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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