Tentukan himpunan penyelesaian dari [tex]\large \left\{\begin{matrix}\frac{3}{x-1}+\frac{4}{y+3}+\frac{2}{z+2}=6\\ \frac{1}{x-1}+\frac{2}{y+3}+\frac{4}{z+2}=4\\ \frac{4}{x-1}+\frac{6}{y+3}+\frac{2}{z+2}=8\end{matrix}\right.[/tex]

Berikut ini adalah pertanyaan dari syakhayaz pada mata pelajaran Matematika untuk jenjang Sekolah Menengah Atas

Tentukan himpunan penyelesaian dari $\large \left\{\begin{matrix}\frac{3}{x-1}+\frac{4}{y+3}+\frac{2}{z+2}=6\\ \frac{1}{x-1}+\frac{2}{y+3}+\frac{4}{z+2}=4\\ \frac{4}{x-1}+\frac{6}{y+3}+\frac{2}{z+2}=8\end{matrix}\right.$

Jawaban dan Penjelasan

Berikut ini adalah pilihan jawaban terbaik dari pertanyaan diatas.

Himpunan penyelesaian = {(x, y, z) | (2, –1, 0)}
________________

Pembahasan

Diketahui

Sistem persamaan:
$\begin{cases}\vphantom{\Bigg|}(1)&\dfrac{3}{x-1}+\dfrac{4}{y+3}+\dfrac{2}{z+2}=6\\\vphantom{\Bigg|}(2)&\dfrac{1}{x-1}+\dfrac{2}{y+3}+\dfrac{4}{z+2}=4\\\vphantom{\Bigg|}(3)&\dfrac{4}{x-1}+\dfrac{6}{y+3}+\dfrac{2}{z+2}=8\end{cases}$

Ditanyakan

Himpunan penyelesaian

Penyelesaian

Persamaan (3) dalam bentuk sederhana adalah:

$\begin{aligned}\frac{2}{x-1}+\frac{3}{y+3}+\frac{1}{z+2}=4\end{aligned}$

Dengan Metode Matriks

Kolom 1: 1/(x–1)
Kolom 2: 1/(y+3)
Kolom 3: 1/(z+2)

Menyelesaikan dengan Operasi Baris Elementer (OBE):

$\begin{aligned}&\left(\,\left.\begin{matrix}\begin{matrix}3 & 4 & 2 \\1 & 2 & 4 \\2 & 3 & 1\end{matrix}\end{matrix}\;\;\right|\;\begin{matrix}6\\4\\4\end{matrix}\,\right)\\&\textsf{---------------------------------------------}\end{aligned}$

$\begin{aligned}&\begin{aligned}R_1\leftrightarrow R_2\end{aligned}\ \Rightarrow \left(\,\left.\begin{matrix}\begin{matrix}1 & 2 & 4 \\3 & 4 & 2 \\2 & 3 & 1\end{matrix}\end{matrix}\;\;\right|\;\begin{matrix}4\\6\\4\end{matrix}\,\right)\\&\textsf{---------------------------------------------}\end{aligned}$

$\begin{aligned}&\begin{aligned}R_2-3R_1\to R_2\\R_3-2R_1\to R_3\end{aligned}\ \Rightarrow \left(\,\left.\begin{matrix}\begin{matrix}1 & 2 & 4 \\0 & -2 & -10 \\0 & -1 & -7\end{matrix}\end{matrix}\;\;\right|\;\begin{matrix}4\\-6\\-4\end{matrix}\,\right)\\&\textsf{---------------------------------------------}\end{aligned}$

$\begin{aligned}&\begin{aligned}\left(-\tfrac{1}{2}\right)R_2\to R_2\\(-1)R_3\to R_3\end{aligned}\ \Rightarrow \left(\,\left.\begin{matrix}\begin{matrix}1 & 2 & 4 \\0 & 1 & 5 \\0 & 1 & 7\end{matrix}\end{matrix}\;\;\right|\;\begin{matrix}4\\3\\4\end{matrix}\,\right)\\&\textsf{---------------------------------------------}\end{aligned}$

$\begin{aligned}&\begin{aligned}R_1-2R_2\to R_1\\R_3-R_2\to R_3\end{aligned}\ \Rightarrow \left(\,\left.\begin{matrix}\begin{matrix}1 & 0 & -6 \\0 & 1 & 5 \\0 & 0 & 2\end{matrix}\end{matrix}\;\;\right|\;\begin{matrix}-2\\3\\1\end{matrix}\,\right)\\&\textsf{---------------------------------------------}\end{aligned}$

$\begin{aligned}&\begin{aligned}\frac{1}{2}R_3\to R_3\end{aligned}\ \Rightarrow \left(\,\left.\begin{matrix}\begin{matrix}1 & 0 & -6 \\0 & 1 & 5 \\0 & 0 & 1\end{matrix}\end{matrix}\;\;\right|\;\begin{matrix}-2\\3\\1/2\end{matrix}\,\right)\\&\textsf{---------------------------------------------}\end{aligned}$

$\begin{aligned}&\begin{aligned}R_1+6R_3\to R_1\\R_2-5R_3\to R_3\end{aligned}\ \Rightarrow \left(\,\left.\begin{matrix}\begin{matrix}1 & 0 & 0 \\0 & 1 & 0 \\0 & 0 & 1\end{matrix}\end{matrix}\;\;\right|\;\begin{matrix}1\\1/2\\1/2\end{matrix}\,\right)\end{aligned}$

Hasil yang diperoleh:

Baris pertama:
$\begin{aligned}\frac{1}{x-1}=1\implies x=\bf2\end{aligned}$
Baris kedua:
$\begin{aligned}\frac{1}{y+3}=\frac{1}{2}\implies y=\bf{-}1\end{aligned}$
Baris ketiga:
$\begin{aligned}\frac{1}{z+2}=\frac{1}{2}\implies z=\bf0\end{aligned}$

HIMPUNAN PENYELESAIAN:
HP = {(x, y, z) | (2, –1, 0)}
_____________________

Dengan Metode Campuran

Sistem persamaan setelah disederhanakan adalah:

$\begin{cases}\vphantom{\Bigg|}(1)&\dfrac{3}{x-1}+\dfrac{4}{y+3}+\dfrac{2}{z+2}=6\\\vphantom{\Bigg|}(2)&\dfrac{1}{x-1}+\dfrac{2}{y+3}+\dfrac{4}{z+2}=4\\\vphantom{\Bigg|}(3)&\dfrac{2}{x-1}+\dfrac{3}{y+3}+\dfrac{1}{z+2}=4\end{cases}$

Eliminasi 1/(x–1) dan 1/(y+3):

$\begin{aligned}(1):\ &\frac{3}{x-1}+\frac{4}{y+3}+\frac{2}{z+2}=6\\(2):\ &\frac{1}{x-1}+\frac{2}{y+3}+\frac{4}{z+2}=4\\&\textsf{--------------------------------------}\ +\\&\frac{4}{x-1}+\frac{6}{y+3}+\frac{6}{z+2}=10\\(3)\!\times\!2:\ &\frac{4}{x-1}+\frac{6}{y+3}+\frac{2}{z+2}=8\\&\textsf{--------------------------------------}\ -\\&\qquad\qquad\qquad\quad\ \,\frac{4}{z+2}=2\\&\Rightarrow \frac{2}{z+2}=1\\&\Rightarrow z+2=2\\&\Rightarrow z=\bf0\end{aligned}$

Substitusi z ke setiap persamaan, dan eliminasi 1/(x-1):

$\begin{aligned}(2):\ &\frac{1}{x-1}+\frac{2}{y+3}+\frac{4}{2}=4\\(3):\ &\frac{2}{x-1}+\frac{3}{y+3}+\frac{1}{2}=4\\&\textsf{--------------------------------------}\ +\\&\frac{3}{x-1}+\frac{5}{y+3}+\frac{5}{2}=8\\(1):\ &\frac{3}{x-1}+\frac{4}{y+3}+\frac{2}{2}=6\\&\textsf{--------------------------------------}\ -\\&\qquad\qquad\!\frac{1}{y+3}+\frac{3}{2}=2\\&\Rightarrow \frac{1}{y+3}=2-\frac{3}{2}\\&\Rightarrow \frac{1}{y+3}=\frac{1}{2}\\&\Rightarrow y+3=2\\&\Rightarrow y=\bf{-}1\end{aligned}$

Substitusi y dan z ke persamaan (1):

$\begin{aligned}&\frac{3}{x-1}+\frac{4}{2}+\frac{2}{2}=6\\&\Rightarrow \frac{3}{x-1}+3=6\\&\Rightarrow \frac{3}{x-1}=3\\&\Rightarrow x-1=1\\&\Rightarrow x=\bf2\end{aligned}$