15分 F.4 maths Algebraic Method

1)let x, x+1 be the two consecutive intergers
1/x+ 1/x+1=7/12 (1 over x + 1 over (x+1))
(2x+1)/(x^2+x)=7/12
7x^2-17x-12=0
x=3, -0.571428571<---(捨去) 因為佢話係個整數
所以ans 係3 同4 ~~

2a) x^2-13x -48=0
(x-16)(x+3)=0

b) let x be the number of calculators that jack purchased and y be the price of each calculator

xy=2400-----(1) (因為jack 買左x部機，每部機y 元)
(x-1)(y+50)=3000 ----(2) (jack 留番一部機比自己，所以係x-1 ，而佢依家將淨番d機用比原價高$50 賣出，所以係y+50 ，佢話total amount received was $3000，所以得出(x-1)(y+50)=3000)

(2):(x-1)(y+50)=3000
xy+50x-y-50=3000
xy+50x-y-3050=0
2400+50x-y-3050=0<----(從1中xy=2400)
50x-y-650=0
y=50x-650--------(3)

將(3) 代入 (1)
x(50x-650)=2400
50x^2-650x-2400=0(全式除此50得到以下結果)
x^2-13x -48=0
(x-16)(x+3)=0

x=16, -3 (由於答案唔會係負數，所以-3 late 個ans 會捨去)

所以jack 買左16 部calculators

xy=2400
16y=2400
y=150

price of each calculator $150

1)
Let the two intergers are x and x+1

1/x + 1/(x+1) = 7/12, solve x
(2x+1)/[x(x+1)] = 7/12
24x + 12 = 7x^2 + 7x
7x^2 - 17x - 12 = 0
(7x + 4)(x - 3) = 0
x = -4/7 (rejected)
or
x = 3

The two intergers are 3 and 4.
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2)
The unit price he originally purchased = 2400/x
Now he set the new price he sells = 2400/x + 50
(2400/x + 50)(x - 1) = 3000
(2400 + 50x)(x - 1) = 3000x
2400x - 2400 + 50x^2 - 50x - 3000x = 0
50x^2 - 650x - 2400 = 0
x^2 - 13x - 48 = 0
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(x + 3)(x - 16) = 0
x = -3 (rejected)
or
x = 16

So, the number of calculators that he purchased is 16
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The price he purchase is 2400/16 = $150