Adding and Subtracting Fractions

When dealing with fractions that have the same denominators, addition and subtraction are straightforward. For addition, simply add the numerators, while the denominator remains the same. For subtraction, subtract the second numerator from the first, and again, the denominator remains the same.

Example: Same Denominators

Addition: \frac{4}{7} + \frac{2}{7} = \frac{4+2}{7} = \frac{6}{7}

Subtraction: \frac{4}{7} - \frac{3}{7} = \frac{4-3}{7} = \frac{1}{7}

If the fractions have different denominators, find a common denominator before adding or subtracting. To do this, multiply the numerators and denominators by appropriate numbers to make the denominators equal.

Example: Different Denominators

1. \frac{5}{12} + \frac{3}{4}

Convert \frac{3}{4} to \frac{3\times3}{4\times3}, making the denominator equal to 12. Now, \frac{3\times3}{4\times3} = \frac{9}{12}.

So, \frac{5}{12} + \frac{3}{4} = \frac{5}{12} + \frac{9}{12} = \frac{5+9}{12} = \frac{14}{12}, which simplifies to \frac{7}{6} or 1\frac{1}{6}.

2. \frac{5}{8} - \frac{2}{7}

Convert \frac{5}{8} and \frac{2}{7} to \frac{5\times7}{8\times7} and \frac{2\times8}{7\times8}, respectively. This yields \frac{35}{56} and \frac{16}{56}.

Now, perform the subtraction: \frac{5}{8} - \frac{2}{7} = \frac{35}{56} - \frac{16}{56} = \frac{19}{56}.

More Challenging Examples

Example 1

Find \frac{2}{7} + \frac{3}{5} - \frac{1}{8}.

First, find the least common multiple (LCM) of the denominators 7, 5,7, 5, and 88:

LCM(7,5,8) = 2^3 \times 5 \times 7 = 280(7,5,8) = 2^3 \times 5 \times 7 = 280.

Now, rewrite the fractions with the denominator 280280:

\frac{2}{7} = \frac{2\times40}{7\times40} = \frac{80}{280}\frac{2}{7} = \frac{2\times40}{7\times40} = \frac{80}{280}

\frac{3}{5} = \frac{3\times56}{5\times56} = \frac{168}{280}\frac{3}{5} = \frac{3\times56}{5\times56} = \frac{168}{280}

\frac{1}{8} = \frac{1\times35}{8\times35} = \frac{35}{280}\frac{1}{8} = \frac{1\times35}{8\times35} = \frac{35}{280}

Therefore, \frac{2}{7} + \frac{3}{5} - \frac{1}{8} = \frac{80}{280} + \frac{168}{280} - \frac{35}{280} = \frac{(80+168-35)}{280} = \frac{213}{280}\frac{2}{7} + \frac{3}{5} - \frac{1}{8} = \frac{80}{280} + \frac{168}{280} - \frac{35}{280} = \frac{(80+168-35)}{280} = \frac{213}{280}.

Example 2

Find \frac{3}{16} + \frac{5}{24} - \frac{1}{30}, giving your answer in its simplest form.

First, find the least common multiple (LCM) of the denominators 16, 24,16, 24, and 3030:

LCM(16, 24, 30) = 2^4 \times 3 \times 5 = 240(16, 24, 30) = 2^4 \times 3 \times 5 = 240.

Now, rewrite the fractions with the denominator 240240:

\frac{3}{16} = \frac{3\times15}{16\times15} = \frac{45}{240}\frac{3}{16} = \frac{3\times15}{16\times15} = \frac{45}{240}

\frac{5}{24} = \frac{5\times10}{24\times10} = \frac{50}{240}\frac{5}{24} = \frac{5\times10}{24\times10} = \frac{50}{240}

\frac{1}{30} = \frac{1\times8}{30\times8} = \frac{8}{240}\frac{1}{30} = \frac{1\times8}{30\times8} = \frac{8}{240}

So, \frac{3}{16} + \frac{5}{24} - \frac{1}{30} = \frac{45}{240} + \frac{50}{240} - \frac{8}{240} = \frac{(45+50-8)}{240} = \frac{87}{240}\frac{3}{16} + \frac{5}{24} - \frac{1}{30} = \frac{45}{240} + \frac{50}{240} - \frac{8}{240} = \frac{(45+50-8)}{240} = \frac{87}{240}.

Since 33 is a common factor of both the numerator 8787 and the denominator 240240, you can simplify the fraction by dividing both the numerator and the denominator by 33:

\frac{87}{240} = \frac{29\times3}{80\times3} = \frac{29}{80}\frac{87}{240} = \frac{29\times3}{80\times3} = \frac{29}{80}

Since 2929 is a prime number, \frac{29}{80}\frac{29}{80} is the simplest form.

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