math f,4 問題

1)
Let X & Y be the length & width of playground respectively,
2(X + Y) = 324
(X + Y) = 162
X = 162 - Y -- (i)

(X + 6)*(Y + 3) = XY(1+1/9)
XY + 6Y + 3X + 18 = 10/9*XY
9XY + 54Y + 27X + 162 = 10XY
54Y + 27X - XY + 162 = 0 -- (ii)
sub (i) into (ii),
54Y + 27(162-Y) - (162-Y)Y + 162 = 0
54Y + 4374 - 27Y - 162Y + Y^2 + 162= 0
Y^2 - 135Y + 4536 = 0
(Y-72)(Y-63) = 0
Y=72 or Y=63

X = 162 - Y
X = 90 or 99

Original dimension = 90*72 or 99*63


2)
Let X be the ten digit place, Y be the unit digit
10X + Y = original no.

10X + Y = 3(XY) - 8
10X + Y = 3XY - 8
10X + Y - 3XY + 8 = 0 -- (i)

X -Y = 2
X = 2 + Y -- (ii)

sub (ii) into (i),
10(2+Y) + Y - 3(2+Y)Y + 8 = 0
20 + 10Y + Y - 6Y - 3Y^2 + 8 = 0
-3Y^2 + 5Y + 28 = 0
(Y-4)(-3Y-7) = 0
Y = 4 or -7/3 (rejected)
Y = 4 and X = 6
Hence, the no. is 64

題1
Let the original length be x m.
so, the original width = (162 - x) m.
Let the original area be y m^2.
y = x(162-x) -----(1)
So, the new area: 10/9y = (x+6)(162-x+3)
y = 9/10(x+6)(165-x) -----(2)
Then, we combine (1) and (2): x(162-x) = 9/10(x+6)(165-x)
10(162x-x^2) = (9x+54)(165-x)
1620x-10x^2 = 1485x-9x^2+8910-54x
x^2-1674x+10395=0
Then, you an find out the length and the width.
題2
Sorry, I don't really understand your question....SORRY!!!