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4.12: Tatua equations na FRACTIONS (Sehemu ya 1)

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    Malengo ya kujifunza
    • Kuamua kama sehemu ni suluhisho la equation
    • Tatua equations na sehemu ndogo kwa kutumia Kuongeza, Kutoa, na Mali ya Idara ya Usawa
    • Tatua equations kwa kutumia Mali ya Kuzidisha ya Usawa
    • Tafsiri sentensi kwa equations na kutatua
    kuwa tayari!

    Kabla ya kuanza, fanya jaribio hili la utayari. Ikiwa umepoteza tatizo, rudi kwenye sehemu iliyoorodheshwa na uhakiki nyenzo.

    1. Tathmini\(x + 4\) wakati\(x = −3\) Kama amekosa tatizo hili, mapitio Mfano 3.2.10.
    2. Kutatua:\(2y − 3 = 9\). Ikiwa umekosa tatizo hili, kagua Mfano 3.5.2.
    3. Tatua:\(y − 3 = −9\) Ikiwa umekosa tatizo hili, kagua Mfano 4.2.10.

    Kuamua Kama Fraction ni Suluhisho la Equation

    Kama tulivyoona katika Kutatua Equations na Ondoa na Kuongeza Mali ya Usawa na Kutatua Equations Kutumia Integers; Mali ya Idara ya Usawa, suluhisho la equation ni thamani ambayo inafanya taarifa ya kweli wakati kubadilishwa kwa kutofautiana katika equation. Katika sehemu hizo, tulipata idadi nzima na ufumbuzi wa integer kwa equations. Sasa kwa kuwa tumefanya kazi na sehemu ndogo, tuko tayari kupata ufumbuzi wa sehemu kwa usawa.

    Hatua tunazochukua ili kuamua kama namba ni suluhisho la equation ni sawa kama suluhisho ni namba nzima, integer, au sehemu.

    JINSI YA: TAMBUA KAMA NAMBARI NI SULUHISHO LA EQUATION

    Hatua ya 1. Badilisha idadi ya kutofautiana katika equation.

    Hatua ya 2. Kurahisisha maneno pande zote mbili za equation.

    Hatua ya 3. Kuamua kama equation kusababisha ni kweli. Ikiwa ni kweli, nambari ni suluhisho. Ikiwa si kweli, nambari sio suluhisho.

    Mfano\(\PageIndex{1}\): find solution

    Kuamua kama kila moja ya yafuatayo ni suluhisho la\(x − \dfrac{3}{10} = \dfrac{1}{2}\).

    1. \(x = 1\)
    2. \(x = \dfrac{4}{5}\)
    3. \(x = − \dfrac{4}{5}\)

    Suluhisho

    Mbadala\(\textcolor{red}{1}\) kwa ajili ya x. \(\textcolor{red}{1} - \dfrac{3}{10} \stackrel{?}{=} \dfrac{1}{2}\)
    Badilisha kwa sehemu ndogo na LCD ya 10. \(\textcolor{red}{\dfrac{10}{10}} - \dfrac{3}{10} \stackrel{?}{=} \dfrac{5}{10}\)
    Ondoa. \(\dfrac{7}{10} \stackrel{?}{=} \dfrac{5}{10}\)

    Tangu\(x = 1\) haina kusababisha equation kweli,\(1\) si ufumbuzi wa equation.

    Mbadala\(\textcolor{red}{\dfrac{4}{5}}\) kwa ajili ya x. \(\textcolor{red}{\dfrac{4}{5}} - \dfrac{3}{10} \stackrel{?}{=} \dfrac{1}{2}\)
      \(\textcolor{red}{\dfrac{8}{10}} - \dfrac{3}{10} \stackrel{?}{=} \dfrac{5}{10}\)
    Ondoa. \(\dfrac{5}{10} = \dfrac{5}{10} \; \checkmark\)

    Kwa kuwa\(x = \dfrac{4}{5}\) matokeo katika equation kweli,\(\dfrac{4}{5}\) ni suluhisho la equation\(x − \dfrac{3}{10} = \dfrac{1}{2}\).

    Mbadala\(\textcolor{red}{- \dfrac{4}{5}}\) kwa ajili ya x. \(\textcolor{red}{- \dfrac{4}{5}} - \dfrac{3}{10} \stackrel{?}{=} \dfrac{1}{2}\)
      \(\textcolor{red}{- \dfrac{8}{10}} - \dfrac{3}{10} \stackrel{?}{=} \dfrac{5}{10}\)
    Ondoa. \(\dfrac{11}{10} \neq \dfrac{5}{10}\)

    Tangu\(x = − \dfrac{4}{5}\) haina kusababisha equation kweli,\(− \dfrac{4}{5}\) si ufumbuzi wa equation.

    Zoezi\(\PageIndex{1}\)

    Kuamua kama kila namba ni suluhisho la equation iliyotolewa. \(x − \dfrac{2}{3} = \dfrac{1}{6}\)

    1. \(x = 1\)
    2. \(x = \dfrac{5}{6}\)
    3. \(x = − \dfrac{5}{6}\)
    Jibu

    hapana

    Jibu b

    ndiyo

    Jibu c

    hapana

    Zoezi\(\PageIndex{2}\)

    Kuamua kama kila namba ni suluhisho la equation iliyotolewa. \(y − \dfrac{1}{4} = \dfrac{3}{8}\)

    1. \(y = 1\)
    2. \(y = - \dfrac{5}{8}\)
    3. \(y = \dfrac{5}{8}\)
    Jibu

    hapana

    Jibu b

    hapana

    Jibu c

    ndiyo

    Kutatua Equations na Fractions kwa kutumia Kuongeza, Kutoa, na Idara ya Mali ya Usawa

    Katika Kutatua Equations na Ondoa na Kuongeza Mali ya Usawa na Kutatua Equations Kutumia Integers; Mali ya Idara ya Usawa, tulitatua equations kwa kutumia Kuongeza, Kutoa, na Mali ya Idara ya Usawa. Tutatumia mali hizi sawa kutatua equations na sehemu ndogo.

    Aidha, Ondoa, na Idara ya Mali ya Usawa

    Kwa idadi yoyote\(a\),\(b\), na\(c\),

    Jedwali\(\PageIndex{1}\)
    kama\(a = b\), basi\(a + c = b + c\). Kuongeza Mali ya Usawa
    kama\(a = b\), basi\(a − c = b − c\). Ondoa Mali ya Usawa
    kama\(a = b\), basi\(a c = b c \),\(c ≠ 0\). Idara ya Mali ya Usawa

    Kwa maneno mengine, unapoongeza au kuondoa kiasi sawa kutoka pande zote mbili za equation, au ugawanye pande zote mbili kwa kiasi sawa, bado una usawa.

    Mfano\(\PageIndex{2}\): solve

    Kutatua:\(y + \dfrac{9}{16} = \dfrac{5}{16}\).

    Suluhisho

    Ondoa\(\dfrac{9}{16}\) kutoka kila upande ili uondoe kuongeza. \(y + \dfrac{9}{16} \textcolor{red}{- \dfrac{9}{16}} = \dfrac{5}{16} \textcolor{red}{- \dfrac{9}{16}}\)
    Kurahisisha kila upande wa equation. \(y + 0 = - \dfrac{4}{16}\)
    Kurahisisha sehemu. \(y = - \dfrac{1}{4}\)

    Angalia:

    Mbadala y = (-\ dfrac {1} {4}\). \(\textcolor{red}{- \dfrac{1}{4}} + \dfrac{9}{16} \stackrel{?}{=} \dfrac{5}{16}\)
    Andika upya kama sehemu ndogo na LCD. \ textcolor {nyekundu} {-\ dfrac {4} {16}} +\ dfrac {9} {16}\ stackrel {?} {=}\ dfrac {5} {16}\)
    Ongeza. \(\dfrac{5}{16} \stackrel{?}{=} \dfrac{5}{16} \; \checkmark\)

    Tangu\(y = − \dfrac{1}{4}\) inafanya\(y + \dfrac{9}{16} = \dfrac{5}{16}\) kauli ya kweli, tunajua tumegundua ufumbuzi wa equation hii.

    Zoezi\(\PageIndex{3}\)

    Kutatua:\(y + \dfrac{11}{12} = \dfrac{5}{12}\).

    Jibu

    \(-\dfrac{1}{2}\)

    Zoezi\(\PageIndex{4}\)

    Kutatua:\(y + \dfrac{8}{15} = \dfrac{4}{15}\).

    Jibu

    \(-\dfrac{4}{15}\)

    Tulitumia Mali ya Kutoa ya Usawa katika Mfano\(\PageIndex{2}\). Sasa tutatumia Mali ya Kuongeza ya Usawa.

    Mfano\(\PageIndex{3}\): solve

    Kutatua: a -\(\dfrac{5}{9}\) =\(− \dfrac{8}{9}\).

    Suluhisho

    Ongeza\(\dfrac{5}{9}\) kutoka kila upande ili uondoe kuongeza. \(a - \dfrac{5}{9} \textcolor{red}{+ \dfrac{5}{9}} = - \dfrac{8}{9} \textcolor{red}{+ \dfrac{5}{9}}\)
    Kurahisisha kila upande wa equation. \(a + 0 = - \dfrac{3}{9}\)
    Kurahisisha sehemu. \(a = - \dfrac{1}{3}\)

    Angalia:

    Badilisha a =\(− \dfrac{1}{3}\). \(\textcolor{red}{- \dfrac{1}{3}} - \dfrac{5}{9} \stackrel{?}{=} - \dfrac{8}{9}\)
    Badilisha kwa denominator ya kawaida. \(\textcolor{red}{- \dfrac{3}{9}} - \dfrac{5}{9} \stackrel{?}{=} - \dfrac{8}{9}\)
    Ondoa. \(- \dfrac{8}{9} = - \dfrac{8}{9} \; \checkmark\)

    Tangu\(a = − \dfrac{1}{3}\) inafanya equation kweli, tunajua kwamba\(a = − \dfrac{1}{3}\) ni ufumbuzi wa equation.

    Zoezi\(\PageIndex{5}\)

    Kutatua:\(a − \dfrac{3}{5} = − \dfrac{8}{5}\).

    Jibu

    \(-1\)

    Zoezi\(\PageIndex{6}\)

    Kutatua:\(n − \dfrac{3}{7} = − \dfrac{9}{7}\).

    Jibu

    \(-\dfrac{6}{7}\)

    Mfano unaofuata hauwezi kuonekana kuwa na sehemu, lakini hebu tuone kinachotokea tunapoitatua.

    Mfano\(\PageIndex{4}\): solve

    Kutatua:\(10q = 44\).

    Suluhisho

    Gawanya pande zote mbili kwa 10 ili uondoe kuzidisha. \(\dfrac{10q}{10} = \dfrac{44}{10}\)
    Kurahisisha. \(q =\dfrac{22}{5}\)

    Angalia:

    Mbadala\(q = \dfrac{22}{5}\) katika equation ya awali. \(10 \left(\dfrac{22}{5}\right) \stackrel{?}{=} 44\)
    Kurahisisha. \(\stackrel{2}{\cancel{10}} \left(\dfrac{22}{\cancel{5}}\right) \stackrel{?}{=} 44\)
    Kuzidisha. \(44 = 44 \; \checkmark\)

    Suluhisho la equation lilikuwa sehemu\(\dfrac{22}{5}\). Tunaondoka kama sehemu isiyofaa.

    Zoezi\(\PageIndex{7}\)

    Kutatua:\(12u = −76\).

    Jibu

    \(-\dfrac{19}{3}\)

    Zoezi\(\PageIndex{8}\)

    Kutatua:\(8m = 92\).

    Jibu

    \(\dfrac{23}{2}\)

    Kutatua Equations na FRACTIONS Kutumia Mali ya Kuzidisha ya Usawa

    Fikiria equation\(\dfrac{x}{4} = 3\). Tunataka kujua nini idadi kugawanywa na\(4\) anatoa\(3\). Ili “kurekebisha” mgawanyiko, tutahitaji kuzidisha na\(4\). Mali ya Kuzidisha ya Usawa itatuwezesha kufanya hivyo. Mali hii inasema kwamba ikiwa tunaanza na kiasi mbili sawa na kuzidisha wote kwa idadi sawa, matokeo ni sawa.

    Ufafanuzi: Mali ya kuzidisha ya Usawa

    Kwa idadi yoyote\(a\),\(b\), na\(c\), ikiwa\(a = b\), basi\(ac = bc\). Ikiwa unazidisha pande zote mbili za equation kwa kiasi sawa, bado una usawa.

    Hebu tumia Mali ya kuzidisha ya Usawa ili kutatua equation\(\dfrac{x}{7} = −9\).

    Mfano\(\PageIndex{5}\): solve

    Kutatua:\(\dfrac{x}{7} = −9\).

    Suluhisho

    Tumia Mali ya Kuzidisha ya Usawa ili kuzidisha pande zote mbili kwa 7. Hii itatenganisha kutofautiana. \(\textcolor{red}{7} \cdot \dfrac{x}{7} = \textcolor{red}{7} (-9)\)
    Kuzidisha. \(\dfrac{7x}{7} = -63\)
    Kurahisisha. \(x = -63\)
    Angalia. Mbadala\(\textcolor{red}{-63}\) ya x katika equation ya awali. \(\dfrac{\textcolor{red}{-63}}{7} \stackrel{?}{=} -9\)
    Equation ni kweli. \(-9 = -9 ; \checkmark\)
    Zoezi\(\PageIndex{9}\)

    Kutatua:\(\dfrac{f}{5} = −25\).

    Jibu

    \(-125\)

    Zoezi\(\PageIndex{10}\)

    Kutatua:\(\dfrac{h}{9} = −27\).

    Jibu

    \(-243\)

    Mfano\(\PageIndex{6}\):solve

    Kutatua:\(\dfrac{p}{−8} = −40\).

    Suluhisho

    Hapa,\(p\) imegawanywa na\(−8\). Lazima tuzidishe na\(−8\) kujitenga\(p\).

    Panua pande zote mbili kwa -8. \(\textcolor{red}{-8} \left(\dfrac{p}{-8}\right) = \textcolor{red}{-8} (-40)\)
    Kuzidisha. \(\dfrac{-8p}{-8} = 320\)
    Kurahisisha. \(p = 320\)

    Angalia:

    Mbadala p = 320. \(\dfrac{\textcolor{red}{320}}{-8} \stackrel{?}{=} -40\)
    Equation ni kweli. \(-40 = -40 \; \checkmark\)
    Zoezi\(\PageIndex{11}\)

    Kutatua:\(\dfrac{c}{−7} = −35\).

    Jibu

    \(245\)

    Zoezi\(\PageIndex{12}\)

    Kutatua:\(\dfrac{x}{−11} = −12\).

    Jibu

    \(132\)

    Kutatua Equations na mgawo wa\(−1\)

    Angalia equation\(−y = 15\). Je! Inaonekana kama y tayari imetengwa? Lakini kuna ishara hasi mbele ya\(y\), hivyo sio pekee.

    Kuna njia tatu tofauti za kutenganisha variable katika aina hii ya equation. Tutaonyesha njia zote tatu katika Mfano\(\PageIndex{7}\).

    Mfano\(\PageIndex{7}\): solve

    Kutatua:\(−y = 15\).

    Suluhisho

    Njia moja ya kutatua equation ni kuandika upya\(−y\) kama\(−1y\), na kisha kutumia Idara ya Mali ya Usawa kutenganisha\(y\).

    Andika upya -y kama -1y. \(-1y = 15\)
    Gawanya pande zote mbili kwa -1. \(\dfrac{-1y}{\textcolor{red}{-1}} = \dfrac{15}{\textcolor{red}{-1}}\)
    Kurahisisha kila upande. \(y = -15\)

    Njia nyingine ya kutatua equation hii ni kuzidisha pande zote mbili za equation na -1.

    Panua pande zote mbili kwa -1. \(\textcolor{red}{-1} (-y) = \textcolor{red}{-1} (15)\)
    Kurahisisha kila upande. \(y = -15\)

    Njia ya tatu ya kutatua equation ni kusoma\(−y\) kama “kinyume cha\(y\).” Nambari gani ina\(15\) kinyume chake? Kinyume cha\(15\) ni\(−15\). Hivyo\(y = −15\).

    Kwa njia zote tatu, sisi pekee\(y\) ni pekee na kutatuliwa equation.

    Angalia:

    Mbadala y = -15. \(-(\textcolor{red}{15}) \stackrel{?}{=} (15)\)
    Kurahisisha. Equation ni kweli. \(15 = 15 \; \checkmark\)
    Zoezi\(\PageIndex{13}\)

    Kutatua:\(−y = 48\).

    Jibu

    \(-48\)

    Zoezi\(\PageIndex{14}\)

    Kutatua:\(−c = −23\).

    Jibu

    \(23\)

    Wachangiaji na Majina