Skip to main content
Global

8.3: Kutatua Equations Kutumia Idara na Kuzidisha Mali ya Usawa

  • Page ID
    173342
  • \( \newcommand{\vecs}[1]{\overset { \scriptstyle \rightharpoonup} {\mathbf{#1}} } \) \( \newcommand{\vecd}[1]{\overset{-\!-\!\rightharpoonup}{\vphantom{a}\smash {#1}}} \)\(\newcommand{\id}{\mathrm{id}}\) \( \newcommand{\Span}{\mathrm{span}}\) \( \newcommand{\kernel}{\mathrm{null}\,}\) \( \newcommand{\range}{\mathrm{range}\,}\) \( \newcommand{\RealPart}{\mathrm{Re}}\) \( \newcommand{\ImaginaryPart}{\mathrm{Im}}\) \( \newcommand{\Argument}{\mathrm{Arg}}\) \( \newcommand{\norm}[1]{\| #1 \|}\) \( \newcommand{\inner}[2]{\langle #1, #2 \rangle}\) \( \newcommand{\Span}{\mathrm{span}}\) \(\newcommand{\id}{\mathrm{id}}\) \( \newcommand{\Span}{\mathrm{span}}\) \( \newcommand{\kernel}{\mathrm{null}\,}\) \( \newcommand{\range}{\mathrm{range}\,}\) \( \newcommand{\RealPart}{\mathrm{Re}}\) \( \newcommand{\ImaginaryPart}{\mathrm{Im}}\) \( \newcommand{\Argument}{\mathrm{Arg}}\) \( \newcommand{\norm}[1]{\| #1 \|}\) \( \newcommand{\inner}[2]{\langle #1, #2 \rangle}\) \( \newcommand{\Span}{\mathrm{span}}\)\(\newcommand{\AA}{\unicode[.8,0]{x212B}}\)

    Malengo ya kujifunza
    • Kutatua equations kwa kutumia Idara na Kuzidisha Mali ya Usawa
    • Tatua milinganyo ambayo inahitaji kuwa rahisi
    kuwa tayari!

    Kabla ya kuanza, fanya jaribio hili la utayari.

    1. Kurahisisha: -7\(\left(\dfrac{1}{−7}\right)\). Ikiwa umekosa tatizo hili, tathmini Mfano 4.3.10.
    2. Je, ni kurudi kwa\(− \dfrac{3}{8}\) nini? Kama amekosa tatizo hili, mapitio Mfano 4.4.11.
    3. Tathmini 9x + 2 wakati x = 1-3. Ikiwa umekosa tatizo hili, tathmini Mfano 3.8.10.

    Kutatua Equations Kutumia Idara na Kuzidisha Mali ya Usawa

    Tulianzisha Mali ya Kuzidisha na Idara ya Usawa katika Kutatua Equations Kutumia Integers; Mali ya Idara ya Usawa na Kutatua Equations na Fractions. Tulielezea jinsi mali hizi zilivyofanya kazi kwa kutumia bahasha na counters na kisha kuzitumia kutatua equations (Angalia Kutatua Equations Kutumia Integers; Mali ya Idara ya Usawa). Tunawapa tena hapa tunapojiandaa kutumia mali hizi tena.

    Ufafanuzi: Idara na Kuzidisha Mali ya Usawa

    Idara ya Mali ya Usawa: Kwa namba zote halisi a, b, c, na c ∙ 0, ikiwa = b, basi\(\dfrac{a}{c} = \dfrac{b}{c}\).

    Kuzidisha Mali ya Usawa: Kwa namba zote halisi a, b, c, ikiwa = b, basi ac = bc.

    Unapogawanya au kuzidisha pande zote mbili za equation kwa kiasi sawa, bado una usawa.

    Hebu tuchunguze jinsi mali hizi za usawa zinaweza kutumika ili kutatua equations. Kumbuka, lengo ni 'kuondoa' operesheni juu ya kutofautiana. Katika mfano hapa chini variable ni kuzidisha kwa 4, hivyo sisi kugawanya pande zote mbili na 4 kwa 'tengua' kuzidisha.

    Mfano\(\PageIndex{1}\):

    Kutatua: 4x = -28.

    Suluhisho

    Tunatumia Mali ya Idara ya Usawa kugawanya pande zote mbili kwa 4.

    Gawanya pande zote mbili kwa 4 ili uondoe kuzidisha. $$\ dfrac {4x} {\ textcolor {nyekundu} {4}} =\ dfrac {-28} {\ textcolor {nyekundu} {4}} $$
    Kurahisisha. $$x = -7$$
    Angalia jibu lako. Hebu x = -7. $$\ kuanza {kupasuliwa} 4x &= -28\\ 4 (\ rangi ya maandishi {nyekundu} {-7}) &\ stackrel {?} {=} -28\\ -28 &= -28\;\ checkmark\ mwisho {mgawanyiko} $$

    Kwa kuwa hii ni kauli ya kweli, x = -7 ni suluhisho la 4x = -28.

    Zoezi\(\PageIndex{1}\):

    Kutatua: 3y = -48.

    Jibu

    y = -16

    Zoezi\(\PageIndex{2}\):

    Kutatua: 4z = -52.

    Jibu

    z = -13

    Katika mfano uliopita, kwa 'kurekebisha' kuzidisha, sisi kugawanywa. Je, unafikiriaje sisi 'tengeneze' mgawanyiko?

    Mfano\(\PageIndex{2}\):

    Tatua:\(\dfrac{a}{−7}\) = -42.

    Suluhisho

    Hapa a imegawanywa na -7. Tunaweza kuzidisha pande zote mbili na -7 kujitenga.

    Panua pande zote mbili kwa -7. $$\ textcolor {nyekundu} {-7}\ kushoto (\ dfrac {a} {-7}\ haki) =\ textcolor {nyekundu} {7} (-42) $$
    Kurahisisha. $$a = 294 $$
    Angalia jibu lako. Hebu = 294. $$\ kuanza {split}\ dfrac {a} {-7} &= -42\\ dfrac {\ textcolor {nyekundu} {294}} {-7} &\ stackrel {?} {=} -42\\ -42 &= -42\;\ checkmark\ mwisho {mgawanyiko} $$
    Zoezi\(\PageIndex{3}\):

    Tatua:\(\dfrac{b}{−6}\) = -24.

    Jibu

    b = 144

    Zoezi\(\PageIndex{4}\):

    Tatua:\(\dfrac{c}{−8}\) = -16.

    Jibu

    c = 128

    Mfano\(\PageIndex{3}\):

    Tatua: -r = 2.

    Suluhisho

    Kumbuka -r ni sawa na -1r.

    Andika upya -r kama -1r. $-1r = $2 $
    Gawanya pande zote mbili kwa -1. $$\ dfrac {-1r} {\ textcolor {nyekundu} {-1}}} =\ dfrac {2} {\ textcolor {nyekundu} {-1}} $$

    Angalia.

    Mbadala r = -2. $-r = $2 $
    Kurahisisha. $$\ kuanza {split} - (\ textcolor {nyekundu} {-2}) &\ stackrel {?} {=} 2\\ 2 &= 2\;\ checkmark\ mwisho {mgawanyiko} $$

    Katika Kutatua Equations na Fractions, tuliona kwamba kuna njia nyingine mbili za kutatua -r = 2.

    1. Tunaweza kuzidisha pande zote mbili kwa -1.
    2. Tunaweza kuchukua kinyume cha pande zote mbili.
    Zoezi\(\PageIndex{5}\):

    Tatua: -k = 8.

    Jibu

    k = -8

    Zoezi\(\PageIndex{6}\):

    Tatua: -g = 3.

    Jibu

    g = -3

    Mfano\(\PageIndex{4}\):

    Kutatua:\(\dfrac{2}{3}\) x = 18.

    Suluhisho

    Kwa kuwa bidhaa ya nambari na usawa wake ni 1, mkakati wetu utakuwa kutenganisha x kwa kuzidisha kwa usawa wa\(\dfrac{2}{3}\).

    Kuzidisha kwa usawa wa\(\dfrac{2}{3}\). $$\ textcolor {nyekundu} {\ dfrac {3} {2}}\ dot\ dfrac {2} {3} x =\ textcolor {nyekundu} {\ dfrac {3} {2}}\ dot $18 $
    Recipurals kuzidisha kwa moja. $1x =\ dfrac {3} {2}\ dot\ dfrac {18} {1} $
    Kuzidisha. $$x = 27 $$
    Angalia jibu lako. Hebu x = 27. $$\ kuanza {split}\ dfrac {2} {3} x &= 18\\ dfrac {2} {3}\ cdot\ textcolor {nyekundu} {27} &\ stackrel {?} {=} 18\\ 18 &= 18\;\ checkmark\ mwisho {mgawanyiko} $$

    Taarifa kwamba tunaweza kuwa na kugawanywa pande zote mbili za equation\(\dfrac{2}{3}\) x = 18 na\(\dfrac{2}{3}\) kujitenga x Wakati hii ingekuwa kazi, kuzidisha kwa usawa inahitaji hatua chache

    Zoezi\(\PageIndex{7}\):

    Tatua:\(\dfrac{2}{5}\) n = 14.

    Jibu

    n = 35

    Zoezi\(\PageIndex{8}\):

    Tatua:\(\dfrac{5}{6}\) y = 15.

    Jibu

    y = 18

    Kutatua equations kwamba haja ya kuwa Kilichorahisishwa

    equations wengi kuanza nje ngumu zaidi kuliko wale tumekuwa tu kutatuliwa. Kwanza, tunahitaji kurahisisha pande zote mbili za equation iwezekanavyo.

    Mfano\(\PageIndex{5}\):

    Kutatua: 8x + 9x - 5x = -3 + 15.

    Suluhisho

    Anza kwa kuchanganya maneno kama ili kurahisisha kila upande.

    Kuchanganya kama maneno. $12x = 12$$
    Gawanya pande zote mbili kwa 12 ili kutenganisha x. $$\ dfrac {12x} {\ textcolor {nyekundu} {12}} =\ dfrac {12} {\ textcolor {nyekundu} {12}} $$
    Kurahisisha. $$x = $1 $
    Angalia jibu lako. Hebu x = 1. $$\ kuanza {kupasuliwa} 8x + 9x - 5x &= -3 + 15\\ 8\ cdot\ textcolor {nyekundu} {1} + 9\ cdot\ textcolor {nyekundu} {1} - 5\ cdot\ textcolor {nyekundu} {1} &\ stackrel {?} {=} -3 + 15\\ 12 &= 12\;\ checkmark\ mwisho {mgawanyiko} $$
    Zoezi\(\PageIndex{9}\):

    Kutatua: 7x + 6x - 4x = -8 + 26.

    Jibu

    x = 2

    Zoezi\(\PageIndex{10}\):

    Tatua: 11n - 3n - 6n = 7 - 17.

    Jibu

    n = -5

    Mfano\(\PageIndex{6}\):

    Tatua: 11 - 20 = 17y - 8y - 6y.

    Suluhisho

    Kurahisisha kila upande kwa kuchanganya maneno kama hayo.

    Kurahisisha kila upande. $-9 = 3y $$
    Gawanya pande zote mbili kwa 3 ili kutenganisha y. $$\ dfrac {-9} {\ textcolor {nyekundu} {3}} =\ dfrac {3y} {\ textcolor {nyekundu} {3}} $$
    Kurahisisha. $-3 = y $$
    Angalia jibu lako. Hebu y = -3. $$\ kuanza {kupasuliwa} 11 - 20 &= 17y - 8y - 6y\\ 11 - 20 &\ stackrel {?} {=} 17 (\ rangi ya maandishi {nyekundu} {-3}) -8 (\ rangi ya maandishi {nyekundu} {-3}) -6 (\ textcolor {nyekundu} {-3})\\ 11 - 20 &\ stackrel {?} {=} -51 + 24 + 18\\ -9 &\ stackrel {?} {=} -9\;\ checkmark\ mwisho {split} $$

    Kumbuka kwamba variable kuishia upande wa kulia wa ishara sawa wakati sisi kutatuliwa equation. Unaweza kupendelea kuchukua hatua moja zaidi ya kuandika suluhisho na kutofautiana upande wa kushoto wa ishara sawa.

    Zoezi\(\PageIndex{11}\):

    Tatua: 18 - 27 = 15c - 9c - 3c.

    Jibu

    c = -3

    Zoezi\(\PageIndex{12}\):

    Tatua: 18 - 22 = 12x - x - 4x.

    Jibu

    \(x = -\frac{4}{7}\)

    Mfano\(\PageIndex{7}\):

    Tatua: -3 (n - 2) - 6 = 21.

    Suluhisho

    Kumbuka-daima kurahisisha kila upande kwanza.

    Kusambaza. $-3n + 6 = 21$$
    Kurahisisha. $-3x = 21$$
    Gawanya pande zote mbili na -3 ili kutenganisha n. $$\ kuanza {split}\ dfrac {-3n} {\ textcolor {nyekundu} {-3}} &=\ dfrac {21} {\ textcolor {nyekundu} {-3}}\ n &= -7\ mwisho {mgawanyiko} $$
    Angalia jibu lako. Hebu n = -7. $$\ kuanza {kupasuliwa} -3 (n - 2) - 6 &= 21\\ -3 (\ textcolor {nyekundu} {-7} - 2) - 6 &\ stackrel {?} {=} 21\\ -3 (-9) - 6 &\ stackrel {?} {=} 21\\ 27 - 6 &\ stackrel {?} {=} 21\\ 21 &= 21\;\ checkmark\ mwisho {mgawanyiko} $$
    Zoezi\(\PageIndex{13}\):

    Tatua: -4 (n - 2) - 8 = 24.

    Jibu

    n = -6

    Zoezi\(\PageIndex{14}\):

    Tatua: -6 (n - 2) - 12 = 30.

    Jibu

    n = -5

    Mazoezi hufanya kamili

    Kutatua Equations Kutumia Idara na Kuzidisha Mali ya Usawa

    Katika mazoezi yafuatayo, tatua kila equation kwa kutofautiana kwa kutumia Mali ya Idara ya Usawa na uangalie suluhisho.

    1. 8x = 32
    2. 7p = 63
    3. -5c = 55
    4. -9x = -27
    5. -90 = 6y
    6. -72 = 12y
    7. -16p = -64
    8. -8m = -56
    9. 0.25z = 3.25
    10. 0.75a = 11.25
    11. -3x = 0
    12. 4x = 0

    Katika mazoezi yafuatayo, tatua kila equation kwa kutofautiana kwa kutumia Mali ya Kuzidisha ya Usawa na uangalie suluhisho.

    1. \(\dfrac{x}{4}\)= 15
    2. \(\dfrac{z}{2}\)= 14
    3. -20 =\(\dfrac{q}{−5}\)
    4. \(\dfrac{c}{−3}\)= -12
    5. \(\dfrac{y}{9}\)= -6
    6. \(\dfrac{q}{6}\)= -8
    7. \(\dfrac{m}{−12}\)= 5
    8. —4 =\(\dfrac{p}{−20}\)
    9. \(\dfrac{2}{3}\)y = 18
    10. \(\dfrac{3}{5}\)r = 15
    11. \(− \dfrac{5}{8}\)w = 40
    12. 24 =\(− \dfrac{3}{4}\) x
    13. \(− \dfrac{2}{5} = \dfrac{1}{10} a\)
    14. \(− \dfrac{1}{3} q = − \dfrac{5}{6}\)

    Kutatua equations kwamba haja ya kuwa Kilichorahisishwa

    Katika mazoezi yafuatayo, tatua equation.

    1. 8a + 3a - 6a = -17 + 27
    2. 6y - 3y + 12y = -43 + 28
    3. -9x - 9x + 2x = 50 - 2
    4. -5m + 7m - 8m = -6 + 36
    5. 100 - 16 = 4p - 10p - p
    6. -18 - 7 = 5t - 9t - 6t
    7. \(\dfrac{7}{8} n − \dfrac{3}{4} n\)= 9 + 2
    8. \(\dfrac{5}{12} q + \dfrac{1}{2} q\)= 25 ÷ 3
    9. 0.25d + 0.10d = 6 ÷ 0.75
    10. 0.05p - 0.01p = 2 + 0.24

    kila siku Math

    1. Balloons Ramona alinunua balloons 18 kwa chama. Anataka kufanya makundi 3 sawa. Kupata idadi ya balloons katika kila kundi, b, kwa kutatua equation 3b = 18.
    2. Kufundisha darasa la chekechea la Connie lina watoto 24. Anataka waingie katika makundi 4 sawa. Pata idadi ya watoto katika kila kikundi, g, kwa kutatua equation 4g = 24.
    3. Bei ya tiketi Daria alilipa $36.25 kwa tiketi za watoto 5 kwenye rink ya skating ya barafu. Pata bei ya kila tiketi, p, kwa kutatua equation 5p = 36.25.
    4. Bei ya kitengo Nishant kulipwa $12.96 kwa pakiti ya chupa 12 za juisi. Pata bei ya kila chupa, b, kwa kutatua equation 12b = 12.96.
    5. Uchumi wa mafuta SUV ya Tania inapata nusu ya maili kwa kila lita (mpg) kama gari la mseto la mumewe. SUV inapata 18 mpg. Pata maili kwa galoni, m, ya gari la mseto, kwa kutatua equation\(\dfrac{1}{2}\) m = 18.
    6. Kitambaa Timu ya kuchimba ilitumia yadi 14 za kitambaa kufanya bendera kwa theluthi moja ya wanachama. Kupata ni kiasi gani kitambaa, f, wangehitaji kufanya bendera kwa timu nzima kwa kutatua equation\(\dfrac{1}{3}\) f = 14.

    Mazoezi ya kuandika

    1. Frida alianza kutatua equation -3x = 36 kwa kuongeza 3 kwa pande zote mbili. Eleza kwa nini njia ya Frida itasababisha suluhisho sahihi.
    2. Emiliano anadhani x = 40 ni suluhisho la equation\(\dfrac{1}{2}\) x = 80. Eleza kwa nini yeye ni makosa.

    Self Check

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

    (b) Baada ya kuchunguza orodha hii, utafanya nini ili uwe na ujasiri kwa malengo yote?

    Wachangiaji na Majina