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P be put =0, we shall have the least value of y==4, and the corresponding, or greatest value of x=92.
And the rest of the answers will be found by adding 21 continually to the least value of y, and subtracting 17
, from the greatest value of x; which being done we shall obtain the six following results :
109 These being all the solutions, in whole numbers, that the question admits of.
Note. 1 When there are three or more unknown quantities, and only one equation by which they can be determined, as
ax+by+cz=d, it will be proper first to find the limit of the quantity that has the greatest coefficient, and then to ascertain the different values of the rest, by separate substitutions of the several values of the former, from 1 up to that extent, as in the following question.
5. Given 3x+5y+7z=100, to find all the different values of x, y, and z, in whole numbers. (0)
Here each of the least integer values of x and y are 1, by the question ; whence it follows, that
(6) If any indeterminate equation, of the kind above given, bas one or more of its coefficients, as c, negative, the equation may be put under the form
ax+by=d+cz, in which case it is evident that an indefinite number of values may be given to the second side of the equation, by means of the
quantity zi and consequently, also, to x and y in the first. And if the coefficients a, b, c, in any such equation, have a common divisor, while d has not, the question, as in the first case, be. comes impossible.
100_5-3 100-8 92
7 Consequently z cannot be greater than 13, which is also the lirit of the number of answers ; though they inay be considerably less.
By proceeding, therefore, as in the former rule, we shall have
3 And, by rejecting 33-y-22, 1-2y—2
3y 1 - 2y—2
And y=3p+:-1; And consequently, putting p=0, we shall have the least value of y=2-!; where z may be any number, from 1 up to 13, that will answer the conditions of the question. When, therefore, 22 we have y=1,
3 Hence, by taking z=2, 3, 4, 5, &c. the corresponding values of x and y, together with those of z, will be found to be as below.
3 4 5 6 in 8
2 3 4 5 6 7
=27 23 19 15 11 3 Which are all the integral values of x, y, and
that can be obtained from the given equation.
Note 2 If there be three unkıown quantities, and only two equations for determining them, as
ax+by+cz=d, and ex+fy+gz=h. exterminate one of these quantities in the usual way, and find the values of the other two from the resulting equation, as before.
Then, if the values, thus found, be separately substituted, in either of the given equations, the corresponding values of the remaining quantities will likewise be determined :
: thus, 6. Let there be given x-2y+z=5, and 2x+y-2=7, to find the values of x, y, and z
Here, by multiplying the first of these equations by 2, and subtracting the second from the product, we shall have
And consequently , or =$=w.=p.
2y 3y ly
Whence y =3p. And, by taking p=1, 2, 3, 4, &c. we shall have y=3, 6, 9, 12, 15, &c. and 2=6, 11, 16, 21, 26, &c. But from the first of the two given equations
x=5+2y-2; whence, by substituting the above values for y and, 2, the results will give x=5, 6, 7, 8, 9, &c
And therefore the first six values of x, y, and 2, are as below :
x=5 6 718 9 10
6 9 12 15 18
2=6 11 16 21 26 31 Where the law by which they can be continued is sufficiently obvious.
EXAMPLES FOR PRACTICE.
1. Given 3x=8y-16, to find the least values of x and y in whole numbers.
Ans. 48, y=5 P
2. Given 14x=5y+7, to find the least values of x and y in whole numbers.
Ans. x=3, y=7 3. Given 27x=1600- 16y, to find the least values of x and y in whole numbers.
Ans. 2=48, y=19 4. It is required to divide 100 into two such parts, that one of them may be divisible by 7,, and the other
Ans. The only parts are 56 and 44 5. Given 9x+ 13y=2000, to find the greatest value of at and the least value of y in whole numbers.
Ans. x=215, y=15
, = 6. Given 11:+5y=254, to find all the possible values of x and y in whole numbers.
Ans. =19, 14, 9, 4; y=9, 20, 31, 42 7. Given 17x+19y+212=400, to find all the answers in whole numbers which the question admits of.
Aps. 10 different answers 8. Given 5x+7y+112=224, to find all the possible values of x, y, and 2, in whole positive numbers.
Ans. The number of answers is 59 9. It is required to find in how many different ways it is possible to pay 201. in half-guineas and half-crowns, without using any other sort of coin?
Áos. 7 different ways 10. I owe my friend a shilling, and bave nothing about me but guineas, and he has nothing but louisd'ors ; how must I contrive to acquit myself of the debt, the louis being valued at 178. a piece, and the guineas at 218. ? Ans. I must give him 13 guineas, and he must
give me 16 louis
" 11. How many gallons of British spirits, at 12s., 158., and 18s. a gallon, must a rectifier of compounds take to make a mixture of 1000 gallons, that shall be worth 17s. a gallop ?
Ans. 1114, at 12s., 111} at 158., and 777 at 185,
To find such a whole number, as, being divided by other giren numbers, shall leave given remainders.
1. Call the number that is to be determined x, the numbers by which it is to be divided a, b, c, &c. and the given remainders f, g, h, &c.
2. Subtract each of the remainders from X, and divide the differences by a, b, c, &c. and there will arise x-f x-g2- -h
&c. = whole numbers. b 3. Put the first of these fractions 1-f=p, and substi
, tute the value of a, as found in terms of
from this equation, in the place of x in the second fraction.
4. Find the least value of p in this second fraction, by the last problem, which put =r, and substitute the value of x, as found in terms of r, in the place of a in the third fraction. .
Find, in like manner, the least value of r, in this third fraction, which put =s, and substitute the value of x, as found in terms of s, in the fourth fraction, as before.
Proceed in the same way with the next following fraction, and so on, to the last ; when the value of x, thus determined, will give the whole number required.
1. It is required to find the least whole number, which, being divided by 17, shall leave a remainder of 7, and when divided by 26, shall leave a remainder of 13.
Let z= the number required.