中二簡單simultaneous equations(20點

1. The length of a rectangle is 2 times of its width. The perimeter is 42cm. Find the area of the rectangle.
Let the width be x cm
Then the length is 2x cm
The perimeter = 2(x + 2x) = 6x = 42
x = 7
The area = (x)(2x) = 2x^2 = 98 cm^2
2. Sandy plans to pay $48 for two kinds of pens. If she buys 5 ballpoints and 7 highlighters, she will need $2 more. If she buys 7 ballpoints and 5 highlighters, she will have $2 left. Find the selling price of each kind of pen.
Let the selling price of a ballpoint and a highlighter be $B and $H respectively.
5B + 7H = 48 + 2 = 50 ... (1)
7B + 5H = 48 - 2 = 46 ... (2)
(1) - (2) -> 2H - 2B = 4
H - B = 2
H = 2 + B ... (3)
Sub into (1), 5B + 7(2 + B) = 50
12B + 14 = 50
12B = 36
B = 3
(3) => H = 2 + 3 = 5
The selling price of a ballpoint and a highlighter are $3 and $5 respectively.
3. The sum of the hundreds digit and the tens digit of a three-digit number is 13. If the hundreds digit and the tens digit are interchanged, the new number is 90 less than the oringinal number. It is given that the units digit of the original number is 4. Find the original number.
Let the hundreds digit be H and the tens digit be T.
H + T = 13
H = 13 - T ... (1)
(100H + 10T + 4) - (100T + 10H + 4) = 90
90H - 90T = 90
H - T = 1 ... (2)
Sub (1) into (2), 13 - T - T = 1
2T = 12
T = 6
(1) => H = 13 - 6 = 7
The original number is 764

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1) Let width be x cm, then length is 2x cm.
2 x (2x+x) = 42cm
x = 7cm
width 7cm, length 14cm, area = width x length = 7 x 14 = 98cm sq

2) Let price of ballpens & highlighters be $x & $y respectively
5x+7y=48+2 ---- (1)
7x+5y=48-2 ----- (2)
from (1)
x = 10-7/5y ---- (3)
sub (3) into (2)
7 (10-7/5y) + 5y = 46
y = 5
x = 3
ballpens $3, highlighter $5