Equations of the First Degree. SECTION III. Solution of Equations of the First Degree, with one unknown quantity. 120. Theorem. Every equation of the first degree, with one unknown quantity, can be reduced to the form Ax + B 0; in which A and B denote any known quantities, whether positive or negative, and x is the unknown quantity. Proof. When an equation of the first degree with one unknown quantity is reduced, as in art. 118, its first member is composed of two classes of terms, one of which contains the unknown quantity, and the other does not. If the unknown quantity, which we may suppose to be x, is taken out as a factor from the terms in which it is contained, and its multiplier represented by A, the aggregate of the first class of terms is represented by Ax; and the aggregate of the terms of the second class may be represented by B; whence the equation is represented by Ax + B = 0. 121. Problem. To solve an equation of the first degree with one unknown quantity. Solution. Having reduced the given equation to the form Ax + B = 0, transpose B to the second member by art. 117, and we have Dividing both members of this equation by A, gives x B Cases in Equations in the First Degree. Hence, to solve an equation of the first degree, reduce it, as in art. 120, and transpose its known terms to the second member, and all its unknown terms to the first member; and the value of the unknown quantity is equal to the quotient arising from the division of the second member by the multiplier of the unknown quantity in the first member. 122. Corollary. When A and B are both positive or both negative, the value of x is, by art. 35, negative; but when A and B are unlike in their signs, one positive and the other negative, a is positive. 123. Corollary. When we have But the smaller a divisor is, the oftener must it be contained in the dividend, that is, the larger must the quotient be; and when the divisor is zero, it must be contained an infinite number of times in the dividend, or the quotient must be infinite. Infinity is represented by the sign o We have, then, in this case, X The given equation is, however, in this case, 0 x x + B = 0, Cases in Equations of the First Degree. which reduces itself to B = 0, an obvious absurdity, unless B is zero. The signo is, therefore, rather to be regarded as the expression of the peculiar species of absurdity which arises from diminishing the denominator of a fraction till it becomes zero. 125. Corollary. When we have which is equal to any quantity whatever, and is called an indeterminate expression. The given equation is, indeed, in this case 0xx+0= 0, an equation which is satisfied by any value whatever of x, and is called an identical equation. Equations of the First Degree with one unknown quantity. 7. Two capitalists calculate their fortunes, and it appears that one is twice as rich as the other, and that together they possess $ 33 700. What is the capital of each? Ans. The one has $ 12 900, the other $25 800. 8. To find two such numbers, that the one may be m times as great as the other, and that their sum = a. 9. The sum of $ 1200 is to be divided between two persons, A and B, so that A's share is to B's as 2 to 7. How much does each receive? Ans. A$ 266, B $9331. Equations of the First Degree with one unknown quantity. 10. To divide a number a into two such parts, that the first part is to the second as m to n. 11. How much money have I, when the 4th and 5th parts of it amount together to $2,25? Ans. $5. 12. Find a number such, that when it is divided successively by m and by n, the sum of the quotients = a. 13. Divide the number 46 into two parts, so that when the one is divided by 7, and the other by 3, the sum of the quotients = 10. Ans. 28 and 18. 14. All my journeyings taken together, says a traveller, amount to 3040 miles; of which I have travelled 3 times as much by water as on horseback, and 24 times as much on foot as by water. How many miles did he travel in each of these three ways? Ans. 240 miles on horseback, 840 miles by water, and 1960 miles on foot. 15. Divide the number a into three such parts, that the second may be m times, and the third n times as great as the first. Ans. 1+m+n'1+ m + n'1+m+n 16. A bankrupt leaves $21000 to be divided among four creditors A, B, C, D, in proportion to their claims. Now A's claim is to B's as 2: 3; B's claim : C's = 4 : 5 ; and C's claim: D's 6:7. How much does each creditor receive? Ans. A $3200, В $4800, C $ 6000, D $ 7000. |