Buku Teks Matematik Tingkatan 1 (KSSM) Bab 6 – Mahir Diri 6.3 (Soalan 2)

Soalan 2:
Selesaikan persamaan linear serentak yang berikut.
(a) \[ \begin{aligned} & x+4 y=14 \\ & 3 x+2 y=12 \end{aligned} \] (b) \[ \begin{aligned} & 3 m-2 n=19 \\ & 5 m+7 n=11 \end{aligned} \] (c) \[ \begin{aligned} & \frac{1}{3} p+q=4 \\ & \frac{p-q}{4}=2 \end{aligned} \] (d) \[ \begin{aligned} & \frac{f}{2}+\frac{g}{5}=3 \\ & 2 g-f=10 \end{aligned} \]


Penyelesaian:
(a)
\(\begin{aligned} & x+4 y=14 \ldots \ldots(1) \\ & 3 x+2 y=12 \ldots \ldots(2) \\ & \operatorname{Dari}(1), x=14-4 y \ldots(3)\end{aligned}\)

Kaedah penggantian:
Gantikan (3) ke dalam (2),

\(\begin{aligned} 3(14-4 y)+2 y & =12 \\ 42-12 y+2 y & =12 \\ -10 y & =12-42 \\ -10 y & =-30 \\ y & =\frac{-30}{-10} \\ & =3\end{aligned}\)

Gantikan y = 3 ke dalam (3),
x = 14 – 4(3)
   = 2

Maka, x = 2 dan y = 3.


(b)
Kaedah penghapusan:
$\begin{array}{}\begin{array}{l}3m-2n=19\dots \dots \left(1\right)\\ 5m+7n=11\dots \dots \left(2\right)\\ \left(1\right)\times 5:15m-10n=95\dots \dots (3)\\ \underset{\_}{(2)\times 3:15m+21n=33\dots \dots (4)}\\ \left(3\right)-\left(4\right) :-31n=62\\ n=\frac{62}{-31}\\ n=-2\\ \\ \text{Gantikan }n=-2\text{ ke dalam }\left(1\right),\\ 3m-2\left(-2\right)=19\\ 3m+4=19\\ 3m=19-4\\ m=\frac{15}{3}\\ m=5\\ \\ \text{Maka},m=5\text{ dan }n=-2.\end{array}\end{array}$


(c)
$\begin{array}{l}\frac{1}{3}p+q=4\dots \dots \left(1\right)\\ \frac{p-q}{4}=2\dots \dots \left(2\right)\\ \text{Dari }(2),p-q=4\times 2\\ p=8+q\dots (3)\\ \\ \text{Gantikan (3) ke dalam }\left(1\right),\\ \frac{1}{3}(8+q)+q=4\\ (\times 3):8+q+3q=12\\ 4q=12-8\\ q=\frac{4}{4}\\ q=1\\ \\ \text{Gantikan }q=1\text{ ke dalam }\left(3\right),\\ p=8+1\\ p=9\\ \\ \text{Maka},p=9\text{ dan }q=1.\end{array}$


(d)
$\begin{array}{l}\frac{f}{2}+\frac{g}{5}=3\dots \dots \left(1\right)\\ 2g-f=10\dots \dots \left(2\right)\\ \text{Dari}\left(2\right),-f=10-2g\\ f=2g-10\dots \dots (3)\\ \\ \text{Gantikan }\left(3\right)\text{ ke dalam }\left(1\right),\\ \frac{2g-10}{2}+\frac{g}{5}=3\\ \left(\frac{2g-10}{\bcancel{2}}\right){\times }^5\sout{1}\sout{0}+\left(\frac{g}{\bcancel{5}}\right){\times }^2\sout{1}\sout{0}=3\times 10\\ 5(2g-10)+2g=30\\ 10g-50+2g=30\\ 12g=30+50\\ 12g=80\\ g=\frac{80}{12}\\ =\frac{20}{3}\\ \\ \text{Gantikan }g=\frac{20}{3}\text{ ke dalam }\left(3\right),\\ f=2\left(\frac{20}{3}\right)-10\\ =\frac{40}{3}-10\\ =\frac{10}{3}\\ \\ \text{Maka},f=\frac{10}{3}\text{ dan }g=\frac{20}{3}.\end{array}$