SPM 2021 Add Maths Trial Paper (Selangor) – Paper 1 (Question 3 & 4)

Question 3:
(a) Find the value of p and q for the equation (√2 – 1)(p + q√2) = 5 – √8. [3 marks]

(b) Find the value of p which satisfies the following equation:
$\frac{3}{{\mathrm{log}}_{p}pq}+\frac{3}{{\mathrm{log}}_{q}pq}+2=5p$ [3 marks]

Solution:
(a)

(b)
$\begin{array}{l}\frac{3}{{\mathrm{log}}_{p}pq}+\frac{3}{{\mathrm{log}}_{q}pq}+2=5p\\ \\ \frac{3}{\frac{{\mathrm{log}}_{pq}pq}{{\mathrm{log}}_{pq}p}}+\frac{3}{\frac{{\mathrm{log}}_{pq}pq}{{\mathrm{log}}_{pq}q}}=5p-2\\ \\ \frac{3{\mathrm{log}}_{pq}p+3{\mathrm{log}}_{pq}q}{{\mathrm{log}}_{pq}pq}=5p-2\\ \\ 3{\mathrm{log}}_{pq}p+3{\mathrm{log}}_{pq}q=5p-2\\ {\mathrm{log}}_{pq}{\left(pq\right)}^{3}=5p-2\\ 3{\mathrm{log}}_{pq}\left(pq\right)=5p-2\\ 3=5p-2\\ 5p=5\\ p=1\end{array}$

Question 4:
Determine whether the following equations are system of linear equations in three variables or not. Explain. [1 mark]
2x + y = z2
4x – 2y + 3z = 6
11y + 5xz = 3

Solve the following system of linear equations and explain the result of the findings. [4 marks]
y – 7z = -2
xy + 5z = 2
-2x + 2y – 10z = 6

Solution
:

(a) No, because the power for one of the equation is 2.

(b)
y – 7z = -2
y = -2 + 7z ………… (1)

xy + 5z = 2
2x – 2y + 10z = 4 ………… (2)

-2x + 2y – 10z = 6 ………… (3)

(2) + (3),
0 = 10

Since 0 ≠ 10, therefore there is no solution for the linear equations