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be found, so that y may be an integer ; and, in the fourth, x is to be found, fo that y may be a rational quantity.
1. When y=4+6*, the folution depends on the rule for deducing a fraction to its lowest terms. See the text, page 125.
it is plain that; in order to have y an integer, b+cx must be a divisor of a. Let d be one of its divifors, then, if b+cx=d, x=;
b; so that among the divisors of a, we must find one, if possible, from which b being taken away, the remainder may be divisible by c: the quotient is a value of x.
if d be a divisor of
b, x will be taken out of the numerator, if we divide it by dxtc; and this form is then reduced to the preceding. But, if d is not a divisor of b, multiply both sides by , or, dividing bdx tad
d', then dy=adtbdx
by dx+c, dy=b+adambi
, and so x is found
Let 2xy+x+y=195; then y (2x+1)=
. Therefore 2y =
4. When y=va+bx+cx?, and x is to
quantity. Here there are four cases, according to the nature of the coefficients a, b, and c.
Imo, If a be a square number, as for instance g?, so that the formula is vg2+bx+cx?. Suppose Vg2+bx+cx2 = 8 + mx, then g2+bx+cx?=g2+2mgx+m2x, or bx+ cx2 2mgx + m2x2, that is, b+cx = 2mg +m’x, and x=
If for x this value of it be substituted in the given formula, its irrationality will dis
and vg2+bx+cxt = cg—bm+gma m may be assumed, therefore, equal to any quantity, positive or negative, integral or fractional, and the corresponding value of it will answer the conditions prescribed.
2do, If c be a square number, as g?, then let Va+bx+g2x2=m+gx. Hence a+bx+g2x2=m2 +2mgx +g4x?, or a+bx
ma =m2 +2mgx. Therefore x =
Here my as before, may be assumed at pleasure.
3tio, Though neither a nor c are square numbers, yet if a+bx+cava can be resolved into two simple factors, as f+gx, and b+kx, the irrationality of the formula may be taken away. For, let Va+bx+cx2
v (f+gx) (b+kx)=m(f+gx), and (f+gx) (b+kx) =m2(f+8x), or h+kx = m2 (f+ex), and x = By the sub
k--gm fiitution of this value of x, (m being afsti
med at pleasure), the irrationality will be removed, as before.
4to, The fourth case, in which the formùla a+bx+cx? may be rendered a complete square, is when it can be divided into two parts, one of which is a complete square, and the other a produd of two simple factors. For atbx+cx? is then of the form pa-qr, , q, and r, being quantities into which there enters no power
of x higher than the first : And, if we assume ✓patar=p+mq, p2 will be exterminated, and the remainder, being divided by 9, will be a simple equation, from which x may be easily determined.
These methods of removing the irrationality of the preceding formula are to be particularly attended to, as being of great use in the higher geometry.