Jun Heng's answer to Stel's Secondary 3 A Maths Singapore question.
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a)
2y+3x = 7 ––– (1)
ky²–kxy = 1 ––– (2)
Sub (a, 1.5) into (1)
2(1.5)+3a = 7
3+3a = 7
3a = 4
a = 4/3
Sub (4/3, 1.5) into (2)
k(1.5)²–k(4/3)(1.5) = 1
2.25k–2k = 1
0.25k = 1
k = 4 (shown)
b)
From (1)
2y+3x = 7
3x = 7–2y
x = (7–2y)/3 ––– (3)
Sub (3) into (2)
4y²–4[(7–2y)/3]y = 1
12y²–4(7–2y)y = 3
12y²–28y+8y² = 3
20y²–28y–3 = 0
(2y–3)(10y+1) = 0
y = 3/2 or y = –1/10
Sub y = –1/10 into (3)
x = [7–2(–1/10)]/3 = 2.4
Coordinates of other point = (2.4, –0.1)
2y+3x = 7 ––– (1)
ky²–kxy = 1 ––– (2)
Sub (a, 1.5) into (1)
2(1.5)+3a = 7
3+3a = 7
3a = 4
a = 4/3
Sub (4/3, 1.5) into (2)
k(1.5)²–k(4/3)(1.5) = 1
2.25k–2k = 1
0.25k = 1
k = 4 (shown)
b)
From (1)
2y+3x = 7
3x = 7–2y
x = (7–2y)/3 ––– (3)
Sub (3) into (2)
4y²–4[(7–2y)/3]y = 1
12y²–4(7–2y)y = 3
12y²–28y+8y² = 3
20y²–28y–3 = 0
(2y–3)(10y+1) = 0
y = 3/2 or y = –1/10
Sub y = –1/10 into (3)
x = [7–2(–1/10)]/3 = 2.4
Coordinates of other point = (2.4, –0.1)
Date Posted:
1 year ago
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