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Saturday, January 27, 2007 7:56 PM

these are the answers to the qns on gc's blog...

Qn 1)

y = 2 – x --------[ 1 ]
x(x+2– x) = 5 – 3(2 – x)²--------[ 2 ]

Substituting [ 1 ] into [ 2 ],
x+2x - x² = 5 – 3(4 – 4x+x²)
2x = 5 – 12+12x – 3x²
3x² - 10x+7 = 0
(x – 1)(3x - 7) = 0
x = 1 or x = 2 1/3
when x = 1, when x = 2 1/3
y = 1 or y = - 1/3

Qn 2)

3x² + y² = 28 -------- [ 1 ]
3x + y = 8 -------- [ 2 ]

From [ 2 ] : y = 8 – 3x
Substituting into [ 1 ],
3x² + (8 – 3x) ² = 28
3x² + 64 – 48x + 9x² = 28
12x² - 48x + 36 = 0
x² - 4x +3 = 0
(x - 1)(x - 3) = 0
x = 1 or x = 3
when x = 1 , when x = 3,
y = 5 or y = -1

Qn 3)

2x + 3y = 5 -----[ 1 ]
x² + y² = 6x – 4y -----[ 2 ]

From [ 1 ] : y = (5 – 2x) / 3
Substituting into [ 2 ],
x² + [(5 – 2x) ²] / 3 = 6x – 4[(5 – 2x) / 3]
x² + (25 - 20x + 4x²) / 9 = 6x - (20 + 8x) / 3
9x² + 25 - 20x + 4x² = 54x - 60 + 24x
13x² - 98x + 85 = 0
(x - 1)(13x - 85) = 0
x = 1 or x = 6 7/13
when x = 1, when x = 6 7/13
y = 1 or y = -2 9/13

Qn 4)

2x - 5y +17 = 0 ------[ 1 ]
xy = 6 ------[ 2 ]

From [ 2 ], y = 6/x
Substituting to [ 2 ],
2x -5[(6/x)] + 17 = 0
2x² - 30 + 17x = 0
(x + 10)(2x - 3) = 0
x = -10 or x = 1 1/2
when x = -10, when x = 1 1/2,
y = -3/5 or y = 4

p.s. i nvr check ans so if there is anything wrong pls contact me.