RULE VI.

:. WHEN there is any analogy or proportion, it is to be changed into an equation, by multiplying the two extreme terms together, and the two means together, and making the one product equal to the other.

Thus, if 2x: 9 :: 3 : 5.

Then, mult. the extremes and means, gives 10x = 27; And dividing by 10, gives a

And if x a :: 5b: 2c.

20.

Then mult. extremes and means gives cx=5ab;

And multiplying by 2, gives 3cx = 10ab;

Lastly, dividing by 3c, gives a =

Also, if 10-x: zx:: 3 : 1.

10ab

3c

Then mult. extremes and means, gives 10-x = 2x;
And transposing x, gives 10 = 3x ;
Lastly, dividing by 3, gives 33 = x.

RULE VII.

WHEN the same quantity is found on both sides of an equation, with the same sign, either plus or minus, it may be left out of both: and when every term in an equation is either multiplied or divided by the same quantity, it may be struck out of them all.

Thus, if 3x + 2a = 2a + b:

Then, by taking away 2a, it is 3x = b.

And, dividing by 3, it is r = b.

Also if there be 4ax + 6ab7ac.

Then striking out or dividing by a, gives 4x + Gb = 7c. Then, by transposing 6b, it becomes 4r7c-6b;

And then dividing by 4 gives x=2c-3b.

Again, if 3x-3= }}.

-7.

Then, taking away the 4, it becomes r

And taking away the 3's, it is 2x = 10;

Lastly, dividing by 2 gives x = 5,

MISCELLANEOUS EXAMPLES,

1. Given 72-18=4x+ 6; to find the value of x.
First, transposing 18 and 5x gives 3x = 24;
Then dividing by 3, gives x 8.

2. Gives

2. Given 20-4x-12=92-10x; to find ✰.

First transposing 20 and 12 and 10x, gives 6x ≈ 84;
Then dividing by 6, gives x = 14.

3. Let 4ax-5b3dx + 2c be given; to find r. First, by trans. 56 and 3dx, it is 4ax-3dx = 5b + 2c;

Then dividing by 4a-3d, gives x =

5b+2c

4a-3d

4. Let 5x2-12r = 9x + 2x2 be given; to find x.
First, by dividing by x, it is 5x − 12 = 9 + 2x ;
Then transposing 12 and 2x, gives 3x = 21;
Lastly, dividing by 3, gives x 7.

5. Given 9ax3- 15abx2 = 6ax3 + 12ax2; to find x. First, dividing by 3ax2, gives 3x-5b = 2x + 4; Then transposing 56 and 2x, gives x 5b + 4.

I 2

उ

First, multiplying by 3, gives .r-3x + 3x = 6;
Then multiplying by 4, gives x + 12 = 24.
Also multiplying by 5, gives 17x = 120;
Lastly, dividing by 17, gives r777.

First, mult. by 3, gives x−5 + 3x = 36−x + 10;
Then transposing 5 and r, gives 2x + 3x = 51;
And multiplying by 2, gives 7x = 102;

Lastly, dividing by 7, gives x=144.

3x 4

XV

8. Let √+7= 10, be given; to find x.

First, transposing 7, gives r = 3;

Then squaring the equation, gives r9;
Then dividing by 3, gives
= 3;.

Lastly, multiplying by 4, gives x = 12.

9. Let 2x + 2√ a2 + x2 =

5a2

√ a2 + x22

be given; to find x.

First, mult. by a2+x2, gives 2x √ a2 + x2 + 2a2 + 2x2

=5a2.

Then transp. 2a2 and 2x2, gives 2x √ a2 + x2 = 3a2 −2.x2; R. VOL. I.

Then

Then by squaring, it is 4a2x a+ x2=3a2 - 2x22;
That is, 4a2x2 + 4x2 = 9aa — 12a2x2 + 4x2;

By taking 4.24 from both sides, it is 4a2x2-9a-12a2x2;
Then transposing 12a2x2, gives 16a2x2 = 4aa;
Dividing by a2, gives 16x24a2;

And dividing by 16, gives x2+a2;
Lastly extracting the root, gives x = a.

EXAMPLES FOR PRACTICE.

1. Given 2x-5 + 16 = 21; to find x.

Ans. x = 5.

Ans. x = 44.

2. Given 9x-15=x+6; to find x.
3. Given 8-3x+12=30-5x+4; to find x. Ans.x=7:
4. Given x+x-13; to find x.
5. Given 3x+x+2=5x-4; to find x.
6. Given 4ax+ ža−2 = ax − bx; to find

Ans. x 12.
Ans. x = 4.

x.

13

7. Given -4x + 3x = 4; to find r.

8. Given √4+x = 4-√x; to find x.

Ans. x=/17/20

Ans. x = 24,

·; to deter. x. Ans. x=-2a.

10. Given 4a2 + x2 = 1/4b + x^; to find x.

13. Given a + x = √ a2 + x √ 4b2 + x2; to find x.

OF REDUCING DOUBLE, TRIPLE, &c. EQUATIONS, CONTAINING TWO, THREE, OR MORE UNKNOWN QUANTITIES.

PROBLEM I.

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

one.

RULE L.

FIND the value of one of the unknown letters, in terms of the other quantities, in each of the equations, by the methods already explained. Then put those two values equal to each other for a new equation, with only one unknown quantity in it, whose value is to be found as before.

Note. It is evident that we must first begin to find the values of that letter which are easiest to be found in the two proposed equations.

17-3y.

In the 1st equat. transp. 3y and div. by 2, gives x

2

14+2y.

In the 2d transp. 2y and div. by 5, gives x =

5

14+2y_17-3y

Putting these two values equal, gives -17-34

5

=

2

Then mult. by 5 and 2, gives 28 + 4y = 85-15y;
Transposing 28 and 15y, gives 19y= 57;
And dividing by 19, gives y = 3.

And hence x = 4.

Or, to do the same by finding two values of y, thus:

In the 1st equat. tr. 2x and div. by 3, gives y

17-2x

3

5x – 14

In the 2d tr. 2y and 14, and div. by 2, gives y = —..

Mult. by 2 and by 3, gives 15x-42. — 34 −4x;

Q2

3

;

Transp.

Transp. 42 and 4x, gives 19x=76;
Dividing by 19, gives x = 4.

Hence y 3, as before.
y=

2. Given

zx+2y=a2 2 zx - 2y = bS

:

; to find x and y.

Áns. x = a + b, and y = 4a-4b.

3. Given 3x + y = 22, and 3y+x= 18; to find x and y.

s2 + ď2

Ans. *=

and y

2$

5. Given + = and +

and y.

6. Given x+2y=s, and 2 - 4y2 = d2;

شر

7. Given x-2y=d, and x: y::a: b; to find x and

45

y.

RULE II.

FIND the value of one of the unknown letters, in only one of the equations, as in the former rule; and substitute this value instead of that unknown quantity in the other equation, and there will arise a new equation, with only one unknown quantity, whose value is to be found as before.

Note. It is evident that it is best to begin first with that letter whose value is easiest found in the given equations.

1. Given This will admit of four ways of solution; thus: First,

in the 1st eq. trans. 3y and div. by 2,

This val. subs. for x in the 2d, gives
Mult. by 2, this becomes 85-15y-4y = 28 ;