4.2.1 Simultaneous Equations Long Questions (Question 1 & 2)


Question 1:
Solve the following simultaneous equations.
y + 2 x = 2 2 x + 1 y = 5

 Solution:
y+2x=2(1) 2 x + 1 y =5(2) y=22x(3) substitute (3) into (2), 2 x + 1 22x =5 2( 22x )+x x( 22x ) =5 44x+x=5x( 22x ) 43x=10x10 x 2 10 x 2 13x+4=0 ( 5x4 )( 2x1 )=0 5x4=0     or     2x1=0 x= 4 5            or     x= 1 2 Substitute values of x into (3), When x= 4 5  ,  y=22( 4 5 )= 2 5 When x= 1 2 y=22( 1 2 )=1 The solutions are x= 4 5 , y= 2 5  and x= 1 2 , y=1


Question 2:
Solve the following simultaneous equations.
x – 3y + 5 = 3y + 5y2– 6 – x = 0

Solution:
x – 3y + 5 = 0
x = 3y – 5 —–(1)
3y + 5y2 – 6 – x = 0 —–(2)

Substitute (1) into (2),
3y + 5y2 – 6 – (3y – 5) = 0
3y + 5y2 – 6 – 3y + 5 = 0
5y2 – 1 = 0
5y2  = 1
y±0.447

Substitute the values of y into (1),
When y = 0.447
x = 3 (0.447) – 5
x = –3.659

When y = – 0.447
x = 3 (–0.447) – 5
x = –6.341

The solutions are x = –3.659, y = 0.447 and x = –6.341, y = – 0.447.

3 thoughts on “4.2.1 Simultaneous Equations Long Questions (Question 1 & 2)”

  1. This site actually good but I want to suggest to provide more questions on this site such as exercise. Well, many student can learn if you provide more question . Thank you…..

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