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with which the same process is to be repeated, and so on till there is only one equation, to be reduced by the rules for a single equation.
2. Or, as in the 2d rule of the same problem, find the value of one of the unknown quantities in one of the equations only; then substitute this value instead of it in the other equations ; which gives a new set of equations to be resolved as before, by repeating the operation.
3. Or, as in the 3d rule, reduce the equations, by multiplying or dividing them, so as to make some of the terms to agree : then, by adding or subtracting them, as the signs may require, one of the letters may be exterminated, &c. before.
x + y + z = 9 1. Given x + 2y + 3 = 16 ; to find x, y, and z,
x + 3y + 4z = 31 1. By the 1st method : Transp. the terms containing y and z in each equa. gives
X = 9
y x = 16
2y 3z, * = 21
Зу Then putting the 1st and 2d values equal, and the 2d and 3d values equal, give 9 y Z= 16
3z = 21 · 3y – 42;
y - Z=4
2dly. By the 2d method :
From the 1st equa. x = 9 - Y -
9 + y + 2z= 16,
9 + 2y + 3z
Trans, z and 21, gives 2 – Z.
y + 2z
y = 5
3dly. By the 3d method ; subtracting the 1st equ. from the 2d, and the ad from the 3d, gives
y + z = 5;
- I = 3, and x = 9 - y - %= 4,
x + y + z = 18
x + 3y + 2z 38 ; to find x, y, and z.
Ans. 2 * 4, y = 6, z = 8.
x + y + zz 20 ; to find x, y, and z.
Ans. x = 1, y = 20, z = 60. 4.
y = 2, X — z = 3, and y Z-l; to find x, y, and z.
Ans. x = 7;y
5 ; 2= 4,
1.7 Given x
2x + 3y + 4z = 34
; to find x, y, and z.
A COLLECTION OF QUESTIONS PRODUCING SIMPLE
Quest. 1. To find two numbers, such, that their sum shall
Then, by the ist condi:ion x + y 10,
Y, and x =
6 + y;
y, gives 2y = 4;
6+ y = 8,
Y = 2.
In all these solutions, as many unknown letters are always used as there are unknown numbers to be found, purposely the better to exercise the modes of reducing the equations : avoiding the short ways of notation, which though giving a shorter solution, are for that reason less useful to the pupil, as affording less exercise in practising the several rules in reducing equations.
z + 10
QUEST. 2. Divide 1001. among A, B, C, so that A may have 201. more than B, and B 101, more than c.
Let x = A's share, y = B's, and z = c's.
x = y + 20,
y = In the 1st substit. y + 20 for x, gives 2y + 2 + 20 = 100; In this substituting 2 + 10 for y, gives 3z + 40
100; By transposing 40, gives
32 And dividing by 3. gives
20. Hence y = 2 + 10 = 30, and x = y + 20 = 50.
Quest. 3. A prize of 5001, is to be divided between two persons, so as their shares may be in proportion as 7 to 8; required the share of each. Put x and y for the two shares; then by the question,
7:8:: x : y, or mult. the extremes and the means, 7y 8.x,
and x + y = 500;
233;. QUEST. 4. What number is that whose 4th part exceeds its 5th part by 10?
Let x denote the number sought.
By mult. by 5 it gives X = 200, the number sought. Quest. 5. What fraction is that to the numerator of which if i be added, the value will be h; but if one be added to the denominator, its value will be ?
y + 1
Let denoté the fraction.
QUEST. 6. A labourer engaged to serve for 30 days on
Let x be the days worked, and y the days idle.
x + y
and 20x 10y 240 ;
= 12, the days idled.
y = 30
QUEST. 7. Out of a cask of wine, which had leaked away 4,
Let it be supposed to have held x gallons,
QUEST. 8. To divide 20 into two such parts, that 3 times the one part added to 5 times the other may make 76.
Let x and y denote the two parts.
2 + y 20,
and 3x + 5y Mult. the 1st by 3, gives
3x + 3y Subtr. the latter from the former, gives 24
16; And dividing by 2, gives
y : Hence, from the 1st,
x = 20
= 8. y = 12.
Quest 9. A market woman bought in a certain number
Let x = number of eggs of each sort.
But 5 : 2 :: 2x (the whole number of eggs): **;
price of both soris. ai 5 for 2 pence ;
Quest. 10. Two persons, A and B, engage at play. Before they begin, A has 8 guineas, and B has 60. Afie, a certain number of games won and lost between them, a rises with three times as many guineas as B. Query, how many guineas did a win of B ?
Let x denote the number of guineas a won.
QUESTIONS FOR PRACTICE.
1 To determine two numbers such, that their difference may be 4, and the difference of their squares 64.
Ans. 6 and 10.
2. To find two numbers with these conditions, viz. that half the first with a 3d part of the second may make 9, and that a 4th part of the first with a 5th part of the second may make 5.
Ans. 8 and 15. 3. To divide the number 20 into two such parts, that a 3d of the one part added to a fifth of the other, may make 6.
Ans. 15 and 5.
4. To find three numbers such, that the sum of the Ist and 2d shall be 7, the sum of the 1st and 3d 8, and the sum of the 2d and 3d 9.
Ans. 3, 4, 5.
5. A father, dying, bequeathed his fortune, which was 28001. to his son and daughter, in this manner; that for every half crown the son might have, the daughter was to have a shilling. What then were their twn shares ?
Ans. The son 20001. and the daughter 8001. 6. Three persons, A, B, C, make a joint contribution, which in the whole amounts to 4001. : of which sum B. con