Co-aontaran co-amail
Dòigh ailseabrach
Bho bhith ag ath-sgrùdadh na puing-trasnaidh, tha fios againn gum feum sinn dà cho-aontar loidhneach le dà chaochladair. Ach faodaidh e ùine mhòr a thoirt a bhith a' fuasgladh cho-aontaran gu grafaigeach.
Nuair a bhios tu a' fuasgladh cho-aontaran san dòigh ailseabrach, 's e a' chiad cheum feuchainn ri aon dhe na teirmean neo-aithnichte a dhubhadh às.
Eisimpleir
Fuasgail na co-aontaran co-amail seo agus obraich a-mach luachan \(x\) agus \(y\).
\(2x + y = 7(Co-aontar\,1)\)
\(3x - y = 8(Co-aontar\,2)\)
Freagairt
Gus aon dhe na teirmean a dhubhadh às, feumaidh sinn an dara cuid an aon àireamh de \(x\) no an aon àireamh de \(y\) a bhith againn sna co-aontaran. An seo, tha an aon cho-èifeachd de \(y\) againn, agus cuideachd tha aon dearbhte agus aon àicheil. Gus gach \(y\) a dhubhadh às, feumaidh sinn co-aontar 1 agus co-aontar 2 a chur-ris gus am faigh sinn \(y + ( - y) = 0\).
\(2x+y=7\)
\(3x-y=8\)
Le bhith a' cur-ris nan co-aontaran seo gheibh sinn:
\(5x = 15\)
\(x = 15 \div 3\)
\(x = 3\)
Aig a' phuing-trasnaidh tha na co-chomharran-x an aon rud agus tha na co-chomharran-y an aon rud. Tha sin a' ciallachadh gum faod sinn an luach seo airson \(x\) ionadachadh a-steach a dh'aon dhe na co-aontaran tùsail gus luach \(y\) obrachadh a-mach.
Mar as trice, tha e nas fhasa \(x\) ionadachadh a-steach dhan cho-aontar sa bheil \(y\) dearbhte.
Ionadaich a-steach do cho-aontar 1: \(2x + y = 7\)
Nuair a tha \(x = 3\)
\(2 \times 3 + y = 7\)
\(6 + y = 7\)
\(y = 7 - 6\)
\(y = 1\)
Mar sin \(x = 3\) agus \(y = 1\)
Mar sin 's e a' phuing-trasnaidh (3,1)
Eisimpleir
Fuasgail na co-aontaran co-amail seo agus obraich a-mach luachan \(x\) agus \(y\).
\(5x+2y=23\)
\(2x-3y=-6\)
Freagairt
Chan eil an aon àireamh de \(y\) againn anns gach sreath.
Airson seo fhaighinn, bidh sinn ag iomadachadh a' chiad cho-aontair le 3 agus an dara co-aontar le 2.
\(15x+6y=69\)
\(4x-6y=-12\)
Faodaidh sinn a-nis cur-ris a dhèanamh mar a rinn sinn roimhe
\(19x=57\)
Bidh sinn a-nis ag ionadachadh \(x=3\) a-steach dhan chiad cho-aontar
\(5(3)+2y=23\)
\(15+2y=23\)
\(2y=8\)
\(y=4\)
Mar sin 's e am fuasgladh \(x=3\), \(y=4\)
(Cuimhnich gu bheil e uaireannan nas luaithe cur às do \(x\) seach do \(y\). Tha e a rèir 's dè a' cheist a gheibh thu.)
Feuch a-nis na ceistean gu h-ìosal.
Question
Fuasgail na co-aontaran co-amail seo agus obraich a-mach luachan \(x\) agus \(y\).
\(2x + 3y = 16(Co-aontar\,1)\)
\(3x - 4y = 7(Co-aontar\,2)\)
Bu chòir gum faic thu nach eil an aon àireamh de \(y\) againn an turas seo.
Gheibh sinn an aon àireamh de \(y\) ma dh'iomadaicheas sinn co-aontar 1 le 4 agus co-aontar 2 le 3, gus am bi \(12y\) san dà cho-aontar. (Cuideachd feumaidh sinn aon 12y dearbhte agus aon àicheil.)
\(8x + 12y = 64(Co-aontar\,3)\)
\(9x - 12y = 21(Co-aontar\,4)\)
Cuir-ris na co-aontaran seo a-nis gus gach y a dhubhadh às.
\(8x + 12y = 64\)
\(9x - 12y = 21\)
Le bhith a' cur-ris nan co-aontaran seo, gheibh sinn:
\(17x = 85\)
\(x = \frac{{85}}{{17}} = 5\)
Ionadaich a-steach do cho-aontar 1: \(2x + 3y = 16\)
Nuair a tha \(x = 5\)
\(2 \times 5 + 3y = 16\)
\(10 + 3y = 16\)
\(3y = 16 - 10\)
\(3y = 6\)
\(y = \frac{6}{3} = 2\)
Mar sin tha \(x = 5\) agus \(y = 2\)
Mar sin 's e a' phuing-trasnaidh (5,2)
Question
Fuasgail am paidhir cho-aontaran co-amail seo agus obraich a-mach luachan:
\(d\) agus \(e\).
\(15d + 2e = 4(Co-aontar\,1)\)
\(2d + e = 2(Co-aontar\,2)\)
Iomadaich co-aontar 2 le -2.
\(-4d - 2e = -4(Co-aontar\,3)\)
Cuir-ris co-aontaran 1 agus 3 oir tha 2c dearbhte agus 2c àicheil againn a-nis.
\(15d+2c=4\)
\(-4d-2c=-4\)
Le bhith a' cur-ris nan co-aontaran gheibh sinn:
\(11d = 0\)
\(d = 0 \div 11\)
\(d = 0\)
Ionadaich a-steach a cho-aontar 2: \(2d + e = 2\)
\(2 \times 0 + e = 2\)
\(0 + e = 2\)
\(e = 2\)
Mar sin \(d = 0\) agus \(e = 2\)
Mar sin 's e a' phuing-trasnaidh (0,2)
