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1. By making the two first terms equal :

Mult. the 1st equ. by 2, gives 10.7 by = 18;
And mult. the 2d by 5, gives 10.6 +25y = 80;
Subtr. the upper from the under, gives 31y = 62;
And dividing by 31, gives

9 + 3y
Hence, from the 1st given equ. X =
2. By making the two 2d terms equal:

Mult. the 1st equat. by 5, gives 25.x - 15y = 45;
And mult. the 2d by 3, gives 6.X + 15y = 48;
Adding these two, gives

31.1. = 93;
And dividing by 31, gives

3.

5.7 9 Hence, from the 1st equ. y =

2.

5

MISCELLANEOUS EXAMPLES.

* +8 1. Given =

5.1 23 4

3 find .x and y.

Ans. x = 4, and y = 3. 3х

^x 2. Given + 10 = 13, and

= 4

2 to find x and yo

Ans. * = 5, and y = 3.

6.7 3. Given

+ = 10, and 5

3

6 to find x and y.

Ans. * = 8, and y 4. Given 3x + 4y = 38, and 4.x - 3y = 9; to find x and y.

Ans. x = 6, and 5.

2y

+

b=14

14;

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PROBLEM II.

To Exterminate Three or More Unknown Quantities; Or, to

Reduce the Simple Equations, containing them, to a Single

one.

RULE.

This may be done by any of the three methods in the last problem: viz.

1. AFTER the manner of the first rule in the last problem, find the value of one of the unknown letters in each of the given equations : next put two of these values equal to each other, and then one of these and a third value equal, and so on for all the values of it ; which gives a new set of equations,

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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, as before.

EXAMPLES

9

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2y

3%,

16 - 2y

4%;

x + y + z= 1. Given * + 2y + 32

=16

; to find x, y, and e. * + 3y + 4z = 21 ) 1. By the 1st method : Transp. the terms containing y and zin each equa. gives

9

y
* 16
X = 21

3y - 42; Then putting the 1st and 2d values equal, and the 2d and 3d values equal, give 9 % = 16

2y 3z,

32 21 3y
In the Ist trans. 9, %, and 2y, gives y = 7 - 22;
In the 2d trans. 16, 3z, and 3y, gives y = 5 - Z;
Putting these two equal, gives 5-23 7-22;

Trans. 5 and 2z, gives z = 2.
Hence y = 5-%= 3, and x = 9-y-z = 4.
2dly. By the 2d method:

From the 1st equa. x = 9-y-z;
This value of x substit. in the 2d and 3d, gives

9 + y + 2z = 16,

9 + 2y + 32 = 21;
In the 1st trans. 9 and 2z, gives y
This substit. in the last, gives 23 z = 21;
Trans. % and 21, gives

2 = 2.
Hence again y = 7--23 = 3, and x = 9-y-z = 4.

3dly. By

2z;

38

3dly. By the 3d method : subtracting the 1st equ. from the 2d, and the 2d from the 3d, gives

y + 2% = 7,

y + z = 5;
Subtr. the latter from the former, gives z = 2.
Hence y 5—% = : 3, and x = - 9-y-2 = 4.

2 + y + z = 18 2. Given x + 3y + 2z s to find x, y, and z. x + ty + z = 10 Ans. x = 4, y

6,= 8. x + y + z 3. Given

x + 3y + z = 20°; to find x, y, and z. x + 4y + šz

Ans. x = 1, y = 20, z = 60. 4. Given x - y = 2, x - x 3, and y - x =l; to find x, y, and z.

Ans. r = 7; y= 5; z = 4. $2,5

2.x + 3y + 4z = 34 5. Given 3.x + 4y + 5z = 46 ; to find x, y, and z.

4x + 5y + 6z = 58

= 272

= 16

A COLLECTION OF QUESTIONS PRODUCING SIMPLE

EQUATIONS.

X

Quest. 1. To find two numbers, such, that their sum shall be 10, and their difference 6. Let x denote the greater number, and y the less *.

Then, by the 1st condition x + y = 10,
And by the 2d

y = 6,
Transp. y in cach, gives

X = 10
and x =

6 + y;
Put these two values equal, gives 6 +y = 10 - Y;
Transpos. 6 and -- y, gives 2y = 4;
Dividing by 2, gives

y = 2.
And hence

X = 6 +y = 8.

Y,

* 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.

QUEST. 2

QUEST. 2. Divide 100). among A, B, C, so that A may have 201. more than B, and B 101. niore than c.

Let x = A's share, y = B's, and z = c's.
Then x +y + z = 100,

2 Iy + 20,

y=% + 10. In the 1st substit. y + 20 for x, gives 2y + x + 20 = 100; In this substituting % + 10 for y, gives 3z + 40 = 100; By transposing 40, gives

37 = 60; And dividing by 3, gives

z = 20. Hence y = x + 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::8:y, or mult. the extremes and the means, 7y

and * 4- 5 = 500;
Transposing y, gives x = 500 - Y;
This substituted in the 1st, gives 7y = 4000 sy;
By transposing 8y, it is 15y = 4000;
By dividing by 15, it gives y = 266};

And hence x = 500- y = 233;. QUEST. 4. What number is that whose 4th part exceeds its 5th part by 10?

Let x denote the number sought.
Then by the question fr - x = 10;
By mult. by 4, it becomes x - x = 40;

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 ; but if 1 be added to the denominator, its value will be ?

8x,

Let

3

denote the fraction. y

X + 1 Then by the quest.

=, and y

y + 1 The 1st mult. by 2 and y, gives 2x + 2 = y; The 2d mult. by 3 and y + 1, is 3x = y +l; The upper taken from the under leaves x

- 2 =l; By transpos. 2, it gives .r = 3. And hence y = 2.x + 2 = 8; and the fraction is

QUEST. 6.

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Quest. 6. A labourer engaged to serve for 30 days on these conditions: that for every day he worked, he was to receive 20d. but for every day he played, or was absent, he was to forfeit 10d. Now at the end of the time he had to receive just 20 shillings, or 240 pence. It is required to find how many days he worked, and how many he was idle ?

Let x be the days worked, and y the days idled.
Then 20x is the pence earned, and 10y the forfeits ;
Hence, by the question x + y = 30,

and 20.x Toy = 240;
The 1st, mult. by 10, gives 10.r + 10y = 300;
These two added give

30.r = 540; This div. by 30, gives

= 18, the days worked; Hence

y = 30– X = 12, the days idled. Quest. 7. Out of a cask of wine, which had leaked away, 30 gallons were drawn; and then, being gaged, it appeared to be half full; how much did it hold?

Let it be supposed to have held x gallons,
Then it would have leaked 2x gallons,
Conseq. there had been taken away 1 + 30 gallons.
Hence x = x + 30 by the question.
Then mult. by 4, gives 2.x = x + 120;

And transposing *, gives x = 120 the contents. 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.
Then by the question

x + y = 20,

and 3x + 5y = 76. Mult, the 1st by 3, gives

3x + 3y

= 60; Subtr. the latter from the former, gives 24 = 16; And dividing by 2, gives

8. Hence, from the 1st,

x = 20 - y = 12.,

Y =

QUEST. 9. A market woman bought in a certain number of eggs at 2 a penny, and as many more at 3 a penny, and sold them all out again at the rate of 5 for two-pence, and by so doing, contrary to expectation, found she lost 3d.; what number of eggs had she?

Let x = number of eggs of each sort.
Then will zx = cost of the first sort,
And fx = cost of the second sort ;

But

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