Bab 4 Persamaan Serentak

Posted on May 10, 2016 by user

4.2 SPM Praktis, Persamaan Serentak, (Soalan panjang)

Soalan 1:

Selesaikan persamaan serentak.

\begin{array}{l} y + 2 x = 2 \\ \frac{2}{x} + \frac{1}{y} = 5 \end{array}

Penyelesaian:

\begin{array}{l} y + 2 x = 2 - - - - (1) \\ \frac{2}{x} + \frac{1}{y} = 5 - - - - (2) \\ y = 2 - 2 x - - - - (3) \\ Gantikan (3) ke dalam (2), \\ \frac{2}{x} + \frac{1}{2 - 2 x} = 5 \\ \frac{2 (2 - 2 x) + x}{x (2 - 2 x)} = 5 \\ 4 - 4 x = 5 x (2 - 2 x) \\ 4 - 4 x = 10 x - 10 x^{2} \\ 10 x^{2} - 14 x + 4 = 0 \\ 5 x^{2} - 7 x + 2 = 0 \\ (5 x - 2) (x - 1) = 0 \\ 5 x - 2 = 0 or x - 1 = 0 \\ x = \frac{2}{5} or x = 1 \\ Gantikan nilai-nilai x ke dalam (3), \\ Apabila x = \frac{2}{5}, \\ y = 2 - 2 (\frac{2}{5}) = 1 \frac{1}{5} \\ Apabila x = 1 \\ y = 2 - 2 (1) = 0 \\ Penyelesaian ialah x = \frac{2}{5}, y = 1 \frac{1}{5} dan x = 1, y = 0 \end{array}

Soalan 2:

Selesaikan persamaan serentak.

x – 3y + 5 = 3y + 5y²– 6 – x = 0

Penyelesaian:

x – 3y + 5 = 0

x = 3y – 5 -----(1)

3y + 5y² – 6 – x = 0 -----(2)

Gantikan (1) ke dalam (2),

3y + 5y² – 6 – (3y – 5) = 0

3y + 5y² – 6 – 3y + 5 = 0

5y² – 1 = 0

5y² = 1

y = 0.447

Gantikan nilai-nilai yke dalam (1),

Apabila y = 0.447

x = 3 (0.447) – 5

x = –3.659

Apabila y = – 0.447

x = 3 (–0.447) – 5

x = –6.341

Penyelesaian ialah x = –3.659, y = 0.447 dan x = –6.341, y = – 0.447.