in the LEFT hand member, we shall then have the first, or GIVEN equation.

Second. TRANSFER the numerator (5), from the left to the right of the FIRST EQUATION, by simply changing its SIGN, or making it a DENOMINATOR, and put a unit (1), in its 1 1.1

place; we shall then have 1=1.5' or which is the same as, for the SECOND equation.

Third. Multiply the SECOND equation (}=},) by 3, and we shall have, which is the third, or REQUIRED equation, the LEFT hand member of which contains the REQUIRED quantity (3), and the RIGHT hand member, contains ITs equivalent, or the answer.

TO READ a solution of this kind, it is necessary to REPEAT the IDENTICAL quantity; hence, to read the above equation, we say, if the ratio of FIVE (5), to five, equals one, then the ratio of ONE to five equals one fifth (}); hence, the ratio of THREE (3), to five equals THREE-FIFTHS, (3). Answer.

2. Required the INVERSE ratio of 3 to 5. Here the given and required quantities are the same as before, and hence the STATEMENT of the equation will be precisely SIMILAR; but the MODE of solving the equation is exactly the REVERSE, i. e., the left hand member of the GIVEN equation is reduced to unity by transferring the quantities WITHOUT changing their signs, and when so reduced, the LEFT hand member is multiplied by the required quantity; but the RIGHT hand member is DIVIDED by it, which is exactly the REVERSE of what has already been done; thus, by the conditions of the problem, we have f=1, for the given equation; now transfer 5 without changing its sign, and we have, for the second equation, and by placing 3 as a NUMERATOR in the left hand member, and also as a DENOMINATOR in the right hand member of the equation (=), we have =, which is the INVERSE ratio required. In solving the above example (2), the successive equations should be written under each other as before; thus,

=, given equation.

1. simple equation.

=5. required equation,

in which, for the sake of distinction, we call the first the GIVEN equation, because it contains the GIVEN quantity in the LEFT hand member; the second, we call the SIMPLE equation, because it has a UNIT in the LEFT hand member, and the third, we call the REQUIRED equation, because it contains the REQUIRED quantity in the LEFT hand member, and its equivalent in the RIGHT hand member of the equation which (equivalent) is the QUANTITY, or RATIO sought.

3. Required the direct ratio of to . Here the terms of the ratio are both fractions; the given quantity is, and the required quantity is ;

Here the given equation contains =, and by transferring the numerator (2), and denominator, (3), and changing their signs, we have =3, which multiplied by, as before, gives

3.1

=2.2=2, which is the ratio required.

4. Required the inverse ratio of to 7. Here is given equal to (one); hence, we have 3, or 17, by transferring 7 and 9 without changing their signs, and by placing as a multiplier in the left hand member, and as a divisor in the right hand member of the equation, we have for the re3 7.5 35

quired equation=9-3-27-17, for the ratio sought. This solution may be written thus:

Given equation.

= Simple equation.

5 7

35

Required equation, ==2;=1, Answer.

27

7

5. Required the direct ratio of 8×6 to 9×12. Here

9x12 is the GIVEN, and 8×6, the REQUIRED quantity, hence the

When the equation has two or more factors in the left hand member, they are all transferred, and the solution conducted according to the same rule as before. In example 5,

6. Required the direct ratio of 3x

and the REQUIRED quantity.

to x.

Here

Hence we

is the GIVEN,

7. What is the inverse ratio of 8x 6 to 9 × 12. Here (as in Example 5,) 9× 12 is the GIVEN quantity, and 8× 6 the REQUIRED quatity; hence we have for the

8. What is the inverse ratio of 3×3 to . Here × is the given quantity, and 2×2 the required quantity, and the ratio being inverse, we proceed thus:

8.6

1 8

6

Hence the inverse ratio of × to ×

equals 1 (one). 9. What is the RATIO of 6x8 to 4× 9, when 8 and 9 are

INVERSE FACTORS.

When any of the factors in the GIVEN quantity, and an EQUAL number in the REQUIRED quantity are inverse, the inverse factors should be dashed (thus 8' 9') to distinguish them from the factors which are DIRECT. Thus in the preceding problem, we have 4×9′ given, and 6× 8' required; where by the conditions, 4 and 6 are direct; but 9' and 8′ are inverse, and are managed in the equation as before directed.

We will now form the equation, for problem 9, and solve it as follows.

Hence, the ratio of 6 x 8 to 4 × 9, when 8 and 9, are INVERSE factors, is 1.

NOTE. In all cases where there are direct and inverse factors in the GIVEN quantity, there must be an equal number of each in the REQUIRED quantity, or the equation is not RATIONAL, and therefore, cannot be solved.

10. Required the ratio of x to X, when and are inverse. The equation is thus:

12. What is the DIRECT ratio ofofofofto of 4 of 7 of 8.*

Solution thus:

* Since the word of denotes multiplication, this problem is the same as the preceding problems.