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:
[3 marks]
Solution:
(a)
(b)
(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:
[3 marks]
Solution:
(a)
(b)
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 + 5x – z = 3
Solve the following system of linear equations and explain the result of the findings. [4 marks]
y – 7z = -2
x – y + 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)
x – y + 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
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 + 5x – z = 3
Solve the following system of linear equations and explain the result of the findings. [4 marks]
y – 7z = -2
x – y + 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)
x – y + 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