Middle School Math & Grammar/ProOneSoftware-MiddleSchoolMath&Grammar-Win31.iso/math/chapter2.4b

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 353
àï2.4ëSubtracting Fractions.
äëPlease Subtract the following Fractions that have
the same denominators.
â

êêë25ë14êï25 - 14êï11
#êêë──ï-ï──è =è ───────è =è ──
êêë37ë37êë37êè 37
éSëIn order to perform the subtraction,ï25/37ï-ï14/37, you
should just write the denominator down once and subtract the tops.
êêë25ë14êï25 - 14êï11
#êêë──ï-ï──è =è ───────è =è ──
êêë37ë37êë37êè 37
Subtraction can also be done in a column.
#êêêêë╪╩═╪
êêêêë 37
#êêêêë╪╔╠╪
#êêêêï╘ ╪╪╦╧╪╪

#êêêêë╪╔╔╪
êêêêë 37
 1êêêè7ë 5
#êêè Subtractë ──ï-ï──
êêêêè 12ë12

êë4êêï1êêè 1
#êA)ï──êëB)ï─êë C)ï──êè D)ïå
êè 12êêï6êêè12
ü

êê7ë 5êï7 - 5êè2êï1
#êë ──ï-ï──è =è ─────è =è ──è =è ─
êë 12ë12êè 12êè12êï6
Ç B
 2êêêï23ë17
#êêè Subtractë ──ï-ï──
êêêêè 30ë30

êè 1êêè 7êêè4
#êA)ï─êë B)ï──êëC)ï──êè D)ïå
êè 5êêè30êêï30
ü

êë 23ë17êï23 - 17êè6êï1
#êë ──ï-ï──è =è ───────è =è ──è =è ─
êë 30ë30êè 30êë30êï5
Ç A
 3êêêï33ë16
#êêè Subtractë ──ï-ï──
êêêêè 45ë45

êè 2êêè2êêè17
#êA)ï─êë B)ï─êë C)ï──êè D)ïå
êè 5êêè9êêè45
ü

êë 33ë16êï33 - 16êï17
#êë ──ï-ï──è =è ───────è =è ──
êë 45ë45êè 45êë45
Ç C
 4êêêï11ë 7
#êêè Subtractë ──ï-ï──
êêêêè 24ë24

êè 1êêè 5êêè5
#êA)ï─êë B)ï──êëC)ï──êè D)ïå
êè 6êêè24êêï12
ü

êë 11ë 7êï11 - 7êè 4êï1
#êë ──ï-ï──è =è ──────ë=è ──è =è ─
êë 24ë24êè 24êë24êï6
Ç A
 5êêêï42ë19
#êêè Subtractë ──ï-ï──
êêêêè 75ë75

êè 2êêè23êêï14
#êA)ï─êë B)ï──êëC)ï──êè D)ïå
êè 5êêè75êêï25
ü

êë 42ë19êï42 - 19êï23
#êë ──ï-ï──è =è ───────è =è ──
êë 75ë75êè 75êë75
Ç B
äëPlease Subtract the following Fractions that have
unlike denominators.
âè Subtractè 7ë3êè7ê 3
#êêê ──ï-ï─è =è ───ï-ï─────è=
êêê 10ë8êï2∙5ë2∙2∙2

 2∙2è 7ê 3ë5êï28ë15êï28 - 15ê13
# ─── ∙ ───ï-ï───── ∙ ─è =è ──ï-ï──è =è ───────è=è──
 2∙2è2∙5ë2∙2∙2è5êï40ë40êë40êï40
éSè To Subtract the fractions,ï7/10ïandï3/8 , it is first
necessary to express the denominators in prime factored form.
êêê7ë3êè7ê 3
#êêë ──ï-ï─è =è ───ï-ï─────
êêë 10ë8êï2∙5ë2∙2∙2
At this point you can see that the first denominator is missing two
factors of "2", and the second denominator is missing a factor of "5".
It is necessary to multiply both the top and bottom of the first frac-
tion by "2∙2" and the second fraction by "5".
êê 7ê 3êè 2∙2è 7ê 3ë5
#êê───ï-ï─────è =è ─── ∙ ───ï-ï───── ∙ ─
êê2∙5ë2∙2∙2êï2∙2è2∙5ë2∙2∙2è5
Now both denominators have all the same factors and you can multiply
them out.ë2∙2è 7ê 3ë5êï28ë15
#êê─── ∙ ───ï-ï───── ∙ ─è =è ──ï-ï──
êê2∙2è2∙5ë2∙2∙2è5êï40ë40
Since ç two fractions have the same denominator, 40, you can write
the denominator down once and subtract the tops.
êêè 28ë15êï28 - 15êï13
#êêè ──ï-ï──è =è ───────è =è ──
êêè 40ë40êë40êè 40
It is also possible to Subtract ç fractions in a column.
#êêë ╪╧êè╪╪êë ╩╨
#êêë 10è ╫ë40ë ╫ë 40
#êêê╦êè╪╪êë ╔═
#êêè╘ ╪╪╨╪╪ï╫è╘╪╠╚╪ë╫è ╘╪╠╚╪

#(The least common denominator of 8 and 10ë ╔╦
is 40.ïThe missing numerators are 28 andë 40
15.)
 6êêêï7ë 4
#êê Subtractêï─ï-ï──
êêêêè 9ë15

êè23êêè7êêï2
#ë A)ï──êëB)ï──êëC)ï─êëD)ïå
êè45êêï15êêï5
üêë7ë 4êè7ê4
#êêè ─ï-ï──è =è ───ï-ï───è=
êêè 9ë15êï3∙3ë3∙5

ï5è 7ê4è 3êï35ë12êï35 - 12êï23
#ï─ ∙ ───ï-ï─── ∙ ─è =è ──ï-ï──è =è ───────è =è ──
ï5è3∙3ë3∙5è3êï45ë45êë45êè 45
Ç A
 7êêêï5ë 3
#êê Subtractêï─ï-ï──
êêêêè 8ë16

êè 1êêè7êêè2
#ë A)ï──êëB)ï──êëC)ï──êè D)ïå
êè16êêï16êêï16
üêë5ë 3êè 5êè3
#êêè ─ï-ï──è =è ─────ï-ï───────è=
êêè 8ë16êï2∙2∙2ë2∙2∙2∙2

ï2ë5êè3êë10ë 3êï10 - 3êè 7
#ï─ ∙ ─────ï-ï───────è =è ──ï-ï──è =è ──────ë=è ──
ï2è2∙2∙2ë2∙2∙2∙2êï16ë16êè 16êë16
Ç B
 8êêêï5ë1
#êê Subtractêï─ï-ï─
êêêêè 9ë6

êè 7êêï2êêè4
#ë A)ï──êëB)ï─êë C)ï─êëD)ïå
êè18êêï9êêè6
üêë5ë1êè5ê1
#êêè ─ï-ï─è =è ───ï-ï───è=
êêè 9ë6êï3∙3ë2∙3

ï2è 5ê1è 3êï10ë 3êï10 - 3êè 7
#ï─ ∙ ───ï-ï─── ∙ ─è =è ──ï-ï──è =è ──────ë=è ──
ï2è3∙3ë2∙3è3êï18ë18êè 18êë18
Ç A
 9êêêï4ë 1
#êê Subtractêï─ï-ï──
êêêêè 5ë15

êè 4êêï13êêï11
#ë A)ï──êëB)ï──êëC)ï──êè D)ïå
êè15êêï15êêï15
üêë4ë 1êï4ë 1
#êêè ─ï-ï──è =è ─ï-ï───è =
êêè 5ë15êï5ë3∙5

ï3è4ë 1êè12ë 1êï12 - 1êè11
#ï─ ∙ ─ï-ï───è =è ──ï-ï──è =è ──────ë=è ──
ï3è5ë3∙5êï15ë15êè 15êë15
Ç C
 10êêê13ë 7
#êê Subtractê ──ï-ï──
êêêêè16ë12

êè11êêè3êêè5
#ë A)ï──êëB)ï──êëC)ï──êè D)ïå
êè48êêï16êêï12
üêè 13ë 7êë13êï7
#êêè──ï-ï──è =è ───────ï-ï─────è=
êêè16ë12êï2∙2∙2∙2ë2∙2∙3

êï3ë 13êï7ë2∙2êï39ë28êï11
#êï─ ∙ ───────ï-ï───── ∙ ───è =è ──ï-ï──è =è ──
êï3è2∙2∙2∙2ë2∙2∙3è2∙2êï48ë48êï48
Ç A
 11êë33êêë2êêè 37
#êêê────êêA)ï─êë B)ï───
èSubtractêè 60êêë5êêè120
êêê 11
#êêè -ï────êêè 11
#êêê 40êê C)ï──êëD)ïå
#êêë ──────êêè40
#üêêè ╦╦êï╪╪╪êï╪╬╬
#êêêè60è ╫è 120è ╫è 120
#êêêè╔╔êï╪╪╪êï╪╦╦
#êêê ╘╪╠╚╪è╫ï╘╪╔╩╚╪è╫ï╘╪╔╩╚╪

#êêêêêêë╪╦╦êï╔╔
#êêêêêêë120è ╫è 40
Ç C
 12êë7êêë 2êêè31
#êêê───êê A)ï─êë B)ï──
èSubtractêè 8êêë 9êêè72
êêê 4
#êêè -ï───êêë3
#êêê 9êêïC)ï─êë D)ïå
#êêë ─────êêè 8
#üêêè ╧êï╪╪êï╬╦
#êêêè8è ╫è 72è ╫è 72
#êêêè╠êï╪╪êï╦╩
#êêê ╘╪╤╪è╫ï╘╪╧╩╪è╫ï╘╪╧╩╪

#êêêêêêè╦╔
êêêêêêè72
Ç B
äè Please Subtract the following Mixed Numbers.
â

êê4ê1êè 12ê 5êë7
#êë5 ─ï-ï2 ─è =è 5 ──ï-ï2 ──è =è 3 ──
êê5ê3êè 15ê15êè 15
éSëIn order to perform the subtraction,ï5 4/5ï-ï2 1/3ï, it
is first necessary to write the fractional parts in equivalent form
with a common denominator of 15.ïYou can then subtract the Whole
Numbers and the fractional parts separately.
êêêë 4êê 12
#êêêè 5 ─ë =ë 5 ──
êêêë 5êê 15
êêêë 1êêï5
#êêêï- 2 ─ë =è - 2 ──
êêêë 3êê 15
#êêêè───────êï───────
êêêêêêï7
#êêêêêë 3 ──
êêêêêê 15
Notice that in the example the fractions were subtracted side-by-side,
and here in the details the fractions were subtracted in a column.
You can see that it is possible to subtract either way.
 13êè 18êêêï1êêï3
#êêè 24 ──êê A)ï14 ──ê B)ï13 ──
èSubtractêè33êêê 33êê 11
êêê 9
#êê -ï11 ──êêê 2
#êêê33êê C)ï12 ─êïD)ïå
#êêè─────────êêë3
ü

êêê 18êï9êë 9êë 3
#êêë24 ──ï-ï11 ──è =è 13 ──è =è 13 ──
êêê 33ê 33êë33êë11
Ç B
 14êè 12êêêï7êêï5
#êêè 47 ──êê A)ï24 ──ê B)ï23 ──
èSubtractêè17êêê 17êê 17
êêê 5
#êê -ï23 ──êêêï5
#êêê17êê C)ï24 ──ê D)ïå
#êêè─────────êêë17
ü

êêê 12êï5êë 7
#êêë47 ──ï-ï23 ──è =è 24 ──
êêê 17ê 17êë17
Ç A
 15êè 1êêê 3êêï1
#êêë5 ─êêïA)ï2 ─êèB)ï2 ─
èSubtractêè8êêê 8êêï8
êêê5
#êê -è3 ─êêê 1
#êêê8êêïC)ï1 ─êèD)ïå
#êêè─────────êêè 2
üè It is necessary in this problem to borrow "1" from the "5"
and add it to the fractional part.ï( 5 1/8 equalsï4 9/8 )
#┌ï1êè 1êê 1êè 8è1êè 9êï9 ┐
#│5 ─è=è5 + ─è=è4 + 1 + ─è=è4 + ─ + ─è=è4 + ─è=è4 ─ │
#└ï8êè 8êê 8êè 8è8êè 8êï8 ┘

êë1ê5êè 9ê5êè 4êè 1
#êè5 ─ï-ï3 ─è =è 4 ─ï-ï3 ─è =è 1 ─è =è 1 ─
êë8ê8êè 8ê8êè 8êè 2
Ç C
 16êè 7êêêï13êê 1
#êêè 15 ─êêïA)ï11 ──ê B)ï12 ─
èSubtractêè8êêêï24êê 8
êêê1
#êê -è4 ─êêêï3
#êêê3êêïC)ï11 ─êïD)ïå
#êêè─────────êêë8
ü

êê7ê1êë21ê 8êë13
#êè 15 ─ï-ï4 ─è =è 15 ──ï-ï4 ──è =è 11 ──
êê8ê3êë24ê24êë24
Ç A
 17êë5êêêï7êê 17
#êêè 15 ──êê A)ï11 ──ê B)ï12 ──
èSubtractêè12êêê 12êê 24
êêê17
#êê -è2 ──êêêï1
#êêê24êê C)ï13 ──ê D)ïå
#êêè─────────êêë12
üêë5ê17êë10ê17
#êê15 ──ï-ï2 ──è =è 15 ──ï-ï2 ──
êêè12ê24êë24ê24
You can see that it will be necessary to borrow "1" from "15" in this
problem since 10 is smaller than 17. ( 15 10/24ïequalsï14 34/24 )
êê10ê17êë34ê17êë17
#êè 15 ──ï-ï2 ──è =è 14 ──ï-ï2 ──è =è 12 ──
êê24ê24êë24ê24êë24
Ç B
 18è How much milk is left in a four gallon container if 2/3
of a gallon has been used?

êë1êêè2êêè2
#ë A)ï3 ─ galêB)ï3 ─ galêC)ï2 ─ galêD)ïå
êë3êêè3êêè3
ü

êêêï2êè 3ë2êè 1
#êêè4ï-ï─è =è 3 ─ï-ï─è =è 3 ─ gal
êêêï3êè 3ë3êè 3
Ç A
 19è If aï2 1/4ïfoot section is cut from aï3 1/2ïfoot
board, how much is left?

êë3êê 3êêë1
#ë A)ï1 ─ ftê B)ï─ ftêèC)ï1 ─ ftê D)ïå
êë4êê 4êêë4
ü

êêè1ê1êè 2ê1êè 1
#êê 3 ─ï-ï2 ─è =è 3 ─ï-ï2 ─è =è 1 ─ ft
êêè2ê4êè 4ê4êè 4
Ç C