<J0 ТИК APPLICATION Of ALGEBRA." 
■»consequently x \ a — -f - a ~. Hence it appears 
that, if the difference of the squares be divided by 
twice the difference of the numbers, and half the dif 
ference of the numbers be subtracted from the quotient, 
■the remainder will be the lesser number; but if half 
the difference of the numbers he added to the quotient, 
the sum will give the greater number. Thus, if the dif 
ference (я) be 4, and the difference (b) of the squares 
40 (as in the case above); then ( L f a ~) the difference 
of the squares, divided-by twice the difference of the 
numbers, ivill be 5; from which subtracting (2) half 
the difference of the numbers, there remains 3, for the 
lesser number sought; and by adding the said half dif 
ference, you will have 7 rr the greater number. In 
the same manner, if the difference of the t wo numbers 
had been given 0, and flic difference of their squares do, 
the numbers themselves would have come out 2 aud 
8: and so of any other. 
PROBLEM XXIX. 
Having given the sum o f two numbers, equal to 30, and 
the difference of their squares, equal to 120; to Jind the 
numbers. 
' Put a — 30, and h — 120, and let x he the lesser num 
ber sought, and then the greater will be a—,r; whose 
square is aa — Stax- f x z \ from which the square of the 
lesser being subtracted, we have a~— 2ax zz h; this re 
duced, giyes x, the lesser number, — — is. 
Therefore the greater [a — r) will be — a | 
~ ~ h — 17. But if the greater number 
2 a 2 '2a, b 
had been first made the object of our inquiry, or been 
put — x, the lesser would have heen a — x, and irs 
square a'— 2dx -f .1". which subtracted from x 1 leaves