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.'. by subtraction x = 6, what A had at first;
Peob. 7. A smuggler had a quantity of brandy which he expected would raise £9. 18s.; after he had sold 10 gallons, a revenue officer seized one third of the remainder, in consequence of which he makes only £8. 2a.; required the number of gallons he had, and the price per gallon.
Let .1? be the number of gallons;
198 . . ...
then — is the price per gallon, in shillings,
the quantity seized,
, .T-10 198 . , , . . . ,
and x— the value of the quantity seized, which
appears by the question to be 36 shillings;
.'. a; = 22, the number of gallons;
. 198 198 ..... . .,
and = = 9 shillings, the price per gallon.
Prob. 8. A and B play at bowls, and A bets B three shillings to two upon every game; after a certain number of games it appears that A has won three shillings; but had he ventured to bet five shillings to two, and lost one game more out of the same number, he would have lost thirty shillings: how many games did they play?
Let at be the number of games A won,
y the number B won,
then 2a? is what A won of B, in shillings,
and 3y what B won of A;
.'. 2a? - 3y = 3, by the question.
Also (a? — I)x2 is what A would win on the 2nd supposition,
and (y + l)x5 what B would win
.-. 5y + 5-2a> + 2 = 30, by the question,
or 5y - 2x = SO - 5 - 2 = 23;
and 2.x — 3y>
.•• a? = 21;
Pkob. 9. A sum of money was divided equally amongst a certain number of persons; had there been three more, each would have received one shilling less, and had there been two fewer, each would have received one shilling more, than he did: required the number of persons, and what each received.
Let x be the number of persons,
y the sum each received, in shillings;
then xy is the sum divided,
and (,r + 3)x(«-l) = xy) , ,
V T , vy J yi , the question;
also (x-2)x(y + l)=.ryj
.•. .ry - ;r + Sy - 3 = ay, or — x + 3y = 3; and <ry + .i?-2y-2 = .ry, or «-2y = 2;
.'. y = 5 shillings, the sum received by each.
•\ x = 12, the number of persons.
200. When the terms of an equation involve the square of the unknown quantity, but the first power does not appear, the value of the square is obtained by the preceding rules*; and by extracting the square root on both sides, the quantity itself is found.
Ex. 1. 5x*- 45 = 0; to find x.
By transp. 5#2 = 45,
.•. (Art. 191), ar-\/9=±S.
The signs + and — are both prefixed to the root, because the square root of a quantity may be either positive or negative (Art. 147). The sign of x may also be negative; but still x will be either equal to +3 or -S+.
* It is obvious that the rules proved in Arts. 186—192, apply to all equations, quadratic, cubic, &c. as well as simple, because they are founded simply upon the Ajnonu (Arts. 79—82). Ed.
•f This may be shewn as follows:—suppose a*=at, then extracting the square root of both sides, since V*"=±ir, arid ^a'=±a, we have
But it is evident that (1) and (4) are in fact the same equation and also (2) and (3); so that x=±a includes all the four equations. Ed.
201. If both the first and second powers of the unknown quantity be found in an equation, arrange the terms according to the dimensions of the unknown quantity, beginning with the highest, and transpose the known quantities to the other side; then, if the square of the unknown quantity be affected with a coefficient, divide all the terms by this coefficient, and if its sign be negative, change the signs of all the terms (Art. 187), that the equation may be reduced to this form, a,!±p#=±<7. Then add to both sides the square of half the coefficient of the first power of the unknown quantity, by which means the first side of the equation is made a complete square, (Art. 152), and the other consists of known quantities; and by extracting the square root of both sides, a simple equation is obtained, from which the value of the unknown quantity may be found.
p« Ex. I. Let .r*+p«r = g; now we know that x'+px + — is
the square of x+- (Art. 152); add therefore— to both sides, and
^ P3 P*
ar+ px + — = q + —;
then by extracting the square root of both sides,
7) / D*
and by transposition, x = ± y q-\— .
In the same manner, if xi—px = q,
Ex.2. xt-lSx + 35 = 0', to find x.
12 By transposition x*-\2.x = -35; and adding the square of — ,
or 6, to both sides of the equation,
x%- 12a? + 36 = S6 - 35 m 1;
then extracting the square root of both sides,
.«. j? = 6 ±1 = 7, or 5; either of which, substituted for x in the original equation, answers the condition, that is, makes the whole equal to nothing.