Transp. 15y and 4y and 28, gives 57 19y;
And dividing by 19, gives 3 =y.

Then =

17-3y
2

= 4.

2dly, in the 2d trans. 2y and div. by 5, gives 14+2y,

5

28+4y

This subst. for x in the 1st, gives

Mult. by 5, gives 28 +4y+ 15y=85;
Transpos. 28, gives 19y= 57;

And dividing by 19, gives y = 3.

+3y=17;

5

14+2y

Then r

=4, as before.

5

17-2x

3dly, in the 1st trans. 2x and div. by 3, gives y=

52-14

4thly, in the 2d tr. 2y and 14 and div. by 2, gives y=

2

2. Given 2x + 3y 29, and 3x-2y= 11; to find x

and y.

x + y =
2

Ans. x=7, and y = 5.

3. Given {14}; ; to find x and y.

=

Ans. x 8, and y = 6.

4. Given

7. Given xy:: 4 : 3, and x3-y3 = 37;

Ans.

to find r and y. = 4, and y = 3.

1+ 4, and 2y+

x

لا

2=

= 6.

RULE III.

LET the given equations be so multiplied, or divided, &c, and by such numbers or quantities, as will make the terms which contain one of the unknown quantities the same in both equations; if they are not the same when first proposed.

Then by adding or subtracting the equations, according as the signs may require, there will remain a new equation, with only one unknown quantity, as before. That is, add the two equations when the signs are unlike, but subtract them when the signs are alike, to cancel that common

term.

Note. To make two unequal terms become equal, as above, multiply each term by the co-efficient of the other.

Here we may either make the two first terms, containing *, equal, or the two 2d terms, containing y, equal. To make the two first terms equal, we must multiply the 1st equation by 2, and the 2d by 5; but to make the two 2d terms equal, we must multiply the 1st equation by 5, and the 2d by 3; as follows.

1. By making the two first terms equal:

Subtr. the upper from the under, gives 31y 62;

Mult. the 1st equ. by 2, gives

And mult. the 2d by 5, gives

10x 6y 18;
10.r+25y80;
31y62;

And dividing by 31, gives

y=

9+ 3y

Hence, from the 1st given equ.

== 3.

5

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

Adding these two, gives

And dividing by 31, gives

Hence, from the 1st equ. y =

6x+15y48;

31x=93;

x= 3.

5.x - 9

=2.

+ 6y = 21, and 3+6 + 5x = 23; to

3x

y

4

Ans.

+10= 13, and

2

4, and ጋ 3.

+ 5 = 12

Ans. x = 5, and y = 3.

6.x

Ans.

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

To Exterminate Three or More Unknown Quantities; Or, to Reduce the Simple Equations, containing them, to a Single

one.

RULE.

THIS may problem: viz.

be done by any of the three methods in the last

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,

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.

1. Given

EXAMPLES.

x + y + z =

9

x + 2y + 3z = 16 ; to find x, y, and z. x + 3y + 4% = 21)

1. By the 1st method:

Transp. the terms containing y and z in each equa. gives

Then putting the 1st and 2d values equal, and the 2d and 3d values equal, give

In the 1st trans. 9, z, and 2y, gives y = 7
In the 2d trans. 16, 3z, and 3y, gives y':
= 5
Putting these two equal, gives 5-z=7—2z;
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

1

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 ≈ = 2.
Hence y =5-2= 3, and x = 9-y-z

2. Given

x + y + z = 18

24.

x + 3y + 2x = 38}; to find x, y, and z. x + y + 1/2 = 10

QUEST. 1. To find two numbers, such, that their sum shall be 10, and their difference 6.

Let r denote the greater number,
Then, by the 1st condition
And by the 2d
Transp. y in each, gives

Put these two values equal,
Transpos. 6 and y, gives
Dividing by 2, gives
And hence

and y the less *.
+ y = 10,

y

x = 10

and r =

6,

y,

6+ yi

* 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.