3x-9+6=3x1−9+6=3−9+6=9—9—0. Hence, the value of the unknown quantity being substituted in the equation, will always reduce it to 0=0. § II. RESOLUTION OF SIMPLE EQUATIONS, Involving only one unknown Quantity. 168. The resolution of simple equations is the disengaging of the unknown quantity, in all such expressions, from the other quantities with which it is connected; and making it stand alone, on one side of the equation, so as to be equal to such as are known on the other side, or, which is the same thing, the value of the unknown quantity cannot be ascertained till we transform the given equation, by the addition, subtraction, multiplication, or divison of equal quantities, so that we may finally arrive at the conclusion, x=n, n being a number, or a formula, which indicates the operations to be performed upon known numbers. This number n being substituted for a in the primitive equation, has the property of rendering the first member equal to the second. And this value of the unknown quantity, as has been already observed, is called the root of the equation, this word has not here the same acceptation as in (Art. 15). 169. In the resolution of simple equations, involving only one unknown quantity, the following rules which are dedueed from the Articles in the preceding Section, are to be observed. RULE I. When the unknown quantity is only connected with known quantities by the signs plus or minus. 170. Transpose the known quantities to one side of the equation, so that the unknown may stand by itself on the other; and then the unknown quantity becomes known. Ex. 1. Given x+8=9, to find the value of x. By transposition, x=9-—8, ... x=1. Ex. 2. Given 3x-4=2x+5, to find the value of x. Ex. 3. Given x+a=a+-5,' to find the value of x. x=5; or by transposition, Ex. 4. Given 9—x=2, to find the value of x. By changing the signs of all the terms, we have -9x=-2, by transposition, x=9-2, ... x=7. It may be remarked, that it is the general practice of Analysts, to make the unknown quantity appear on the left-hand side of the equation, which is principally the reason for changing the signs. Ex. 5. Given—b-x-a-c to find x in terms of a, b, and c. (161. Cor. 1), by changing the signs of all the terms, we have b+x-c-a; .. by transposition, a=c-b-a. Ex. 6. Given 2x-4+7=3x-2, to find the value of x. (161.) by transposition, 2x-3x=4-7-2, and (161. Cor. 1), by changing the signs, 3x-2x=7+2-4; but 3x-2x= x, and 7+2—4—5; ... x=5. Ex. 7. Given 7x+3-5-6x-2+7, to find the value of x. Ans. x=7. Ex. 8. Given 3x+5—2—2x-7=0, to find the value of x. Ans. x4. Ex. 9. Given x-3+4-6-0, to find the value of x. Ans. x=5. Ex. 10. Given 7+x=2x+12, to find the value of x. Ans. x=-5. Ex. 11. Given 12-3x=9-2x, to find the value of x. Ans. x 3. Ex. 12. Given x-a+b-c=0, to find the value of x in terms of a, b, and c. Ans. x=a--b+c. Ex. 13. Given x-a+b=2x-2a+b, to find the value of x in terms of a and b. Ans. x=a. Ex. 14. Given 2x+a=x+b, to find a in terms of a and b. Ans. x-b-ɑ. RULE II. 171. Transpose the known quantities to one side of the equation, and the unknown to the other, as in the last Rule; then, if the unknown quantity has a coefficient, its value may be found by dividing each side of the equation by the coefficient, or by the sum of the coefficients. Ex. 1. Given 3x+9=18, to find the value of x. By transposition, 3x=18-9, or 3x=9; dividing both sides 3x 9 of the equation by 3, the coefficient of x, we have 3 =3. Ex. 2. Given 2x-3=9—x, to find the value of x. by collecting the terms, 3x=12, by division, 3x 12 3 3 Ex. 3. Given 7-4x=3x-7, to find the value of x. by division, 7 Ex. 4. Given 6x+10=3x+22, to find the value of x. by collecting the terms, 3x=12, Ex. 5. Given ax+b=c to find the value of x in terms of a, a The value of x is equal to c— b divided by a, which may be positive or negative, according as c is greater or less than 9-5 b; thus, if c=9, b=5, a=2, then x= =2; if c=12, b= Ex. 6. Given 3x-4-7x-16, to find the value of x. Ans. x=3. Ex. 7. Given 9-2x=3x-6, to find the value of x. Ex. 8. Given ar2+bx=9x2+cx, to find the value of x in terms of a, b, &c. Ans. x= Ans. x=-3. Ex. 9. Given x-9=4x, to find the value of x. Ex. 10. Given 5ax-c=b-3ax, to find the value of x in terms of a, b, and c. Ex. 11. Given 3x-1+9--5x=0, to find the value of x. Ans. x=4. Ex. 12. Given azab―ac, to find the value of x. Ans. x-b-c. Ex. 13. Given x2+2x=(x+a)2, to find the value of x. Ex. 14. Given (x-1)=x+1, to find the value of x. Ans. x=3. Ex. 15. Given x+2x2+x=(x2+3x) ×(x-1)+16, to find the value of x. Ans. x=4. RULE III. 172. If in the equation there be any irreducible fractions, in which the unknown quantity is concerned, multiply every term of the equation by the denominators of the fractions in succession, or by their least common multiple; and then proceed according to Rules I, and II. 2x Ex. 1. Given +1=x-9, to find the value of x. 4 Multiplying by 4, 2x+4=4x-36, by transposition, 2x-4x=-36—4, by collecting the terms, -2x=-40, by changing the signs, 2x=40, 6.x 4 Multiplying by 2, x-2 +6=10 by 3, 3x-2x+18=30 by 4, 12x-8r+72-120-6x, by transposing, and collecting, 10x=48, Or, it is more concise and simple to multiply the equation by the least common multiple of the denominators; because, then the equation is reduced to its lowest terms; thus, Multiplying by 12, the least common multiple of 2, 3, and 4, we have, 6x-4x+36=60-3r, Here 30 is the least common multiple of 3, 5, and 6 ; 30x 30x -30=-2 + 5 6 ; .. x=3}. 3 -3, to find the value of x. Here 20, the product of 4 and 5, being their least common by collecting the coefficients, (a-b)x=2a, Here 2ac, the product of 2, a, and c, being the least common multiple, Multiplying by 2ac, 4a2x+3abcx=10cx+6ac, by transposition, and collecting the coefficients, we shall have (4a2+3abc-10c)x=6ac, Multiplying by 12, the least common multiple, we have 36x-3x+12—48—20x+56—1, by transposition, 36x-3x-20r-56--1+48--12, |