Solve the following simultaneous equations algebraically.
x^2=y
2y=x+3
MATHS牙
2006-11-01 2:41 am
回答 (2)
2006-11-01 2:49 am
✔ 最佳答案
x^2=y -------------- (1)2y=x+3 ---------------- (2)
From (1) and (2), 2x^2=x+3
2x^2-x-3=0
(2x-3)(x+1)=0
x=3/2 or x =-1
When x =3/2, y= 9/4
When x=-1, y=1.
Therefore, {(3/2, 9/4), (-1, 1)} is the required solution set.
2006-11-01 2:57 am
y=x^2......(1)
2y=x+3......(2)
(2):y=(x+3)/2
Therefore x^2=(x+3)/2
2x^2=x+3
2x^2-x-3=0
(x+1)(2x-3)=0
x=-1 or x=3/2
y=(-1)^2=1 or y=(3/2)^2=9/4
Therefore the solution is
x=-1,y=1 or x=3/2,y=9/4
2y=x+3......(2)
(2):y=(x+3)/2
Therefore x^2=(x+3)/2
2x^2=x+3
2x^2-x-3=0
(x+1)(2x-3)=0
x=-1 or x=3/2
y=(-1)^2=1 or y=(3/2)^2=9/4
Therefore the solution is
x=-1,y=1 or x=3/2,y=9/4
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