to subtract a from both sides, to obtain the equation x=b—a, which indicates the value of x. 494. If the equation which we have found is x-ab, we add a to both sides, and obtain the value of x=b+a. We proceed in the same manner, if the equation has this form, x-aaa+1; for we shall have immediately a a +a+1. In this equation, x 8 a 206 a, we find x = 20 — 6 a +8 a, or x = 20 + 2 a. And in this, x+6a= 20 + 3 a, we have x = 20 + 3 a 6 a, or x=20 3 a. 495. If the original equation has this form, x-a+b=c, we may begin by adding a to both sides, which will give x + b=c+a; and then subtracting b from both sides, we shall find x=c+a-b. But we might also add +ab at once to both sides; by this we obtain immediately x = c + a − b. So in the following examples, If x 2a+3b= 0, we have x2a-3 b. If x-sa+2 b = 25+ a +2 b, we have x 25+ 4 a. 4 a. If x 496. When the equation which we have found has the form a b a xb, we only divide the two sides by a, and we have x= -, But if the equation has the form ax+b-c=d, we must first make the terms that accompany ax vanish, by adding to both sides-b+c; and then dividing the new equation, a x = d— d-b+c b+c, by a, we shall have x= α We should have found the same value by subtracting+b-c from the given equation; that is, we should have had, in the d-b+c same form, a x=d-b+c, and x = a If 2x+5=17, we have 2 x = 12, and x = 6. 87, we have 3 x 15, and x=5. Hence, ·5—3a=15+9a, we have 4x=20+ 12 a, and, con sequently, x=5+ 3 a. x 497. When the first equation has the form = b, we multiply a both sides by a, in order to have xa b. x But if we have +b➡c=d, we must first make = d b a Letr Let x a = +c; after which, we find a = (d-b+c) a ad—ab+ac 3=4, we have 1 x :7, and x = 14. 1+2 a 3 + a, we have x = 4➡a, and x = ax 498. When we have arrived at such an equation as = c, b we first multiply by b, in order to have a xbc, and then divid ing by a, we find x = bc a c=d, we begin by giving the equation this form =d+c, after which we obtain the value of ax = b d + bc, and Let us suppose x 4 = 1, we shall have x = 5, and 2 x = 15; wherefore x = 15, or 74. If x+1=5, we have x=5-=; wherefore 3 x = 18, and x 6. 499. Let us now consider the case, which may frequently occur, in which two or more terms contain the letter a, either on one side of the equation or on both. If those terms are all on the same side, as in the equation x + x+5=11, we have x+x=6, or Sx= 12, and, lastly, x=4. Let x+x+x=44, and let the value of x be required: if we first multiply by 3, we have 4x+x=132; then multiplying by 2, we have 11 x = 264 ; wherefore x = 24. We might have proceeded more shortly, beginning with the redaction of the three terms which contain x, to the single term x; and then dividing the equation x 44 by 11, we should have had x 4, wherefore x = 24. = Let x-x+x=1, we shall have, by reduction, x = 1, and x = 2}. Let, more generally, a xbx + cx=d; this is the same as (a−b+c) x=d, whence we derive x = a d b+c 500. When there are terms containing a on both sides of the equation, we begin by making such terms vanish from the side from which it is most easily done; that is to say, in which there are fewest of them. If we have, for example, the equation 3x+2=x+10, we must first subtract x from both sides, which gives 2 x +2= 10; wherefore 2x=8, and x = 4. Let x+4=20-x; it is evident that 2x+4= 20; and consequently 2 x 16, and x = 8. Let x+8= 32 =24, and x=6. Let 15 x = 5. x=20 3 x, we shall have 4x + 8 = 32: then 4x 2x, we shall have 15 x 20, and Let 1+x=5-x, we shall have 1+3x=5; after that 3x=4; 3x=8; lastly, x=&= 2 }. -- If } — } x = }} −x, we must add x, which gives = + x; subtracting, there remains1⁄2 x = 1, and multiplying by 12, we obtain x = 2. 1 If 11 − x = +x, we add x, which gives 14 = ÷ +7x. Subtracting, we have 7x=11, whence we deduce x = 1= 15, by multiplying by 6, and dividing by 7. 501. If we have an equation, in which the unknown number x is a denominator, we must make the fraction vanish, by multiplying the whole equation by that denominator. Suppose that we have found 100 have = 20; then multiplying by x, we have 100 20 x ; x and dividing by 20, we find a = 5. If we multiply by x1, we have 5x+3=7x-7. Adding 7, we have 2 x 10. Wherefore x = 5. 502. Sometimes, also, radical signs are found in equations of the first degree. For example, a number x below 100 is required, and such, that the square root of 100-x may be equal to 8, or (100-x)=8; the square of both sides will be 100 -x=64, and adding x we have 100 64+ x; whence we obtain x 100-6436. Or, since 100 x64, we might have subtracted 100 from both sides; and we should then have had X== 36; whence multiplying by — 1, x = 36. - CHAPTER III. Of the Solution of Questions relating to the preceding chapter. 50s. Question I. To divide 7 into two such parts, that the greater may exceed the less by 3. Let the greater part, the less will be 7-x; so that x=7 x+3, or x = 10 x; adding x, we have 2 x = 10; and, dividing by 2, the result is x 5. Answer. The greater part is therefore 5, and the less is 2. Question II. It is required to divide a into two parts, so that the greater may exceed the less by b. Let the greater part =x, the other will be a x; so that x=a-x+b; adding x, we have 2x=a+b; and dividing a+b by 2, x = 2 Another Solution. Let the greater part; and, as it exceeds the less by b, it is evident that the less is smaller than the other by b, and therefore must be = x Now these two parts, taken together, ought to make a; so that 2x-ba; b. adding b, we have 2 x = a + b, wherefore x = the value of the greater part; that of the less will be 504. Question III. A father, who has three sons, leaves them 1600 crowns. The will specifies, that the eldest shall have 200 crowns more than the second, and that the second shall have 100 crowns more than the youngest. Required the share of each ? Let the share of the third son x; then, that of the second will be = x+100, and that of the first = x+300. Now these three shares make up together 1600 crowns. We have, therefore. Answer. The share of the youngest is 400 crowns; that of the second is 500 crowns; and that of the eldest is 700 crowns. 505. Question IV. A father leaves four sons, and 8600 livres; according to the will, the share of the eldest is to be double that of the second, minus 100 livres; the second is to receive three times as much as the third, minus 200 livres; and the third is to receive four times as much as the fourth, minus 300 livres. Required, the respective portions of these four sons. Let us call x the portion of the youngest; that of the third son will be = 4 x - 300; that of the second = 12 x that of the eldest = 24 x must make 8600 livres. 3700 8600, or 41 x = 1100, and 2300. The sum of these four shares We have, therefore, the equation 41 x 12300, and x = 300. Answer. The youngest must have 300 livres, the third son 900, the second 2500, and the eldest 4900. 506. Question. V. A man leaves 11000 crowns to be divided between his widow, two sons, and three daughters. He intends that the mother should receive twice the share of a son, and each son to receive twice as much as a daughter. Required, how much each of them is to receive? Suppose the share of a daughter=x, that of a son is consequently=2x, and that of the widow = 4 x; the whole inheritance is therefore 3x+4x+4x; so that 11x = 11000, and x = 1000. Answer. Each daughter receives So that the three receive in all 1000 crowns, 3000 So that both the sons receive 4000 And the mother receives 4000 Sum 11000 crowns. |