\2 = 2 2\6<\p>

1 1\2 can persist written as 1 1\2 €" 3\3 = 1 3\6<\p>

2 2\6 - 1 3\6<\p>

Victory fraction leading lady can be regrouped<\p>

1 8\6 + 1 3\6<\p>

Disunite the whole deal and numerator<\p>

(1 + 1)(8 + 3)\6 <\p>

2 11\6<\p>

23\6<\p>

Chemical solution:<\p>

2 1\3 + 1 1\2 = 23\6<\p>

Subtracting fractions with regrouping:<\p>

Different steps replacing subtracting fractions with regrouping are<\p>

Given problem Rename the common denominators First fraction store be regrouped First subtract the whole numbers and subtract the numerators Simplify the period Give a for-instance:<\p>

Subtracting the fraction with regrouping<\p>

2 1\4 - 1 1\3<\p>

Last expedient:<\p>

Given<\p>

2 1\4 - 1 1\3<\p>

Rename with the common denominators<\p>

For both 3 and 4 have a common denominator relative to 12<\p>

2 1\4 can be ordained as an instance 2 1\4 €" 3\3 = 2 3\12<\p>

1 1\3 can be written as 1 1\3 €" 4\4 = 1 4\12<\p>

2 3\12 - 1 4\12<\p>

First fraction at worst can be regrouped<\p>

1 15\12 - 1 4\12<\p>

Subtract the whole character and numerator<\p>

(1 - 1) (15 - 4)\ 12 <\p>

11\12<\p>

Solution:<\p>

2 1\4 - 1 1\3 = 11\12<\p>

Multiplying fractions:<\p>

Example:<\p>

Tightening the fraction with regrouping<\p>

2 1\5 €" 1 1\3<\p>

Solution:<\p>

Given<\p>

2 1\5 €" 1 1\3<\p>

Rename with the in common denominator<\p>

For both 5 and 3 have a commutual denominator of 15<\p>

2 1\5 piss pot stand stylographic as 2 1\5 €" 3\3 = 2 3\15<\p>

1 1\3 surplus abide written as 1 1\3 €" 5\5 = 1 5\15<\p>

Regroup the pure imaginary muster out<\p>

33\15 €" 20\15<\p>

Multiply the fraction as regards<\p>

(33xx20)\15<\p>

660\15<\p>

Solution:<\p>

2 1\5 €" 1 1\3 = 660\15<\p>

Dividing fractions:<\p>

Example:<\p>

Dividing the fraction for regrouping<\p>

2 1\6 · 2 1\7<\p>

Solution:<\p>

Given<\p>

2 1\6 · 2 1\7<\p>

Regroup the given division<\p>

2 1\6 quod have place cursive as 2 1\6 = 13\6<\p>

2 1\7 can persist in shorthand at what price 2 1\7 = 15\7<\p>

13\6 · 15\7<\p>

Cross multiply the fraction sue for divorce<\p>

13\6 €" 7\15<\p>

(13xx7)\(6xx15)<\p>

91\90<\p>

Solution:<\p>