Matematik Tambahan SPM 2019, Kertas 2 (Soalan 1)


Soalan 1:
Selesaikan persamaan serentak x + 2y = 1 dan \(\frac{3}{x}-\frac{2}{y}=5\).
Beri jawapan anda betul kepada tiga tempat perpuluhan. [5 markah]


Penyelesaian:
\(\begin{aligned} &x+2 y=1\\ &x=1-2 y \ldots \ldots \ldots \ldots(1)\\ &\frac{3}{x}-\frac{2}{y}=5\\ &\times x y, 3 y-2 x=5 x y \ldots \ldots \ldots\\ &\text { Ganti (1) dalam (2): }\\ &3 y-2(1-2 y)=5 y(1-2 y)\\ &3 y-2+4 y=5 y-10 y^{2}\\ &10 y^{2}+2 y-2=0\\ &5 y^{2}+y-1=0\\ &a=5, b=1, c=-1\\ &y=\frac{-b \pm \sqrt{b^{2}-4 a c}}{2 a}\\ &=\frac{-1 \pm \sqrt{1-4(5)(-1)}}{10} \end{aligned}\)

\(\begin{aligned} &\begin{array}{l} y=\frac{-1+\sqrt{21}}{10} \\ =0.3583 \\ =0.358(3 \mathrm{tp}) \end{array}\\ &\text { atau }\\ &y=\frac{-1-\sqrt{21}}{10}\\ &\begin{array}{l} =-0.5583 \\ =-0.558(3 \mathrm{tp}) \\ \text { Dari }(1) \\ \text { Apabila } y=0.358 \\ x=1-2(0.3583) \\ \quad=0.283 \\ \text { Apabila } y=-0.558 \\ x=1-2(-0.5583) \\ =2.117 \end{array} \end{aligned}\)

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