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This shows that x was used in all cases with the wrong sign, therefore to insert m in place of x we must change the sign in each term where x is found.

Take the quantity first without x, thus,

First insert

Now insert

If

-

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m be inserted by the rules found above, the same result will be produced.

When a negative value has been found for the unknown quantity, we have observed it shows that there was some inconsistency in the question. If then the unknown quantity be put again into the same equation, with the contrary sign, as we introduced-m above, that is, if the unknown quantity be taken with the negative sign, and introduced by the above rules into all the terms where it was found before, a new equation will be produced, differing from the former only in some of the signs. Then if the conditions of the question be altered so as to correspond with the new equation, it will be consistent, and a positive value will be obtained for the unknown quantity. The new value of the unknown quantity however will be the same as the former, with the exception of the sign. Therefore, when once we are accustomed to interpret this kind of results, it will be unnecessary to go through the calculation a second time.

The following examples are intended to exercise the learner in interpreting these results.

1. A father is 55 years old, and his son is 16. In how many years will the son be one fourth as old as the father?

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Here x has a negative value, consequently it entered into the equation with the wrong sign. Putting now - instead of x into the equation, it becomes

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This shows that something must be subtracted from the present age; that is, the son was a fourth part as old as the father some years before.

This equation gives

x = 3.

Therefore he was one fourth part as old 3 years before, when the father was 52, and the son 13.

2. A man when he was married was 45 years old, and his wife 20. How many years before, was he twice as old as she?

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There is a wrong supposition in this question. Putting — x into the equation it becomes

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This shows that she was not half as old as he when they were married, but that it was to happen 5 years afterward, when the man was 50, and the wife 25.

3. A labourer wrought for a man 15 days, and had his wife and son with him the first 9 days, and received $14.25. He afterwards wrought 12 days, having his wife and son with him 5 days, and received $13.50. How much did he receive per day himself, and how much for his wife and son?

4. A labourer wrought for a man 11 days, and had his wife with him 4 days, and received $17.82. He afterwards wrought 23 days, having his wife with him 13 days, and received $38.78. How much did he receive per day for himself, and how much for his wife?

5. A labourer wrought for a gentleman 7 days, having his wife with him 4 days, and his son 3 days, and received $7.89. At another time he wrought 10 days, having his wife with him 7 days, and his son 5 days, and received $11.65. At a third time he wrought 8 days, having his wife with him 5 days, and his son 8 days, and received $7.54. How much did he receive per day himself, and how much for his wife and son severally?

6. What number is that, whose fourth part éxceeds its third part by 16?

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The question as it was proposed involves some contradiction. Putting in -x it becomes

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This shows that the question should have been as follows; What number is that, whose third part exceeds its fourth part by 16?

7. What number is that, of which exceeds of it by 18?

S. What fraction is that, to the numerator of which if 1 be added, its value will be 3, but if 1 be added to its denominator, its value will be ?

9. What fraction is that, from the denominator of which, if 2 be subtracted, its value will be 1, but if 2 be subtracted from its numerator, its value will be ?

10. It is required to divide the number 20 into two such parts, that if the larger be multiplied by 3, and the smaller by 5, the sum of the products will be 125.

11. It is required to find two numbers, whose difference is 25, and such that if the larger be multiplied by 7, and the smal er by 5, the difference of their products shall be 215?

XXV. Explanation of Negative Exponents.

It was observed above, that when the dividend and the divisor were different powers of the same letter, division is performed by subtracting the exponent of the divisor from that of the dividend: thus

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That is, any quantity having zero for its exponent, is equal

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as

Hence it appears that a1 has the same value as

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and a

a

The quantities a3, a2, a', ao, a ̄1, a2, a3, &c. have the same value as a3, a, a', 1,

*

1 1

&c.

Exponents may be used for compound quantities as well as for simple; and multiplication and division may be performed on those which are similar, by adding and subtracting the exponents.

On this principle the denominator of a fraction, or any factor of the denominator may be written in the numerator by giving its exponent the sign- This mode of notation is often very convenient; I shall therefore give a few examples of its application.

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2 a b c2

by

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3 -2

2 a b1 c2 × b3 c=2 ab−1+3 c―2+' = 2 a b2 c—1 =

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2ab2

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2

By the above method 3 a c3÷c2 = 3 a c→→

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Or thus, to divide 3 a c3 by c', is the same as to multiply it

1

by

or

-2

or c2, which gives the same result.

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