Addition or Subtraction of Fractions With Different Denominators

Definition

When the denominators of any fractions are unequal or are different those fractions are called unlike fractions.

Operations like addition and subtraction cannot be done directly on unlike fractions.

These unlike fractions are first converted into like fractions by finding the least common denominator of these fractions and rewriting the fractions into equivalent fractions with same denominators (LCD)

Adding unlike fractions − Formula

When fractions with different or unlike fractions are to be added, first the least common denominator of the fractions is found. The equivalent fractions of given fractions are found with LCD as the common denominator. The numerators are now added and the result is put over the LCD to get the sum of fractions.

• We find the least common denominator of all the fractions.
• We rewrite the fractions to have the denominators equal to the LCD obtained in first step .
• We add the numerators of all the fractions keeping the denominator value equal to the LCD obtained in first step.
• We then express the fraction in lowest terms.

Subtracting unlike fractions − Formula

When fractions with different or unlike fractions are to be subtracted, first the least common denominator of the fractions is found. The equivalent fractions of given fractions are found with LCD as the common denominator. The numerators are now subtracted and the result is put over the LCD to get the difference of the given fractions.

• We find the least common denominator of all the fractions.
• We rewrite the fractions to have the denominators equal to the LCD obtained in step 1.
• We subtract the numerators of all the fractions keeping the denominator value equal to the LCD obtained in step 1.
• We express the fraction in lowest terms.

Problem 1:

Add 15$\frac{1}{5}$ + 27$\frac{2}{7}$

Solution

Step 1:

Add 15$\frac{1}{5}$ + 27$\frac{2}{7}$

Here the denominators are different. As 5 and 7 are prime the LCD is their product 35.

Step 2:

Rewriting

15$\frac{1}{5}$ + 27$\frac{2}{7}$ = (1×7)(5×7)$\frac{(1\times 7)}{(5\times 7)}$ + (2×5)(7×5)$\frac{(2\times 5)}{(7\times 5)}$ = 735$\frac{7}{35}$ + 1035$\frac{10}{35}$

STEP 3:

As the denominators have become equal

735$\frac{7}{35}$ + 1035$\frac{10}{35}$ = (7+10)35$\frac{(7+10)}{35}$ = 1735$\frac{17}{35}$

STEP 4:

So, 15$\frac{1}{5}$ + 27$\frac{2}{7}$ = 1735$\frac{17}{35}$

Problem 2:

Subtract 215$\frac{2}{15}$ − 110$\frac{1}{10}$

Solution

Step 1:

Subtract 215$\frac{2}{15}$ − 110$\frac{1}{10}$

Here the denominators are different. The LCM of 10 and 15 is 30.

Step 2:

Rewriting

215$\frac{2}{15}$ − 110$\frac{1}{10}$ = (2×2)(15×2)$\frac{(2\times 2)}{(15\times 2)}$ − (1×3)(10×3)$\frac{(1\times 3)}{(10\times 3)}$ = 430$\frac{4}{30}$ − 330$\frac{3}{30}$

STEP 3:

As the denominators have become equal

430$\frac{4}{30}$ − 330$\frac{3}{30}$ = (4−3)30$\frac{(4-3)}{30}$ = 130$\frac{1}{30}$

STEP 4:

So, 215$\frac{2}{15}$ − 110$\frac{1}{10}$ = 130$\frac{1}{30}$

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Add or Subtract Fractions With Different Denominators: Advanced

Introduction

When we have addition or subtraction of fractions with unlike denominators, we first find the Least Common Denominator (LCD) of the fractions. We then rewrite all fractions as equivalent fractions with LCD as the denominator. Now that all denominators are alike, we add or subtract the numerators and put the result over the common denominator to get the answer. If necessary, we express the fraction in lowest terms.

Problem 1:

Add 35$\frac{3}{5}$ + 38$\frac{3}{8}$

Solution

Step 1:

Add 35$\frac{3}{5}$ + 38$\frac{3}{8}$

Here the denominators are different. The LCD is 40 (product of 5 and 8) as 5 and 8 are co-prime numbers.

Step 2:

Rewriting

35$\frac{3}{5}$ + 38$\frac{3}{8}$ = (3×8)(5×8)$\frac{(3\times 8)}{(5\times 8)}$ + (5×5)(8×5)$\frac{(5\times 5)}{(8\times 5)}$ = 2440$\frac{24}{40}$ + 2540$\frac{25}{40}$

As the denominators have become equal

2440$\frac{24}{40}$ + 2540$\frac{25}{40}$ = (24+25)40$\frac{(24+25)}{40}$ = 4940$\frac{49}{40}$

Step 3:

So, 35$\frac{3}{5}$ + 38$\frac{3}{8}$ = 4940$\frac{49}{40}$

Problem 2:

Subtract 58$\frac{5}{8}$ − 712$\frac{7}{12}$

Solution

Step 1:

58$\frac{5}{8}$ − 712$\frac{7}{12}$

Here the denominators are different. The LCD here is 24.

Step 2:

Rewriting

58$\frac{5}{8}$ − 712$\frac{7}{12}$ = (5×3)(8×3)$\frac{(5\times 3)}{(8\times 3)}$ − (7×2)(12×2)$\frac{(7\times 2)}{(12\times 2)}$ = 1524$\frac{15}{24}$ − 1424$\frac{14}{24}$

As the denominators have become equal

1524$\frac{15}{24}$ − 1424$\frac{14}{24}$ = (15−14)24$\frac{(15-14)}{24}$ = 124$\frac{1}{24}$

Step 3:

So, 58$\frac{5}{8}$ − 712$\frac{7}{12}$ = 124

Word Problem Involving Add or Subtract Fractions With Different Denominators

Problem 1:

Jamie bought a box of fruit weighing 325$\frac{2}{5}$ kilograms. If she bought a second box that weighed 713$\frac{1}{3}$ kilograms, what is the combined weight of both boxes?

Solution

Step 1:

Weight of the first box of fruit = 325$\frac{2}{5}$ kilograms

Weight of the second box of fruit = 713$\frac{1}{3}$ kilograms

The combined of the two boxes of fruit = 325$\frac{2}{5}$ + 713$\frac{1}{3}$ = 175$\frac{17}{5}$ + 223$\frac{22}{3}$

Step 2:

The denominators are different. So the LCD of the fractions or LCM of denominators 3 and 5 is 15.

Rewriting to get equivalent fractions with LCD as denominator

17×35×3$\frac{17\times 3}{5\times 3}$ + 22×53×5$\frac{22\times 5}{3\times 5}$ = 5115$\frac{51}{15}$ + 11015$\frac{110}{15}$ = (51+110)15$\frac{(51+110)}{15}$ = 16115$\frac{161}{15}$ = 101115$\frac{11}{15}$

Problem 2:

During the weekend, Nancy spent a total 513$\frac{1}{3}$ hours studying. If she spent 314$\frac{1}{4}$ hours studying on Saturday, how long did she study on Sunday?

Solution

Step 1:

Time spent studying on the weekend = 513$\frac{1}{3}$ hours

Time spent studying on Saturday = 314$\frac{1}{4}$ hours

Time spent studying on Sunday =

Time spent studying on the weekend − Time spent studying on Saturday

= 513$\frac{1}{3}$ − 314$\frac{1}{4}$ = 163$\frac{16}{3}$ − 134$\frac{13}{4}$

Step 2:

LCD of the fractions or the LCM of the denominators 3 and 4 is 12

Rewriting to get equivalent fractions with LCD as denominator

16×43×4$\frac{16\times 4}{3\times 4}$ − 13×34×3$\frac{13\times 3}{4\times 3}$ = 6412$\frac{64}{12}$ − 3912$\frac{39}{12}$ = 64−3912$\frac{64-39}{12}$ = 2512$\frac{25}{12}$ = 2112$\frac{1}{12}$ hours

So, the time spent studying on Sunday = 2112$\frac{1}{12}$ hours

Problem 3:

Marcos bought apples that weighed 623$\frac{2}{3}$ kilograms. If he gave away 315$\frac{1}{5}$kilograms of apples to his friends, how many kilograms of apples does he have left?

Solution

Step 1:

Weight of the apples bought = 623$\frac{2}{3}$ kilograms

Weight of the apples given to friends = 315$\frac{1}{5}$ kilograms

Weight of the apples left =

Weight of the apples bought − Weight of the apples given to friends

= 623$\frac{2}{3}$ − 315$\frac{1}{5}$ = 203$\frac{20}{3}$ − 165$\frac{16}{5}$

Step 2:

LCD of the fractions or LCM of the denominators 3 and 5 is 15

Rewriting to get equivalent fractions with LCD as denominator

20×53×5$\frac{20\times 5}{3\times 5}$ − 16×35×3$\frac{16\times 3}{5\times 3}$ = 10015$\frac{100}{15}$ − 4815$\frac{48}{15}$ = 100−4815$\frac{100-48}{15}$ = 5215$\frac{52}{15}$ = 3715$\frac{7}{15}$ kilograms

So, the weight of the apples left = 3715$\frac{7}{15}$ kilograms

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