\

3.2.1 Simultaneous Equations Example 1 & 2


Example 1:
Solve the simultaneous equations.
x + 1 4 y = 1  and  y 2 8 = 4 x .

Solution:
x + 1 4 y = 1 ( 1 ) y 2 8 = 4 x ( 2 ) x = 1 1 4 y ( 3 )

Substitute (3) into (2),
y 2 8 = 4 ( 1 1 4 y ) y 2 8 = 4 4 4 y
y2 + y – 12 = 0
(y + 4)(y – 3) = 0
y = –4 or y = 3 

Substitute the values of y into (3),
when y=4,  x=1 1 4 (4)=2 when y=3,  x=1 1 4 (3)= 1 4 The solutions are x=2, y=4 and x= 1 4 , y=3


Example 2:
Solve the simultaneous equations 2x + y = 1 and 2x2+ y2 + xy = 5.
Correct your answer to three decimal places.

Solution:
2x + y = 1—–(1)
2x2 + y2+ xy = 5—–(2)

From (1),
y = 1 – 2x—–(3)

Substitute (3) into (2).
2x2 + (1 – 2x)2 + x(1 – 2x) = 5
2x2 + (1 – 2x)(1 – 2x) + x – 2x2 = 5
1 – 2x – 2+ 4x2 + x – 5 = 0
4x2 – 3x – 4 = 0


From  x = b ± b 2 4 a c 2 a a = 4 ,   b = 3 c = 4 x = ( 3 ) ± ( 3 ) 2 4 ( 4 ) ( 4 ) 2 ( 4 ) x = 3 ± 73 8 x = 0.693  or  1.443

Substitute the values of x into (3).
When x = –0.693,
y = 1 – 2 (–0.693) = 2.386 (correct to 3 decimal places)

When x = 1.443,
y = 1 – 2 (1.443) = –1.886 (correct to 3 decimal places)

The solutions are x = –0.693, y  = 2.386 and x = 1.443, y = –1.886.

1 thought on “3.2.1 Simultaneous Equations Example 1 & 2”

Leave a Comment