quadratic equation幾題問題

1)Let x be the original number of childern in the class
240 / x = [240 / (x﹣6)]﹣9
multiply both sides bt x(x﹣6),
(240 / x)[x(x﹣6)] = [240 / (x﹣6)][x(x﹣6)]﹣9x(x﹣6)
240(x﹣6) = 240x﹣9x2 + 54x
240x﹣1440 = 240x﹣9x2 + 54x
9x2 ﹣54x﹣1440 = 0
x2 ﹣6x﹣160 = 0
(x﹣16)(x + 10) = 0
x = 16 or -10 (rejected)
∴the original number of childern in the class is 16.

2)Let the original price of a dozen bottles of wine be $x
then the new price of a dozen bottles of wine be $(x + 150),
900 / x = [900 / (x + 150)] + 1
multiply both sides bt x(x + 150),
(900 / x)[x(x + 150)] = [900 / (x + 150)][x(x + 150)] + x(x + 150)
900(x + 150) = 900x + x2 + 150x
900x + 135000 = 900x + x2 + 150x
x2 + 150x﹣135000 = 0
(x﹣300)(x + 450) = 0
x = 300 or -450(rejected)
∴the original price per dozen bottles of wine is $300

3)The born produces number of each day = 800/n
[(800/n) + 10](n﹣4) = 800
[(800/n) + (10n/n)](n﹣4) = 800
[(800 + 10n)/n](n﹣4) = 800
(800 + 10n)(n﹣4) = 800n
800n﹣3200 + 10n2﹣40n = 800n
10n2﹣40n﹣3200 = 0
n2﹣4n﹣320 = 0
(n﹣20)(n + 16) = 0
n = 20 or -16 (rejected)
∴the value of n is 20

4)Let x be the smaller integer,
(1 / x) + [1 / (x + 1)] = 11 / 30
multiply both sides bt x(x + 1),
(1 / x)[x(x + 1)] + [1 / (x + 1)][x(x + 1)] = (11 / 30)x(x + 1)
x + 1 + x = 11x(x + 1) / 30
30(2x + 1) = 11x2 + 11x
60x + 30 = 11x2 + 11x
11x2﹣49x﹣30 = 0
(11x + 6)(x﹣5) = 0
x = -11/6 (rejected) or 5
∴the two consecutive positive integers is 5 and 6.

5)Let y be the speed of the man in still water,
[9 / (y﹣2)] + [9 / (y + 2)] = 6
multiply both sides bt (y﹣2)(y + 2),
[9 / (y﹣2)](y﹣2)(y + 2) + [9 / (y + 2)](y﹣2)(y + 2) = 6(y﹣2)(y + 2)
9(y + 2) + 9(y﹣2) = 6(y﹣2)(y + 2)
9y + 18 + 9y﹣18 = 6(y2﹣4)
18y = 6y2﹣24
6y2﹣18y﹣24 = 0
y2﹣3y﹣4 = 0
(y﹣4)(y + 1) = 0
y = 4 or -1 (rejected)
∴the speed of the man in still water is 4 km/h

1.
240/(x-6) - 9 = 240/x
x^2 - 6x - 160 = 0
x=16 or -10 (reject)

2.
900/(x+150) = 900/x - 1
x^2 + 150x - 135000 = 0
x = 300 or -450 (reject)

3.
800/n = 800/(n-4) - 10
n^2 - 4n - 320 = 0
x = 20 or -16 (reject)

4.
1/x + 1/(x+1) = 11/30
x = 5 or -6/11 (reject)
the two numbers are 5 and 6

5.
9/ (x-2) + 9/ (x+2) = 6
x = 4 or -1 (reject)

I only can solve Q.no.1

1.240 sweets are distributed evenly to a class of children.if six children are absent,then each child will get nine more sweets.find the original number of childern in the class.

240/x=y
240/(x-6)=(y+9)

240=(x-6)(y+9)
240=xy+9x-6y-54
240+6y+54=xy+9x
294+6y=x(y+9)
(294+6y)/(y+9)=x
X=(294+6y)/(y+9)

2.when the price of a dozen bottles of wine is raised by $150,12 fewer bottles of wine can be bought at $900.find the original price per dozen bottles of wine.