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R ﻲﻓ ﻦﻴﺗﻮﻄﺨﻟﺍ ﺕﺍﺫ ِﺔﻳﺮﺒﺠﻟﺍ ِﺕﺎﻨﻳﺎﺒﺘﻤﻟﺍ ﻞﺣ ُ ﺱﺭﺪﻟﺍ
ﱡ

Solving Two-steps Algebraic Inequalities in R [4-4]

ﱠ
ﻢﻠَﻌَﺗ ﺱﺭﺪﻟﺍ ُﺓﺮﻜﻓ
ِ
ﺔﻨﺳ
ﺏﺭﺪﺘﻳ

ﻰﻠﻋ ُ ﺏﺭﺪﺘﻳ ﺔﻨﺳ 14 ﻩﺮﻤﻋ ﻦﻴﺳﺎﻳ ﺕﺍﺫ ِﺔﻳﺮﺒﺠﻟﺍ ِﺕﺎﻨﻳﺎﺒﺘﻤﻟﺍ ﻞَﺣ
ﻰﻠﻋ ُ
ﱡ
ﺮﻜﻔﻳﻭ
ِﺔﻛﺭﺎﺸﻤﻟﺍ ﻲﻓ ﺮﻜﻔﻳﻭ ،ﻡﺪﻘﻟﺍ ِﺓﺮﻛﺓﺮﻛ ِﺕﺎﻴﻠﻤﻌﻟﺍ ﻝﺎﻤﻌﺘﺳﺎﺑ ﻦﻴﺗﻮﻄﺨﻟﺍ
ﻡﺪﻘﻟﺍ
ﻲﻓ
ﺔﻛﺭﺎﺸﻤﻟﺍ

،
ِ
ِ
ِِ
ِ
. ﻲﻨﻁﻮﻟﺍ ﻖﻳﺮﻔﻟﺍ ﻲﻓ ﻰﻠﻋ ﻞﺤﻟﺍ ﻞﻴﺜﻤﺗﻭ ﻊﺑﺭﻷﺍ
ﱢ
ِ
ًً
ﺪﻌﺑ ِﺪﻳﺪﺤﺘﻟ ﺎﻬﻠﺣﻭ ﺔﻨﻳﺎﺒﺘﻣ ْ ﺐﺘﻛﺃ ﱢ .ﺩﺍﺪﻋﻷﺍ ﻢﻴﻘﺘﺴﻣ
ﺔﻨﻳﺎﺒﺘﻣ ْ

ﺐﺘﻛﺃ

ﺪﻳﺪﺤﺘﻟ
ﺎﻬﻠﺣﻭ
ﺪﻌﺑ ِ
ﻖﻳﺮﻔﻠﻟ ﻡﺎﻤﻀﻧﻷﺍ ﻪﻨﻜﻤﻳ ٍﺔﻨﺳ ﻢﻛ
ﻖﻳﺮﻔﻠﻟ ﻡﺎﻤﻀﻧﻷﺍ ﻪﻨﻜﻤﻳ ٍ ﺔﻨﺳ ﻢﻛ ﺕﺍﺩﺮﻔﻤﻟﺍ
ِِ
.ﻲﻨﻁﻮﻟﺍ .ﺔﻳﺮﺒﺠﻟﺍ ﺔﻨﻳﺎﺒﺘﻤﻟﺍ
ﱢ
ُ
ُ
،(ﻕﻮﻓ ﺎﻤﻓ 27ﺮﻤﻌﻟﺍ) ﻲﻨﻁﻮﻟﺍ ﻖﻳﺮﻔﻟﺍ ، (16 - 21ﺮﻤﻌﻟﺍ) ﻦﻴﺌﺷﺎﻨﻟﺍ ﻖﻳﺮﻓ .ﻞﺤﻟﺍ ﺔﻋﻮﻤﺠﻣ
ُ
.(22 - 26ﺮﻤﻌﻟﺍ) ﺏﺎﺒﺸﻟﺍ ﻖﻳﺮﻓ ﱢ
ﺡﺮﻄﻟﺍﻭ ﻊﻤﺠﻟﺍ ﻝﺎﻤﻌﺘﺳﺎﺑ ﻦﻴﺗﻮﻄﺨﻟﺍ ﺕﺍﺫ ِﺔﻳﺮﺒﺠﻟﺍ ِﺕﺎﻨﻳﺎﺒﺘﻤﻟﺍ ﻞﺣ [4-4-1]
ّ
ِ
ِ
Solving Two-steps Algebraic Inequalities by Using addition and subtraction
َ
ً
ً
ُ
ً
ﻮﻫ ﺔﺤﻴﺤﺻ ﺔﻨﻳﺎﺒﺘﻤﻟﺍ ﻞﻌﺠﻳ ٍﺩﺪﻋ ﻞﻛﻭ ،ﺔﻳﺮﺒﺟ ﺔﻨﻳﺎﺒﺘﻣ ﺮﺜﻛﺃ ﻭﺃ ﺮﻴﻐﺘﻣ ﻰﻠﻋ ﻱﻮﺘﺤﺗ ﻲﺘﻟﺍ ﺔﻨﻳﺎﺒﺘﻤﻟﺍ ﻰﻤﺴﺗ
ٍ
ُ
ِﺩﺍﺪﻋﻷﺍ ﻢﻴﻘﺘﺴﻣ ﻰﻠﻋ ﺎﻬﻠﻴﺜﻤﺗ ﻦﻜﻤﻳﻭ ، ﻞﺤﻟﺍ ِﺔﻋﻮﻤﺠﻤﺑ ﺔﻨﻳﺎﺒﺘﻤﻠﻟ ﻝﻮﻠﺤﻟﺍ ﺔﻋﻮﻤﺠﻣ ﻰﻤﺴﺗﻭ ، ِﺔﻨﻳﺎﺒﺘﻤﻠﻟ ﻞﺣ
ِ
ﱞ
ﱢ
. ِﺔﻴﻘﻴﻘﺤﻟﺍ
: ﺔﻴﻘﻴﻘﺤﻟﺍ ﺩﺍﺪﻋﻷﺍ ﻰﻠﻋ ﺕﺎﻨﻳﺎﺒﺘﻤﻟﺍ ﺹﺍﻮﺧ ﻦﻣ
ُ
a + c ≥ b + c ﻥﺈﻓ a ≥ b ﻥﺎﻛ ﺍﺫﺇ ، a, b ,c ˥ R ﻞﻜﻟ : ﻊﻤﺠﻟﺍ ﺔﻴﺻﺎﺧ (1
ِ
ُ
a - c ≥ b - c ﻥﺈﻓ a ≥ b ﻥﺎﻛ ﺍﺫﺇ ، a, b ,c ˥ R ﻞﻜﻟ : ﺡﺮﻄﻟﺍ ﺔﻴﺻﺎﺧ (2
ِ
ً
( > ، < ، ≤ ) ﺕﺎﻗﻼﻌﻟﺎﺑ ≥ ﻝﺍﺪﺒﺘﺳﺍ ِﺔﻟﺎﺣ ﻲﻓ ﺔﺤﻴﺤﺻ ﻰﻘﺒﺗ (2ﻭ (1 ﺕﺎﻗﻼﻌﻟﺍ
ً
ﻦﻣ ﻦﻜﻤﺘﻴﻟ ﻦﻴﺳﺎﻳ ﺎﻫﺮﻈﺘﻨﻳ ﻲﺘﻟﺍ ﺕﺍﻮﻨﺴﻟﺍ ِﺩﺪﻋ ِﺩﺎﺠﻳﻹ ﺎﻬﻠﺣﻭ ﺔﻟﺄﺴﻤﻟﺍ ﻞﺜﻤﺗ ﺔﻨﻳﺎﺒﺘﻣ ْ ﺐﺘﻛﺃ (1) ﻝﺎﺜﻣ
.ﻲﻨﻁﻮﻟﺍ ﻖﻳﺮﻔﻠﻟ ﻡﺎﻤﻀﻧﻻﺍ
x + 14 ≥ 27 ﻲﻫ ﺔﻟﺄﺴﻤﻟﺍ ﻞﺜﻤﺗ ﻲﺘﻟﺍ ﺔﻨﻳﺎﺒﺘﻤﻟﺍ
x +14 -14 ≥ 27 - 14 ﺔﻨﻳﺎﺒﺘﻤﻟﺍ ﻲﻓﺮﻁ ﻰﻟﺍ -14 ﻒﺿﺍ
x ≥ 13 .ﻲﻨﻁﻮﻟﺍ ﻖﻳﺮﻔﻟﺍ ﻰﻟﺍ ﻡﺎﻤﻀﻧﻻﺍ ﻦﻣ ﻞﻗﻷﺍ ﻲﻓ ﺔﻨﺳ 13 ﺪﻌﺑ ﻦﻴﺳﺎﻳ ُﻦﻜﻤﺘﻳ
ّ
ِ
: ِﺩﺍﺪﻋﻷﺍ ﻢﻴﻘﺘﺴﻣ ﻰﻠﻋ ﻪّﻠﺜﻣﻭ ﺡﺮﻄﻟﺍﻭ ﻊﻤﺠﻟﺍ ﱢ ﺹﺍﻮﺧ ﻝﺎﻤﻌﺘﺳﺎﺑ R ﻲﻓ ِﺔﻴﻟﺎﺘﻟﺍ ِﺕﺎﻨﻳﺎﺒﺘﻤﻟﺍ ﻞﺣ (2) ﻝﺎﺜﻣ
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ِ
ِ
ِ
i) 3x -12 ≤ 2x -6 ⇒ 3x -2x ≤ 12 -6 ⇒ x ≤ 6
.......... 3 4 5 6 7 8
5
5 12
ii) 2z - > z - 12 ⇒ 2z -z > - ⇒ z > -1
7 7 7 7 -3 -2 -1 0 1 2 3 4 5 6 ..........
: ﺡﺮﻄﻟﺍﻭ ﻊﻤﺠﻟﺍ ﱢ ﺹﺍﻮﺧ ﻝﺎﻤﻌﺘﺳﺎﺑ R ﻲﻓ ِﺔﻴﻟﺎﺘﻟﺍ ِﺕﺎﻨﻳﺎﺒﺘﻤﻟﺍ ﻞﺣ
ّ
(3) ﻝﺎﺜﻣ
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ِ
i) 3(y - 2 ) < 2y + 2 ⇒ 3y -3 2 <2y + 2 ⇒ 3y- 2y < 2 +3 2 ⇒ y<4 2
ii) 5t + −8 ≥ 6t - 3 27 ⇒ 5t -2 ≥ 6t - 3 ⇒ 3 -2 ≥ 6t – 5t ⇒ t ≤ 1
3
1
3
1
3
iii) 8( h+ 3 ) < 0 ⇒ 8× h + 8× 3 < 0 ⇒ h + < 0 ⇒ h < -
8 16 8 16 2 2
iv) 11( m + 3) > 10( m - 2) ⇒ 11m + 33 > 10 m - 20 ⇒ m > - 53
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