# Pre Algebra homework help

**Equivalent fractions and reducing fractions**

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**Problem:**
Find the missing numerator or denominator:

**(a)** ^{3}⁄_{5} = ^{?}⁄_{30}
**(b)** ^{2}⁄_{3} = ^{22}⁄_{?}
__Solution__:

**(a)** Comapre the denominators on the two given fractions: 5

*x* 6 = 30

Do the same with the given numerator

⇒ Missing numerator is 3

*x* 6 = 18

To do that in a single step,

Missing numerator = (

^{3}⁄_{5})

*x* 30 = 18

**Or** ^{3 x 30}⁄_{5} = 18

**Or** 3

*x* 30 ÷ 5 = 18

**(b)** The numerators are given: 2

*x* 11 = 22

Do the same with the given denominator

⇒ Missing denominator is 3

*x* 11 = 33

To do that in a single step,

Missing denominator = 22

*x* (

^{3}⁄_{2}) = 33

**Or** ^{22 x 3}⁄_{2} = 33

**Or** 22

*x* 3 ÷ 2 = 33

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**Problem:**
Reduce the following fractions to their lowest terms:

**(a)** ^{30}⁄_{45}
**(b)** ^{32}⁄_{40}
**(c)** ^{18}⁄_{21}
**(d)** ^{10}⁄_{33}
__Solution__:

**(a)** ^{30}⁄_{45}
Find a number (except 1) that can divide into both 30 and 45

5 is such a number

**∴** ^{30}⁄_{45} = ^{30÷ 5}⁄_{45÷ 5} = ^{6}⁄_{9}
Find a number (except 1) that can divide into both 6 and 9

3 is the only such number

**∴** ^{6}⁄_{9} = ^{6÷ 3}⁄_{9÷ 3} = ^{2}⁄_{3}
There is no number (except 1) that can divide into both 2 and 3 and therefore,

^{30}⁄_{45} reduced to its lowest term is ^{2}⁄_{3}
__Note__: If you could figure out that 15 can divide into 30 and 45 both, you would need one less step:

^{30}⁄_{45} = ^{30÷ 15}⁄_{45÷ 15} = ^{2}⁄_{3}
*Lesson: Try to find as large a number as you can to divide*
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