Imibuzo Eyisibonelo Exoxa Ngezinhlelo Zezilinganiso Eziqondile Nokungalingani
Izinhlelo zezibalo eziqondile kanye nokungalingani ziyisihloko esibalulekile kwizibalo ezisetshenziswa kabanzi emikhakheni eyahlukene, njengezomnotho, isayensi, kanye nobunjiniyela. Kulesi sihloko, sizoxoxa ngezinkinga eziyisibonelo ezihilela izinhlelo zezibalo eziqondile kanye nokungalingani kanye nendlela yokuzixazulula ngokuningiliziwe.
Incazelo Yesistimu Yezilinganiso Eziqondile
Uhlelo lwezibalo eziqondile luqukethe izibalo ezimbili noma ngaphezulu eziqondile ezihlobene. Izibonelo yilezi:
\[
\begin{cases}
2x + 3y = 5 \\
4x – y = 1
\ukuphela{amacala}
\]
Umgomo wokuxazulula lolu hlelo ukuthola amanani ka-\(x\) kanye no-\(y\) ahlangabezana nezilinganiso zombili ngasikhathi sinye.
Izindlela Zokuxazulula Izinhlelo Zezibalo Eziqondile
Kunezindlela eziningana zokuxazulula izinhlelo zezibalo eziqondile, okuhlanganisa:
1. Indlela Yokufaka Esikhundleni
2. Indlela Yokususa
3. Indlela ye-Matrix (ephambene noma eGauss-Jordan)
Isibonelo Umbuzo 1: Indlela Yokufaka Esikhundleni
Ake sixazulule uhlelo olulandelayo sisebenzisa indlela yokufaka esikhundleni:
\[
\begin{cases}
x + 2y = 10 \\
3x – y = 5
\ukuphela{amacala}
\]
I-Langkah-langkah:
1. Hlukanisa enye yezinto eziguquguqukayo kwenye yezibalo.
Kusukela ku-equation yokuqala, sihlukanisa \(x\):
\[
x = 10 – 2y
\]
2. Faka isisho esitholiwe kwenye i-equation.
Faka u-\(x = 10 – 2y\) esilinganisweni sesibili:
\[
3(10 – 2y) – y = 5
\]
Xazulula i-\(y\):
\[
30 – 6y – y = 5
\]
\[
30 – 7y = 5
\]
\[
-7y = -25
\]
\[
y = \frac{25}{7}
\]
3. Sebenzisa amanani atholiwe ukuthola ezinye iziguquguquko.
Faka esikhundleni se-\(y = \frac{25}{7}\) esisho se-\(x\):
\[
x = 10 – 2\kwesobunxele(\frac{25}{7}\kwesokudla)
\]
\[
x = 10 – \frac{50}{7}
\]
\[
x = \frac{70}{7} – \frac{50}{7}
\]
\[
x = \frac{20}{7}
\]
Ngakho-ke, izixazululo zesistimu yilezi \( x = \frac{20}{7} \) kanye \( y = \frac{25}{7} \).
Isibonelo Umbuzo 2: Indlela Yokususa
Okulandelayo, ake sisebenzise indlela yokususa ukuxazulula uhlelo olulandelayo:
\[
\begin{cases}
2x + 3y = 12 \\
4x + 6y = 24
\ukuphela{amacala}
\]
Kulesi simo, sibona ukuthi i-equation yesibili iyi-multiple ye-equation yokuqala. Ukuze sinciphise uhlelo, singaphindaphinda i-equation yokuqala ngo-2 bese siyisusa ku-equation yesibili:
1. Phindaphinda isibalo sokuqala ngo-2:
\[
2(2x + 3y) = 2 \cdot 12
\]
\[
4x + 6y = 24
\]
2. Susa isibalo sokuqala esiphindaphindiwe kusibalo sesibili:
\[
(4x + 6y) – (4x + 6y) = 24 – 24
\]
\[
= 0 0
\]
Lokhu kunikeza \(0 = 0\), okubonisa ukuthi uhlelo lunezixazululo ezingenamkhawulo futhi lezi zibalo zincike.
Isibonelo Umbuzo 3: Ukungalingani Okuqondile
Ukungalingani okuqondile kulandela izimiso ezifanayo nezibalo eziqondile, kodwa kuhilela izimpawu zokungalingani ezifana ne-\(<, \leq, >, \geq\). Ake sibheke isibonelo esilula:
\[
\begin{cases}
3x – y < 7 \\ 2x + y \geq 4 \end{cases} \] Izinyathelo: 1. Sisebenzisa indlela yesithombe ukuthola indawo yesisombululo salolu hlelo. Faka igrafu yokungalingani ngakunye. 2. Guqula ukungalingani kube yi-equation ukuthola umugqa womngcele: Ku-\(3x - y < 7\), umugqa womngcele ngu-\(3x - y = 7\)
