RULE 1. When the unknown quantity is only connected with known quantities by the signs plus or minus. 195. 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. By transposition, 3x-2x=5+4,.. x=9. Ex. 3. Given x+a=a+5, to find the value of x. By taking a from both sides, we have x=5; or by transposition, x=a—a+5; but a—a=0, .'. x=5. Ex. Given 9-x=2, to find the value of x. By changing the signs of all the terms, we have −9+x=-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. (186. Cor. 1), by changing the signs of all the terms, we have b+x=c-a; .. by transposition, x=c-b-a. Ex. 6. Given 2x-4+7=3x-2, to find the value of x. (186.) by transposition, 2x-3=4—7—2, and (186. 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. x=4. Ex. 9. Given x-3+4-6=0, to find the value Ans. x=5. of x. Ex. 10. Given 7+x=2x+12, to find the value Ans. x=-5. of x. 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 x in terms of a and b. Ans. x=b-a. RULE II. 196. 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 of the equation by 3, the coefficient of x, we have 3x 9 x=3. 3 3' Ex. 2. Given 2x-3=9-x, to find the value of x. By transposition, 2x+x=9+3, by collecting the terms, 3x=12, 3x 12 by division, = x=4. Ex. 3. Given 7-4x-3x-7, to find the value of x. By transposition, -4x-3x=-7-7, by collecting the terms, -7x=-14, by changing the signs, 7x=14, 7x 14 by division, -= ; · x=2. Ex. 4. Given 6x+10=3x+22, to find the value of x. By transposition, 6x-3x=22-10, by collecting the terms, 3x=12, Ex. 5. Given ax+b=c, to find the value of x in 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 b; thus, if c=9, b=5, a=2, 2 2 then 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 Ans. x=3. of x. Ex. 8. Given ax2+bx=9x2+cx, to find the value of x in terms of a, b, &c. " c-b Ex. 9. Given x-9=4x, to find the value of x. Ans. x=-3. Ex. 10. Given 5ax-c-b-3ax, to find the value of x in terms of a, b, and c. b+c Ans. x= 8a Ex. 11. Given 3x-1+9-5x=0, to find the value of x. Ans. x=2. Ex. 12. Given ax=ab-ac, to find the value of x. Ans. x=b-c. Ex. 13. Given x2+2x=(x+a)2, to find the vaа2 lue of x. Ans. x= 2-2α Ex. 14. Given (x-1)2=x+1, to find the value of x. Ans. x=3. Ex. 15. Given x3+2x2+x=(x2+3x) × (x-1) +16, to find the value of x. RULE III. Ans. x=4. 197. 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, 6x 4' by 3, 3x-2x+18=30 by 4, 12x-3x+72-120-6x, by transposing, and collecting, 10x=48, by division, 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+36-60-3x, by transposition, 5x=24, and 6; -- 1 56, to find the value Here 30 is the least common multiple of 3, 5, 30x 30x 30x 5 6 Multiplying by 30, 30x- 30=- + .. 30x-10x-30-6x+5x, by transposition, 9x=30, Here 20, the product of 4 and 5, being their least common multiple, |