Lalu subs nilai x di pers iv ke pers. i dan pers. iii [tex]\begin{pmatrix}\frac{2}{x}+\frac{2}{y}+\frac{3}{z}=4~~(pers~i)\\\\ \frac{3}{x}+\frac{3}{y}+\frac{2}{z}=1~~(pers~iii)\end{pmatrix}[/tex] [tex]\begin{pmatrix}\frac{2}{\frac{yz}{4(y-z)}}+\frac{2}{y}+\frac{3}{z}=4~~(pers~i)\\\\ \frac{3}{\frac{yz}{4(y-z)}}+\frac{3}{y}+\frac{2}{z}=1~~(pers~iii)\end{pmatrix}[/tex] [tex]\begin{pmatrix}\frac{2(4(y-z))}{yz}+\frac{2}{y}+\frac{3}{z}=4\\\\ \frac{3(4(y-z))}{yz}+\frac{3}{y}+\frac{2}{z}=1\end{pmatrix}[/tex] [tex]\begin{pmatrix}\frac{8(y-z)}{yz}+\frac{2}{y}+\frac{3}{z}=4\\\\ \frac{12(y-z)}{yz}+\frac{3}{y}+\frac{2}{z}=1\end{pmatrix}[/tex] [tex]\begin{pmatrix}\frac{8y-8z}{yz}+\frac{2z}{yz}+\frac{3y}{yz}=4\\\\ \frac{12y-12z}{yz}+\frac{3z}{yz}+\frac{2y}{yz}=1\end{pmatrix}[/tex] [tex]\begin{pmatrix}\frac{8y-8z+2z+3y}{yz}=4\\\\ \frac{12y-12z+3z+2y}{yz}=1\end{pmatrix}[/tex] [tex]\begin{pmatrix}\frac{11y-6z}{yz}=4\\\\ \frac{14y-9z}{yz}=1\end{pmatrix}[/tex] [tex]\begin{pmatrix}4yz=11y-6z\\\\ yz=14y-9z\end{pmatrix}[/tex] yz = 14y - 9z, maka 4(14y-9z) = 11y-6z 56y-36z = 11y-6z 56y-11y = 36z-6z 45y = 30z (45/15)y = (30/15)z 3y = 2z (pers v.) y = 2z/3
yz = 14y - 9z, maka (2z/3)z = 14(2z/3) - 9z 2z²/3 = (28z/3) - 9z 2z²/3 = (28z/3) - (27z/3) 2z² = 28z - 27z 2z² = z 2z² - z = 0 2z(z - 1/2) = 0 z = 0, z = 1/2, krn syarat pecahan penyebut z ≠ 0 maka ---> z = 1/2
Substitusi untuk mencari y 3y = 2z (pers v.) 3y = 2(1/2) 3y = 1 ---> y = 1/3
Substitusi untuk mencari x [tex]\boxed{x=\frac{yz}{4(y-z)}}~~\sf(pers.~ iv)[/tex] [tex]\sf x=\frac{\frac{1}{3}\times\frac{1}{2}}{4(\frac{1}{3}-\frac{1}{2})}\\x=\frac{\frac{1}{6}}{4(\frac{2}{6}-\frac{3}{6})}\\x=\frac{\frac{1}{6}}{4(-\frac{1}{6})}[/tex] ---> x = -1/4
Jawab:
Hp = {-1/4, 1/3, 1/2}
Penjelasan:
[tex]\sf\frac{1}{x}+\frac{4}{y}-\frac{4}{z}=0~~(pers.~ii)\\\frac{1}{x}=\frac{4}{z}-\frac{4}{y}\\\frac{1}{x}=\frac{4y}{yz}-\frac{4z}{yz}\\\frac{1}{x}=\frac{4y-4z}{yz}\\\frac{1}{x}=\frac{4(y-z)}{yz}\\\boxed{x=\frac{yz}{4(y-z)}}~~(pers.~ iv)[/tex]
Lalu subs nilai x di pers iv
ke pers. i dan pers. iii
[tex]\begin{pmatrix}\frac{2}{x}+\frac{2}{y}+\frac{3}{z}=4~~(pers~i)\\\\ \frac{3}{x}+\frac{3}{y}+\frac{2}{z}=1~~(pers~iii)\end{pmatrix}[/tex]
[tex]\begin{pmatrix}\frac{2}{\frac{yz}{4(y-z)}}+\frac{2}{y}+\frac{3}{z}=4~~(pers~i)\\\\ \frac{3}{\frac{yz}{4(y-z)}}+\frac{3}{y}+\frac{2}{z}=1~~(pers~iii)\end{pmatrix}[/tex]
[tex]\begin{pmatrix}\frac{2(4(y-z))}{yz}+\frac{2}{y}+\frac{3}{z}=4\\\\ \frac{3(4(y-z))}{yz}+\frac{3}{y}+\frac{2}{z}=1\end{pmatrix}[/tex]
[tex]\begin{pmatrix}\frac{8(y-z)}{yz}+\frac{2}{y}+\frac{3}{z}=4\\\\ \frac{12(y-z)}{yz}+\frac{3}{y}+\frac{2}{z}=1\end{pmatrix}[/tex]
[tex]\begin{pmatrix}\frac{8y-8z}{yz}+\frac{2z}{yz}+\frac{3y}{yz}=4\\\\ \frac{12y-12z}{yz}+\frac{3z}{yz}+\frac{2y}{yz}=1\end{pmatrix}[/tex]
[tex]\begin{pmatrix}\frac{8y-8z+2z+3y}{yz}=4\\\\ \frac{12y-12z+3z+2y}{yz}=1\end{pmatrix}[/tex]
[tex]\begin{pmatrix}\frac{11y-6z}{yz}=4\\\\ \frac{14y-9z}{yz}=1\end{pmatrix}[/tex]
[tex]\begin{pmatrix}4yz=11y-6z\\\\ yz=14y-9z\end{pmatrix}[/tex]
yz = 14y - 9z, maka
4(14y-9z) = 11y-6z
56y-36z = 11y-6z
56y-11y = 36z-6z
45y = 30z
(45/15)y = (30/15)z
3y = 2z (pers v.)
y = 2z/3
yz = 14y - 9z, maka
(2z/3)z = 14(2z/3) - 9z
2z²/3 = (28z/3) - 9z
2z²/3 = (28z/3) - (27z/3)
2z² = 28z - 27z
2z² = z
2z² - z = 0
2z(z - 1/2) = 0
z = 0, z = 1/2, krn
syarat pecahan penyebut
z ≠ 0 maka
---> z = 1/2
Substitusi untuk mencari y
3y = 2z (pers v.)
3y = 2(1/2)
3y = 1
---> y = 1/3
Substitusi untuk mencari x
[tex]\boxed{x=\frac{yz}{4(y-z)}}~~\sf(pers.~ iv)[/tex]
[tex]\sf x=\frac{\frac{1}{3}\times\frac{1}{2}}{4(\frac{1}{3}-\frac{1}{2})}\\x=\frac{\frac{1}{6}}{4(\frac{2}{6}-\frac{3}{6})}\\x=\frac{\frac{1}{6}}{4(-\frac{1}{6})}[/tex]
---> x = -1/4
Hp = {x, y, z}
Hp = {-1/4, 1/3, 1/2}
(xcvi)