Basic Algebra/Working with Numbers/Subtracting Rational Numbers

Lesson
Subtracting fractions when the denominators are equal is also easy
 * $$ \frac{3}{7} - \frac{2}{7} = \frac{1}{7}$$

Notice that when the denominators are the same for either adding or subtracting fractions we only add or subtract the numerators. With this in mind we can now use the techniques learned in Lesson 4 to subtract more complex fractions by finding the factors of the following fractions
 * $$ \frac{121}{456} - \frac{61}{570} = \frac{11 \times 11}{2 \times 2 \times 2 \times 3 \times 19} - \frac{61}{2 \times 3 \times 5 \times 19}$$

We can now see that multiplying 456 by 5 and 570 by 4 gives the smallest possible denominator
 * $$\frac{121 \times 5}{5 \times 456} - \frac{4 \times 61}{4 \times 570} = \frac{605 - 244}{2280} = \frac{361}{2280}$$

Now the question is, is this fraction irreducible? We know the factors in the denominator so checking if 361 is divisible by the numbers 2, 3, 5, 19 will show that 361/19 = 19 therefore
 * $$\frac{361}{2280} = \frac{19}{120}$$

Which is now an irreducible fraction.

Example Problems
$$\frac{20(8-3)-4(10-3)}{10(2-6)} + 2(7+4)$$

$$=\frac{20(5)-4(7)}{10(-4)}+2(11)$$

$$=\frac{100-28}{-40}+22$$

$$=-\frac{72}{40}+22$$

$$=-\frac{9}{5}+\frac {5(22)}{5}$$

$$=-\frac{9}{5}+\frac {110}{5}$$

$$=\frac{101}{5}$$

$$=20\frac{1}{5}$$

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Practice Problems
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