Expert: Sherman D. - 4/13/2007

Question
i must use substitution method and not sure how to solve these. i tried and im not getting the correct answer plus i looked for examples in my book but it didnt have any for this kind of problem. please help or at least help me start it.

1)xy=6
 2x-y=1


2)y^2=x^-9
 2y=x-3

Answer
1.)
xy = 6
2x - y = 1

2x - y = 1
-y = -2x + 1
y = 2x - 1

(2x - 1)x = 6
2x^2 - x = 6
2x^2 - x - 6 = 0

x = (-b ± sqrt(b^2 - 4ac))/(2a)

x = (-(-1) ± sqrt((-1)^2 - 4(2)(-6)))/(2(2))
x = (1 ± sqrt(1 + 48))/4
x = (1 ± sqrt(49))/4
x = (1 ± 7)/4
x = 8/4 or (-6/4)
x = 2 or (-3/2)

y = 2x - 1
y = 2(2 or (-3/2)) - 1
y = (4 or -3) - 1
y = 3 or -4

ANS : (2,3) or ((-3/2),-4)

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Assuming you meant
2.)
y^2 = x^2 - 9
2y = x - 3

2y = x - 3
x = 2y + 3

y^2 = (2y + 3)^2 - 9
y^2 = ((2y + 3)(2y + 3)) - 9
y^2 = (4y^2 + 6y + 6y + 9) - 9
y^2 = 4y^2 + 12y + 9 - 9
3y^2 + 12y = 0
3y(y + 4) = 0
y = 0 or -4

x = 2y + 3
x = 2(0 or -4) + 3
x = (0 or -8) + 3
x = 3 or -5

ANS : (3,0) or (-5,-4)