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8.6: Tatua equations na Fraction au Coefficients Decimal

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    Malengo ya kujifunza
    • Tatua equations na coefficients sehemu
    • Tatua equations na coefficients decimal
    kuwa tayari!

    Kabla ya kuanza, fanya jaribio hili la utayari.

    1. Kuzidisha: 8 •\(\dfrac{3}{8}\). Kama amekosa tatizo hili, mapitio Mfano 4.3.10.
    2. Kupata LCD ya\(\dfrac{5}{6}\) na\(\dfrac{1}{4}\). Ikiwa umekosa tatizo hili, kagua Mfano 4.8.1.
    3. Kuzidisha: 4.78 na 100. Ikiwa umekosa tatizo hili, kagua Mfano 5.3.8.

    Tatua equations na Coefficients Fraction

    Hebu kutumia Mkakati Mkuu wa Kutatua Equations Linear ilianzisha mapema ili kutatua equation\(\dfrac{1}{8}x + \dfrac{1}{2} = \dfrac{1}{4}\).

    Ili kutenganisha neno la x, toa\(\dfrac{1}{2}\) kutoka pande zote mbili. $$\ dfrac {1} {8} x +\ dfrac {1} {2}\ textcolor {nyekundu} {-\ dfrac {1} {2}} =\ dfrac {1} {4}\ textcolor {nyekundu} {-\ dfrac {1} {2}} $$
    Kurahisisha upande wa kushoto. $$\ dfrac {1} {8} x =\ dfrac {1} {4} -\ dfrac {1} {2} $$
    Badilisha constants kwa sehemu sawa na LCD. $$\ dfrac {1} {8} x =\ dfrac {1} {4} -\ dfrac {2} {4} $$
    Ondoa. $$\ dfrac {1} {8} x = -\ dfrac {1} {4} $$
    Kuzidisha pande zote mbili kwa usawa wa\(\dfrac{1}{8}\). $$\ textcolor {nyekundu} {\ dfrac {8} {1}}\ cdot\ dfrac {1} {8} x =\ textcolor {nyekundu} {\ dfrac {8} {1}}\ kushoto (-\ dfrac {1} {4}\ haki) $$
    Kurahisisha. $$x = -2 $$

    Njia hii ilifanya kazi nzuri, lakini wanafunzi wengi hawana ujasiri sana wanapoona sehemu zote hizo. Hivyo sisi ni kwenda kuonyesha njia mbadala ya kutatua equations na FRACTIONS. Njia hii mbadala huondoa sehemu ndogo.

    Tutatumia Mali ya Kuzidisha ya Usawa na kuzidisha pande zote mbili za equation na denominator ya kawaida ya sehemu zote katika equation. Matokeo ya operesheni hii itakuwa equation mpya, sawa na ya kwanza, lakini bila sehemu ndogo. Utaratibu huu unaitwa kusafisha equation ya sehemu ndogo. Hebu tufanye equation sawa tena, lakini wakati huu utumie njia inayofuta sehemu ndogo.

    Mfano\(\PageIndex{1}\):

    Kutatua:\(\dfrac{1}{8} x + \dfrac{1}{2} = \dfrac{1}{4}\).

    Suluhisho

    Kupata denominator angalau ya kawaida ya sehemu zote katika equation. $$\ dfrac {1} {8} x +\ dfrac {1} {2} =\ dfrac {1} {4}\ quad LCD = 8$$
    Kuzidisha pande zote mbili za equation na LCD kwamba, 8. Hii inafuta sehemu ndogo. $$\ textcolor {nyekundu} {8}\ kushoto (\ dfrac {1} {8} x +\ dfrac {1} {2}\ haki) =\ textcolor {nyekundu} {8}\ kushoto (\ dfrac {1} {4}\ haki) $$
    Tumia Mali ya Mgawanyo. $8\ cdot\ dfrac {1} {8} x + 8\ cdot\ dfrac {1} {2} = 8\ cdot\ dfrac {1} {4} $
    Kurahisisha - na taarifa, hakuna sehemu zaidi! $$x + 4 = $2 $
    Kutatua kutumia Mkakati Mkuu wa Kutatua Equations Linear. $$x + 4\ textcolor {nyekundu} {-4} = 2\ textcolor {nyekundu} {-4} $$
    Kurahisisha. $$x = -2 $$
    Angalia: Hebu x = -2. $$\ kuanza {split}\ dfrac {1} {8} x +\ dfrac {1} {2} &=\ dfrac {1} {4}\\ dfrac {1} {8} (\ textcolor {nyekundu} {-2}) +\ dfrac {1} {2} &\ stackrel {?} {=}\ dfrac {1} {4}\\ -\ dfrac {2} {8} +\ dfrac {1} {2} &\ stackrel {?} {=}\ dfrac {1} {4}\\ -\ dfrac {2} {8} +\ dfrac {4} {8} &\ stackrel {?} {=}\ dfrac {1} {4}\\ dfrac {2} {4} &\ stackrel {?} {=}\ dfrac {1} {4}\\ dfrac {1} {4} &=\ dfrac {1} {4}\;\ alama\ mwisho {mgawanyiko} $$
    Zoezi\(\PageIndex{1}\):

    Kutatua:\(\dfrac{1}{4} x + \dfrac{1}{2} = \dfrac{5}{8}\).

    Jibu

    \(x = \frac{1}{2}\)

    Zoezi\(\PageIndex{2}\):

    Kutatua:\(\dfrac{1}{6} y - \dfrac{1}{3} = \dfrac{1}{6}\).

    Jibu

    y = 3

    Taarifa katika Mfano 8.37 kwamba mara moja sisi akalipa equation ya FRACTIONS, equation ilikuwa kama wale sisi kutatuliwa mapema katika sura hii. Tulibadilisha tatizo kwa moja tuliyojua jinsi ya kutatua! Kisha tulitumia Mkakati Mkuu wa Kutatua Ulinganisho wa Linear.

    JINSI YA: KUTATUA EQUATIONS NA COEFFICIENTS SEHEMU KWA KUSAFISHA SEHEMU NDOGO

    Hatua ya 1. Kupata denominator angalau ya kawaida ya sehemu zote katika equation.

    Hatua ya 2. Kuzidisha pande zote mbili za equation na LCD kwamba. Hii inafuta sehemu ndogo.

    Hatua ya 3. Kutatua kutumia Mkakati Mkuu wa Kutatua Equations Linear.

    Mfano\(\PageIndex{2}\):

    Tatua: 7 =\(\dfrac{1}{2} x + \dfrac{3}{4} x − \dfrac{2}{3} x\).

    Suluhisho

    Tunataka kufuta sehemu ndogo kwa kuzidisha pande zote mbili za equation na LCD ya sehemu zote katika equation.

    Kupata denominator angalau ya kawaida ya sehemu zote katika equation. $7 =\ dfrac {1} {2} x +\ dfrac {3} {4} x -\ dfrac {2} {3} x\ quad LCD = 12$$
    Kuzidisha pande zote mbili za equation na 12. $$\ textcolor {nyekundu} {12} (7) =\ textcolor {nyekundu} {12}\ dot\ dfrac {1} {2} x +\ dfrac {3} {4} x -\ dfrac {2} {3} x $$
    Kusambaza. $12 (7) = 12\ cdot\ dfrac {1} {2} x + 12\ cdot\ dfrac {3} {4} x - 12\ dot\ dfrac {2} {3} x $$
    Kurahisisha - na taarifa, hakuna sehemu zaidi! $84 = 6x + 9x - 8x$$
    Kuchanganya kama maneno. $84 = 7x$$
    Gawanya na 7. $$\ dfrac {84} {\ textcolor {nyekundu} {7}} =\ dfrac {7x} {\ textcolor {nyekundu} {7}} $$
    Kurahisisha. $12 = x $$
    Angalia: Hebu x = 12. $$\ kuanza {kupasuliwa} 7 &=\ dfrac {1} {2} x +\ dfrac {3} {4} x -\ dfrac {2} {3} x\\ 7 &\ stackrel {?} {=}\ dfrac {1} {2} (\ textcolor {nyekundu} {12}) +\ dfrac {3} {4} (\ textcolor {nyekundu} {12}) -\ dfrac {2} {3} (\ textcolor {nyekundu} {12})\\ 7 &\ stackrel {?} {=} 6 + 9 - 8\\ 7 &= 7\;\ alama\ mwisho {kupasuliwa} $$
    Zoezi\(\PageIndex{3}\):

    Tatua: 6 =\(\dfrac{1}{2} v + \dfrac{2}{5} v − \dfrac{3}{4} v\).

    Jibu

    v = 40

    Zoezi\(\PageIndex{4}\):

    Tatua: -1 =\(\dfrac{1}{2} u + \dfrac{1}{4} u − \dfrac{2}{3} u\).

    Jibu

    u = -12

    Katika mfano unaofuata, tutaweza kuwa na vigezo na FRACTIONS pande zote mbili za equation.

    Mfano\(\PageIndex{3}\):

    Kutatua:\(x + \dfrac{1}{3} = \dfrac{1}{6} x − \dfrac{1}{2}\).

    Suluhisho

    Find LCD ya FRACTIONS wote katika equation. $$x +\ dfrac {1} {3} =\ dfrac {1} {6} x -\ dfrac {1} {2}\ quad LCD = $6 $
    Panua pande zote mbili na LCD. $$\ textcolor {nyekundu} {6}\ kushoto (x +\ dfrac {1} {3}\ haki) =\ textcolor {nyekundu} {6}\ kushoto (\ dfrac {1} {6} x -\ dfrac {1} {1} {1} {2}\ haki) $$
    Kusambaza. $6\ cdot x + 6\ cdot\ dfrac {1} {3} = 6\ cdot\ dfrac {1} {6} x - 6\ cdot\ dfrac {1} {2} $$
    Kurahisisha - hakuna sehemu zaidi! $6x + 2 = x - $3 $
    Ondoa x kutoka pande zote mbili. $6x\ textcolor {nyekundu} {-x} + 2 = x\ textcolor {nyekundu} {-x} - $3 $
    Kurahisisha. $5x + 2 = -3 $$
    Ondoa 2 kutoka pande zote mbili. $5x + 2\ textcolor {nyekundu} {-2} = -3\ textcolor {nyekundu} {-2} $$
    Kurahisisha. $5x = -$5 $
    Gawanya na 5. $$\ dfrac {5x} {\ textcolor {nyekundu} {5}} =\ dfrac {-5} {\ textcolor {nyekundu} {5}} $$
    Kurahisisha. $$x = -$1 $
    Angalia: Mbadala x = -1. $$\ kuanza {mgawanyiko} x +\ dfrac {1} {3} &=\ dfrac {1} {6} x -\ dfrac {1} {2}\\ (\ textcolor {nyekundu} {-1}) +\ dfrac {1} {3} &\ stackrel {?} {=}\ dfrac {1} {6} (\ textcolor {nyekundu} {-1}) -\ dfrac {1} {2}\\ (-1) +\ dfrac {1} {3} &\ stackrel {?} {=} -\ dfrac {1} {6} -\ dfrac {1} {2}\\ -\ dfrac {3} {3} +\ dfrac {1} {3} &\ stackrel {?} {=} -\ dfrac {1} {6} -\ dfrac {3} {6}\\ -\ dfrac {2} {3} &\ stackrel {?} {=} -\ dfrac {4} {6}\\ -\ dfrac {2} {3} &= -\ dfrac {2} {3}\;\ alama\ mwisho {mgawanyiko} $$
    Zoezi\(\PageIndex{5}\):

    Kutatua:\(a + \dfrac{3}{4} = \dfrac{3}{8} a − \dfrac{1}{2}\).

    Jibu

    a = -2

    Zoezi\(\PageIndex{6}\):

    Kutatua:\(c + \dfrac{3}{4} = \dfrac{1}{2} c − \dfrac{1}{4}\).

    Jibu

    c = -2

    Katika Mfano 8.40, tutaanza kwa kutumia Mali ya Usambazaji. Hatua hii itafuta sehemu ndogo mara moja!

    Mfano\(\PageIndex{4}\):

    Tatua: 1 =\(\dfrac{1}{2}\) (4x + 2).

    Suluhisho

    Kusambaza. $1 =\ dfrac {1} {2}\ dot 4x +\ dfrac {1} {2}\ dot $2
    Kurahisisha. Sasa hakuna sehemu ndogo za kufuta! $1 = 2x + $1 $
    Ondoa 1 kutoka pande zote mbili. $1\ rangi ya maandishi {nyekundu} {-1} = 2x + 1\ rangi ya maandishi {nyekundu} {-1} $$
    Kurahisisha. $0 = 2x$$
    Gawanya na 2. $$\ dfrac {0} {\ textcolor {nyekundu} {2}} =\ dfrac {2x} {\ textcolor {nyekundu} {2}} $$
    Kurahisisha. $0 = x $$
    Angalia: Hebu x = 0. \ [kuanza {mgawanyiko} 1 &=\ dfrac {1} {2} (4x + 2)\\ 1 &\ stackrel {?} {=}\ dfrac {1} {2} [4 (\ textcolor {nyekundu} {0}) + 2]\\ 1 &\ stackrel {?} {=}\ dfrac {1} {2} (2)\\ 1 &\ stackrel {?} {=}\ dfrac {2} {2}\\ 1 &= 1\;\ alama\ mwisho {mgawanyiko} $$
    Zoezi\(\PageIndex{7}\):

    Tatua: -11 =\(\dfrac{1}{2}\) (6p + 2).

    Jibu

    p = -4

    Zoezi\(\PageIndex{8}\):

    Tatua: 8 =\(\dfrac{1}{3}\) (9q + 6).

    Jibu

    q = 2

    Mara nyingi, bado kutakuwa na sehemu ndogo, hata baada ya kusambaza.

    Mfano\(\PageIndex{5}\):

    Tatua:\(\dfrac{1}{2}\) (y - 5) =\(\dfrac{1}{4}\) (y - 1).

    Suluhisho

    Kusambaza. $$\ dfrac {1} {2}\ dot y -\ dfrac {1} {2}\ dot 5 =\ dfrac {1} {4}\ dot y -\ dfrac {1} {4}\ dot $1
    Kurahisisha. $$\ dfrac {1} {2} y -\ dfrac {5} {2} =\ dfrac {1} {4} y -\ dfrac {1} {4} $
    Kuzidisha kwa LCD, 4. $$\ textcolor {nyekundu} {4}\ kushoto (\ dfrac {1} {2} y -\ dfrac {5} {2}\ haki) =\ textcolor {nyekundu} {4}\ kushoto (\ dfrac {1} {4} {4}\ haki) $$
    Kusambaza. $4\ cdot\ dfrac {1} {2} y - 4\ cdot\ dfrac {5} {2} = 4\ cdot\ dfrac {1} {4} y - 4\ cdot\ dfrac {1} {4} $
    Kurahisisha. $2y - 10 = y - $1 $
    Kukusanya maneno y kwa upande wa kushoto. $2y - 10\ textcolor {nyekundu} {-y} = y - 1\ textcolor {nyekundu} {-y} $$
    Kurahisisha. $y - 10 = -$1 $
    Kukusanya constants na haki. $$y - 10\ textcolor {nyekundu} {+10} = -1\ textcolor {nyekundu} {+10} $$
    Kurahisisha. $y = $9 $
    Angalia: mbadala 9 kwa y. $$\ kuanza {kupasuliwa}\ dfrac {1} {2} (y - 5) &=\ dfrac {1} {4} (y - 1)\\ dfrac {1} {2} (\ textcolor {nyekundu} {9} - 5) &\ stackrel {?} {=}\ dfrac {1} {4} (\ textcolor {nyekundu} {9} - 1)\\ dfrac {1} {2} (4) &\ stackrel {?} {=}\ dfrac {1} {4} (8)\\ 2 &= 2\;\ alama\ mwisho {mgawanyiko} $$
    Zoezi\(\PageIndex{9}\):

    Tatua:\(\dfrac{1}{5}\) (n + 3) =\(\dfrac{1}{4}\) (n + 2).

    Jibu

    n = 2

    Zoezi\(\PageIndex{10}\):

    Tatua:\(\dfrac{1}{2}\) (m - 3) =\(\dfrac{1}{4}\) (m - 7).

    Jibu

    m = -1

    Tatua equations na Coefficients ya Decimal

    Baadhi ya milinganyo ina decimals ndani yao. Aina hii ya equation itatokea wakati sisi kutatua matatizo kushughulika na fedha na asilimia. Lakini decimals ni kweli njia nyingine ya kuwakilisha sehemu ndogo. Kwa mfano, 0.3 =\(\dfrac{3}{10}\) na 0.17 =\(\dfrac{17}{100}\). Hivyo, wakati tuna equation na decimals, tunaweza kutumia mchakato huo sisi kutumika wazi sehemu-kuzidisha pande zote mbili za equation na denominator angalau kawaida.

    Mfano\(\PageIndex{6}\):

    Tatua: 0.8x - 5 = 7.

    Suluhisho

    Decimal tu katika equation ni 0.8. Tangu 0.8 =\(\dfrac{8}{10}\), LCD ni 10. Tunaweza kuzidisha pande zote mbili kwa 10 ili kufuta decimal.

    Panua pande zote mbili na LCD. $$\ textcolor {nyekundu} {10} (0.8x - 5) =\ textcolor {nyekundu} {10} (7) $$
    Kusambaza. $10 (0.8x) - 10 (5) = 10 (7) $$
    Kuzidisha, na tazama, hakuna decimals zaidi! $8x - 50 = $70 $
    Ongeza 50 ili kupata vipindi vyote kwa haki. $8x - 50\ textcolor {nyekundu} {+50} = 70\ textcolor {nyekundu} {+50} $$
    Kurahisisha. $8x = 120$$
    Gawanya pande zote mbili kwa 8. $$\ dfrac {8x} {\ textcolor {nyekundu} {8}} =\ dfrac {120} {\ textcolor {nyekundu} {8}} $$
    Kurahisisha. $$x = $15 $
    Angalia: Hebu x = 15. $$\ kuanza {kupasuliwa} 0.8 (\ textcolor {nyekundu} {15}) - 5 &\ stackrel {?} {=} 7\\ 12 - 5 &\ stackrel {?} {=} 7\\ 7 &= 7\;\ checkmark\ mwisho {mgawanyiko} $$
    Zoezi\(\PageIndex{11}\):

    Tatua: 0.6x - 1 = 11.

    Jibu

    x = 20

    Zoezi\(\PageIndex{12}\):

    Tatua: 1.2x - 3 = 9.

    Jibu

    x = 10

    Mfano\(\PageIndex{7}\):

    Tatua: 0.06x + 0.02 = 0.25x - 1.5.

    Suluhisho

    Angalia decimals na fikiria sehemu ndogo sawa.

    \[0.06 = \dfrac{6}{100}, \qquad 0.02 = \dfrac{2}{100}, \qquad 0.25 = \dfrac{25}{100}, \qquad 1.5 = 1 \dfrac{5}{10}\]

    Angalia, LCD ni 100. Kwa kuzidisha na LCD tutaondoa decimals.

    Panua pande zote mbili kwa 100. $$\ textcolor {nyekundu} {100} (0.06x + 0.02) =\ textcolor {nyekundu} {100} (0.25x - 1.5) $$
    Kusambaza. $100 (0.06x) + 100 (0.02) = 100 (0.25x) - 100 (1.5) $$
    Kuzidisha, na sasa hakuna decimals zaidi. $6x + 2 = 25x - 150$$
    Kukusanya vigezo na haki. $6x\ textcolor {nyekundu} {-6x} + 2 = 25x\ textcolor {nyekundu} {-6x} - $150 $
    Kurahisisha. $2 = 19x - $150$
    Kukusanya constants upande wa kushoto. $2\ textcolor {nyekundu} {+150} = 19x - 150\ textcolor {nyekundu} {+150} $$
    Kurahisisha. $152 = 19x$$
    Gawanya na 19. $$\ dfrac {152} {\ textcolor {nyekundu} {19}} =\ dfrac {19x} {\ textcolor {nyekundu} {19}} $$
    Kurahisisha. $8 = x $$
    Angalia: Hebu x = 8. $$\ kuanza {kupasuliwa} 0.06 (\ rangi ya maandishi {nyekundu} {8}) + 0.02 &= 0.25 (\ rangi ya maandishi {nyekundu} {8}) - 1.5\\ 0.48 + 0.02 &= 2.00 - 1.5\\ 0.50 &= 0.50\;\ alama\ mwisho {mgawanyiko} $$
    Zoezi\(\PageIndex{13}\):

    Tatua: 0.14h + 0.12 = 0.35h - 2.4.

    Jibu

    h = 12

    Zoezi\(\PageIndex{14}\):

    Tatua: 0.65k - 0.1 = 0.4k - 0.35.

    Jibu

    k = -1

    Mfano unaofuata unatumia equation ambayo ni ya kawaida ya wale tutaona katika maombi ya fedha katika sura inayofuata. Kumbuka kwamba sisi kusambaza decimal kwanza kabla ya wazi decimals wote katika equation.

    Mfano\(\PageIndex{8}\):

    Tatua: 0.25x + 0.05 (x + 3) = 2.85.

    Suluhisho

    Kusambaza kwanza. $0.25x + 0.05x + 0.15 = 2.85$$
    Kuchanganya kama maneno. $0.30x + 0.15 = 2.85$$
    Ili kufuta decimals, kuzidisha kwa 100. $$\ textcolor {nyekundu} {100} (0.30x + 0.15) =\ textcolor {nyekundu} {100} (2.85) $$
    Kusambaza. $30x + 15 = 285$$
    Ondoa 15 kutoka pande zote mbili. $30x + 15\ textcolor {nyekundu} {-15} = 285\ textcolor {nyekundu} {-15} $$
    Kurahisisha. $30x = 270 $$
    Gawanya na 30. $$\ dfrac {30x} {\ textcolor {nyekundu} {30}} =\ dfrac {270} {\ textcolor {nyekundu} {30}} $$
    Kurahisisha. $$x = $9 $
    Angalia: Hebu x = 9. $$\ kuanza {kupasuliwa} 0.25x + 0.05 (x + 3) &= 2.85\\ 0.25 (\ textcolor {nyekundu} {9}) + 0.05 (\ textcolor {nyekundu} {9} + 3) &\ stackrel {?} {=} 2.85\\ 2.25 + 0.05 (12) &\ stackrel {?} {=} 2.85\\ 2.25 + 0.60 &\ stackrel {?} {=} 2.85\\ 2.85 &= 2.85\;\ checkmark\ mwisho {mgawanyiko} $$
    Zoezi\(\PageIndex{15}\):

    Tatua: 0.25n + 0.05 (n + 5) = 2.95.

    Jibu

    n = 9

    Zoezi\(\PageIndex{16}\):

    Tatua: 0.10d + 0.05 (d - 5) = 2.15.

    Jibu

    d = 16

    PATA RASILIMALI ZA ZIADA

    Tatua Equation na Fractions na Masharti ya kutofautiana kwa pande zote mbili

    Ex 1: Tatua Equation na FRACTIONS na Masharti Variable Pande zote mbili

    Ex 2: Tatua Equation na FRACTIONS na Masharti Variable Pande zote mbili

    Kutatua Ulinganifu wa Hatua nyingi Kuhusisha Decimals

    Ex: Tatua Equation ya Linear Na Decimals na Vigezo Pande zote mbili

    Ex: Tatua Equation na Decimals na Mabano

    Mazoezi hufanya kamili

    Tatua equations na coefficients sehemu

    Katika mazoezi yafuatayo, tatua equation kwa kufuta sehemu ndogo.

    1. \(\dfrac{1}{4} x − \dfrac{1}{2} = − \dfrac{3}{4}\)
    2. \(\dfrac{3}{4} x − \dfrac{1}{2} = \dfrac{1}{4}\)
    3. \(\dfrac{5}{6} y − \dfrac{2}{3} = − \dfrac{3}{2}\)
    4. \(\dfrac{5}{6} y − \dfrac{1}{3} = − \dfrac{7}{6}\)
    5. \(\dfrac{1}{2} a + \dfrac{3}{8} = \dfrac{3}{4}\)
    6. \(\dfrac{5}{8} b + \dfrac{1}{2} = − \dfrac{3}{4}\)
    7. 2 =\(\dfrac{1}{3} x − \dfrac{1}{2} x + \dfrac{2}{3} x\)
    8. 2 =\(\dfrac{3}{5} x − \dfrac{1}{3} x + \dfrac{2}{5} x\)
    9. \(\dfrac{1}{4} m − \dfrac{4}{5} m + \dfrac{1}{2} m\)= -1
    10. \(\dfrac{5}{6} n − \dfrac{1}{4} n − \dfrac{1}{2} n\)= —2
    11. \(x + \dfrac{1}{2} = \dfrac{2}{3} x − \dfrac{1}{2}\)
    12. \(x + \dfrac{3}{4} = \dfrac{1}{2} x − \dfrac{5}{4}\)
    13. \(\dfrac{1}{3} w + \dfrac{5}{4} = w − \dfrac{1}{4}\)
    14. \(\dfrac{3}{2} z + \dfrac{1}{3} = z − \dfrac{2}{3}\)
    15. \(\dfrac{1}{2} x − \dfrac{1}{4} = \dfrac{1}{12} x + \dfrac{1}{6}\)
    16. \(\dfrac{1}{2} a − \dfrac{1}{4} = \dfrac{1}{6} a + \dfrac{1}{12}\)
    17. \(\dfrac{1}{3} b + \dfrac{1}{5} = \dfrac{2}{5} b − \dfrac{3}{5}\)
    18. \(\dfrac{1}{3} x + \dfrac{2}{5} = \dfrac{1}{5} x − \dfrac{2}{5}\)
    19. 1 =\(\dfrac{1}{6}\) (12x - 6)
    20. 1 =\(\dfrac{1}{5}\) (15x - 10)
    21. \(\dfrac{1}{4}\)(p - 7) =\(\dfrac{1}{3}\) (p + 5)
    22. \(\dfrac{1}{5}\)(q + 3) =\(\dfrac{1}{2}\) (q - 3)
    23. \(\dfrac{1}{2}\)(x + 4) =\(\dfrac{3}{4}\)
    24. \(\dfrac{1}{3}\)(x + 5) =\(\dfrac{5}{6}\)

    Tatua equations na Coefficients ya Decimal

    Katika mazoezi yafuatayo, tatua equation kwa kufuta decimals.

    1. 0.6y + 3 = 9
    2. 0.4y - 4 = 2
    3. 3.6j - 2 = 5.2
    4. 2.1k + 3 = 7.2
    5. 0.4x + 0.6 = 0.5x - 1.2
    6. 0.7x + 0.4 = 0.6x + 2.4
    7. 0.23x + 1.47 = 0.37x - 1.05
    8. 0.48x + 1.56 = 0.58x - 0.64
    9. 0.9x - 1.25 = 0.75x + 1.75
    10. 1.2x - 0.91 = 0.8x + 2.29
    11. 0.05n + 0.10 (n + 8) = 2.15
    12. 0.05n + 0.10 (n + 7) = 3.55
    13. 0.10d + 0.25 (d + 5) = 4.05
    14. 0.10d + 0.25 (d + 7) = 5.25
    15. 0.05 (q - 5) + 0.25q = 3.05
    16. 0.05 (q - 8) + 0.25q = 4.10

    kila siku Math

    1. Sarafu Taylor ina $2.00 katika dimes na pennies. Idadi ya pennies ni 2 zaidi ya idadi ya dimes. Tatua equation 0.10d + 0.01 (d + 2) = 2 kwa d, idadi ya dimes.
    2. Mihuri Travis kununuliwa $9.45 yenye thamani ya mihuri 49-asilimia na mihuri 21-asilimia. Idadi ya mihuri ya asilimia 21 ilikuwa 5 chini ya idadi ya mihuri 49-cent. Tatua equation 0.49s + 0.21 (s - 5) = 9.45 kwa s, ili kupata idadi ya mihuri 49-cent Travis kununuliwa.

    Mazoezi ya kuandika

    1. Eleza jinsi ya kupata denominator angalau ya kawaida ya\(\dfrac{3}{8}, \dfrac{1}{6}\), na\(\dfrac{2}{3}\).
    2. Ikiwa equation ina sehemu ndogo, kuzidisha pande zote mbili na LCD hufanya iwe rahisi kutatua?
    3. Kama equation ina sehemu tu upande mmoja, kwa nini una kuzidisha pande zote mbili za equation na LCD?
    4. Katika equation 0.35x + 2.1 = 3.85, LCD ni nini? Unajuaje?

    Self Check

    (a) Baada ya kukamilisha mazoezi, tumia orodha hii ili kutathmini ujuzi wako wa malengo ya sehemu hii.

    CNX_BMath_Figure_AppB_050.jpg

    (b) Kwa ujumla, baada ya kuangalia orodha, unafikiri wewe ni vizuri tayari kwa ajili ya Sura ya? Kwa nini au kwa nini?

    Wachangiaji na Majina