Kopertina e librit Matematika 10 - 11: Pjesa I

Zgjidhja e ushtrimit 2

Zgjidhja e ushtrimit 2 të mësimit 10.1A në librin Matematika 10 - 11: Pjesa I nga shtëpia botuese Botime Pegi me autorë Steve Fearnley, June Haighton, Steve Lomax, Peter Mullarkey, James Nicholson dhe Matt Nixon.

Pyetja

Zgjidhni ekuacionet e mëposhtme.

  1. 4(m+7)=444(m+7)=44
  2. 5(m+6)=455(m+6)=45
  3. 8(p−3)=488(p-3)=48
  4. 7(p−15)=637(p-15)=63
  5. 20=2(f+6)20=2(f+6)
  6. 36=4(h−7)36=4(h-7)
  7. −10=2(v+6)-10=2(v+6)
  8. −12=4(x−7)-12=4(x-7)

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Zgjidhja

  1. 4m+28=44⇒4m=44−28⇒4m=16⇒m=164=44m+28=44 \rArr 4m=44-28 \rArr 4m=16 \rArr m=\dfrac{16}{4}=4
  2. 5m+30=45⇒5m=45−30⇒5m=15⇒m=155=35m+30=45 \rArr 5m=45-30 \rArr 5m=15 \rArr m=\dfrac{15}{5}=3
  3. 8p−24=48⇒8p=48+24⇒8p=72⇒p=728=98p-24=48 \rArr 8p=48+24 \rArr 8p=72 \rArr p=\dfrac{72}{8}=9
  4. 7p−105=63⇒7p=63+105⇒7p=168⇒p=1687=247p-105=63 \rArr 7p=63+105 \rArr 7p=168 \rArr p = \dfrac{168}{7}=24
  5. 2(f+6)=20⇒2f+12=20⇒2f=20−12⇒2f=8⇒f=82=42(f+6)=20 \rArr 2f+12=20 \rArr 2f=20-12 \rArr 2f=8 \rArr f=\dfrac{8}{2}=4
  6. 4(h−7)=36⇒4h−28=36⇒4h=36+28⇒4h=64⇒h=644=164(h-7)=36 \rArr 4h-28=36 \rArr 4h=36+28 \rArr 4h=64 \rArr h=\dfrac{64}{4}=16
  7. 2(v+6)=−10⇒2v+12=−10⇒2v=−10−12⇒2v=−22⇒v=−222=−112(v+6)=-10 \rArr 2v+12=-10 \rArr 2v=-10-12 \rArr 2v=-22 \rArr v=\dfrac{-22}{2}=-11
  8. 4(x−7)=−12⇒4x−28=−12⇒4x=−12+28⇒4x=16⇒x=164=44(x-7)=-12 \rArr 4x-28=-12 \rArr 4x=-12+28 \rArr 4x=16 \rArr x=\dfrac{16}{4}=4