MATHS...simultaneous equation

1.Jack purchased x calculators for $2400.He left one for himself, and sold all the rest ,each at a price of $50 higher than the original purchase price.
The total amount thus received was $3000.

(a)Show that x^2-13x-48=0, and hence find out the number of calculators that
Jack purchased.
New price of each calculator sold out = $[3000/(x-1)]
Original purchase price of each calucator= $(2400/x)
3000/(x-1) - 2400/x = 50
3000x - 2400(x-1) = 50x(x-1)
3000x - 2400x + 2400 = 50x^2 - 50x
50x^2 - 650x - 2400 = 0
x^2 - 13x - 48 = 0
(x-16)(x+3) = 0
x = 16 or x = -3(rej.)


(b)At what price did Jack sell each calculator?

The price = $[(2400/16) + 50] = $200

2. If the sum of two numbers is 1, and the difference between their their
reciprocals is 3/2, find the larger of the two number.

Let x be the one of the number, then the other number is (1-x)
1/x - 1/(1-x) = 3/2
(1-x-x)/x(1-x) = 3/2
(1-2x)/x(1-x) = 3/2
3x(1-x) = 2(1-2x)
3x - 3x^2 = 2 - 4x
3x^2 - 7x + 2 = 0
(x-2)(3x-1) = 0
x = 2 or x = 1/3
When x = 2, the other number is (1-2) = -1
When x = 1/3, the other number is (1-1/3) = 2/3
Hence, the larger number is 2 or 2/3