ponent minus 1, that is, by we have the sum required, = 20(23)"b-20b. Wherefore the 1 20 principal sought is (3)"a+ 20(3})" b — 20b = (}})" × (a + 20b) — 206. 805. To resolve this formula we must separately calculate its first term (}})" × (a + 20b), which is nL. + L. (a + 20b), for the number which answers to this logarithm in the tables will be the first term, and if from this we subtract 206 we have the principal sought. 806. Suppose a principal of 1000l. placed out at 5 per cent. compound interest, and to it there be annually added 100%. besides its compound interest, and it be required to know to what it will amount at the end of 25 years. Here a=1000, b=100, n=25; and the operation is as follows: : The first part or number which answers to this logarithm is 10159·17.; from which if we subtract 206=2000 we find the principal in question to be after 25 years 8159·11. 807. If it be required to know in how many years a principal of 1000l. under the above conditions would amount to 1,000,0007.; let n be the number of years required, and since a=1000, b = 100, the principal at the end of n years will be (}})" (3000) — 2000, which sum must make 1,000,000l. ; whence results this equation: 3000 ()" - 2000 = 1000000 Adding to both sides 2000 we have 3000 (2)" =1002000 Using logarithms we have nL.-L.334, and dividing by L., we obtain n= L.334=2.5237465 and L. =0·0211893, wherefore n= 2-5237465 If, lastly, the two terms 2:5237465, of this fraction be multiplied by 10000000, we shall have n = 211893 equal to one hundred and nineteen years one month and seven days, which is the time wherein the principal of 1000l. will be increased to 1,000,000l. In the case of an annual decrease instead of increase of the capital by a certan sum, we shall have the following gradations as the values of a, year after year, the interest being at 5 per cent., and, representing by b the sum annually abstracted from the principal, This principal evidently consists of two parts, one whereof is (2)a, and the other to be subtracted therefrom, taking the terms inversely, forms a geometrical progression, as follows: b+ (3})b+ (3} )2b+ (31)3b + The sum of this progression has already been found = 20 (3)"b-20b; if, therefore, this be subtracted from ()"a, we have the principal required after n years =(2)(a−20b) +20b. 808. For a less period than a year, the exponent ʼn becomes a fraction; for example, 1 day =53, 2 days=383, and so on. It often happens that we wish to know the present value of a sum of money payable at the end of a number of years. Thus, as 20 pounds in ready money amount in a twelvemonth to 21 pounds, so, reciprocally, 21 pounds payable at the end of a year can be worth only 20 pounds. Therefore, if a be a sum payable at the end of a year, the present value of it is a. Hence, to find the present value of a principal a at the end of a year, we must multiply by ; to find its present value at the end of two years, it must be multiplied by (2)2a; and, in general, its value n years before the time of payment will be expressed by (29)"a. 809. Thus, suppose a rent of 1007. receivable for 5 years, reckoning interest at 5 per cent., if we would know its value in present money, we have For £100 due after 1 year, the present value is £95-239 after 2 years 90.704 after 3 years 86.385 after 4 years 82.272 after 5 years 78.955 Sum of the five terms £432.955 So that in present money, the value is 4321. 19s. 1d. It 810. But for a great number of years such a calculation would become laborious. may be facilitated as follows: Let the annual rent a, commencing directly and con tinuing » years, it will be worth a+ ( 29 ) a + ( 29 )2 a + ( 39 )3a + (??)*a . · +(39)"a, which is a geometrical progression whose sum is to be found. We have therefore only to multiply the last term by the exponent, the product whereof is (29)"+1a, then subtract the first term, and the remainder is (3)"+ a-a. Lastly, dividing by the exponent minus 1, that is, or, which is the same, multiplying by -21, we have the sum required,= −21 (39)"+1a +21a, or 21a-21()*+1a, the value of which second term is easily calculated by logarithms. SOLUTION OF PROBlems. 811. The object of algebra, as well as of mathematics generally, being the determination of quantities which were before unknown, this is obtained by an attentive consideration of the conditions given, which are always expressed in known numbers. 812. When a question is to be resolved, the numbers sought are usually represented by the last letters of the alphabet, and the object is then to find, under the conditions, an equality between two quantities. This equality, represented by a formula, is called an equation, and enables us to determine the value of the number sought, and thence to resolve the equation. More than one number is often sought, but they are found by equations in the same manner. 813. To illustrate this, let us take the following example: - Twenty persons, men and women, go to a tavern. The men spend 24 shillings, and the women as much; but each man, it appears, has spent 1 shilling more than each woman. What was the number of men and the number of the women? 24 shillings. Now, these x men having spent 24 shillings, each man's share must be Again, the 20-x women having also spent 24 shillings, the share of each woman is 24 shillings. 20x But we know that each woman's share is 1 shilling less than that of each man; if, therefore, we subtract 1 from each man's share, we must obtain that of each of the From this last equation we have to find the value of x. We shall hereafter see that x=8, which value will correspond to the equation, for -1}, includes the equality 2=2. 814. It is thus seen that an equation consists of two parts separated by the sign of equality, showing that the two quantities are equal to one another. It is often necessary to submit them to a great number of transformations, in order to deduce the value of the unknown quantity, and these are founded on the following principles : — That two quantities remain equal, whether we add to them or subtract from them equal quantities. That the same obtains whether we multiply or divide them by the same number, or extract their roots of the same degree. And lastly, if we take the logarithms of the quantities, as in the preceding section. 815. The equations most easily resolved are those in which the unknown quantity does not exceed the first power after the terms of the equation have been properly arranged. These are called simple equations, or of the first degree. If after the reduction and ordering of an equation, the second power of an unknown quantity is still found, it is called an equation of the second degree, and is more difficult to resolve. When the cube of the unknown quantity appears in an equation, it is called one of the third degree, and so on. RESOLUTION OF SIMPLE EQUATIONS, OR OF THE FIRST DEGREE. 816. When the number sought, or unknown quantity represented by x, is such that one side only contains that letter, and the other a known number, as x=12, the value of x is already found. The object is therefore to arrive at that form, however complicated the equation may be when first formed. 817. To begin with the simplest cases: suppose we have brought an equation to the form x+9=16; inspection alone here shows us that x=7; and, in general, if we find +ab, where a and b express known numbers, we have only to subtract a from both sides to obtain the equation a=b-a, which indicates the value of x. 818. If the equation found be x-a=b, by adding a to both sides we obtain the value of x= b + a. 819. So, if the equation has the form x-a=aa +1, by adding we have x=aa + a + 1. 820. In the equation a-8a=20-6a, we find x = 20−6a + 8a, or x=20+2a. And in x + 6a= 20 + 3a, we have x = 20 + 3a - 6a, or x=20-3a. 821. If the original equation has the form x-a+b=c, we may begin by adding a to both sides, which gives x+b=c+a; and then subtracting b from both sides, we have x=c+a-b. Or we might add +a-b to both sides, by which we immediately obtain x=c+a-b. So in the following examples: If x-2a+3b=0, we have x=2a-3b. If x-3a+2b=25 + a + 2b, we have x=25 + 4a. If x-9+6a=25 + 2a, we have x=34-4a. When the equation found has the form ax=b, it is only necessary to divide the two sides by a, and we have x=. But when the equation has the form ax+b-c=d, the terms that accompany ar must be made to vanish by adding to both sides -b+c, and then, d-b+c dividing the new equation ax=d-b+c by a, we have x= The same value would have been found by subtracting +b-c from the given equation, for we should have had in the same form ar=d-b+c and x = b+c. Hence, a If 2x+5=17, we have 2x=12 and x=6. If 3x-8=7, we have 3x=15 and x=5. a If 4x-5-3a=15+ 9a, we have 4x=20+12a, consequently x=5+ 3a. a When the equation has the form=b, multiply both sides by a, and we have x=ab. But if+b-c=d, we first make =d-b+c, and then x=(d—b+c)a=ad—ab + ac. Let x-3=4; then fr =(4+3)7 and x = 14. Let jx-1+2a=3+a, we have jx=4-a, and x = 12-3a. Let1-1=a, we have = a + 1, and x=aa−1. bc απ ar a When we haye such an equation asc, multiply first by b, which gives ax = bc, and then dividing by a, we have x= If c=d, the equation must first be made to take the form =d+c; after which, multiplying by b, we have ax=bd + bc, and then x= ar Let fr-4=1, we have 3x=5 and 2x=15; whence x= , or 7 bd+bc a In the case of two or more terms containing the letter r either on one or both sides of the equation, the process is as follows: 822. First. If they are on the same side, as in the equation x + 1x+5=11, we have x+x=6, or 3x=12; and, lastly, r=4. Let x+x+x=44, to find the value of x. Multiplying by 3 we have 4x + 3x =132. Multiply both sides by 2, and we have 11x=264; whence x = 24. This might have been effected more shortly by beginning with the reduction of the three terms which contain a to the single term r, and then dividing the equation Yx= 44 by 11, we should have had fr=4, whence x = 24. Generally, let ax-bx + cx=d. It is the same as (a−b+c) x=d, whence x=a-b+c 823. Second. If there be terms containing x on both sides of the equation, they must be made to vanish from that side in which it can most easily be done, that is to say, in which there are fewest of them; thus, in the equation 3x+2=x+10, r must be first subtracted from both sides, which gives 2x+2=10; whence 2x=8, and r = 4. d Let x+4=20-x, it is evident that 2x+4=20, and thence 2x=16, and x=8. Let 15-x=20-2r, we have then 15+x=20, and x=5. x=}=2}. Let 1+x=5-r, we have 1 + x=5, and 1x=4; 3x=8; and, lastly, x= If an equation occurs wherein the unknown number x is a denominator, we must make the fraction vanish by multiplying the whole equation by that denominator. Thus in the equation 100-8=12, we must first add 8, and we have 100 = 20; then, multiplying by x, we have 100=20x, and dividing by 20, x=5, 5x+3 Let I = 7. Multiplying by x-1, we have 5x+3=7x−7. Subtracting 5x, there remains 3=2x−7. Adding 7, we have 2r=10; whence x=5. Radical signs are not unfrequently found in equations of the first degree. For example, a number below 100 is required such that the square root of 100-x=8, or √(100−x)=8; the square of both sides is 100-x=64; adding x we have 100=64+x, whence we have x=100-64=36. 824. The unknown number x is sometimes found in the exponents; in this case, recourse must be had to logarithms. Thus 2512; taking the logarithms on both sides we have *L.2L.512, and dividing by L.2, we find x= We shall here subjoin a few examples of the resolutions of simple equations. L.512 (1.) Divide 7 into two such parts that the greater may exceed the less by 3. Let the greater part = x, The less will be =7-x. So that x=7−x + 3, or x=10-x. Adding x, we have 2x=10, and dividing by 2, x=5. The greater part is 5, and the less is 2. (2.) Divide the number 1600 into three such parts that the greatest shall be 200 more than the second, and the second 100 more than the third. Let the third part =x, then the second will be =x+100, and the greatest = x + 300. 3x + 400=1600; 3x=1200; and x=400. The third part, therefore, is 400, the second 500, and the greatest 700. (3.) Divide 32 into two such parts that if the less be divided by 6, and the greater by 5, the two quotients taken together may make 6. Let the less of the two parts sought =x. The greater will be 32-x. 32-x The first, divided by 6, gives; the second, divided by 5, gives 5 z 32-6 5 Now,+ =6; multiplying them by 5, we have x + 32-x=30, or — fx + 32=30. Subtracting 30, there remains 2=x. Multiplying by 6, we have x=12. Wherefore the lesser part =12, the greater (4.) Divide 25 into two such parts that the greater may contain the less 49 times. Let the less part =x, then the greater will be 25-x. =20. The latter, divided by the former, ought to give the quotient 49; therefore =49. 25-I Multiplying by x we have 25-x=49x. Adding x, 25=50x. Dividing by 50, x=. Hence the less of the two numbers sought is }, and the greater 24. (5.) To find such a number that if 1 be subtracted from its double, and the remainder be doubled, 2 subtracted, and the remainder divided by 4, the number resulting from these operations shall be 1 less than the number sought. Suppose the number to be = r; the double = 2.x. Subtracting 1, the remainder is 2x-1; doubling this, we have 4x-2. Subtracting 2, the remainder is 4x-4; dividing by 4 we have x-1, and this must be 1 less than r, so that x-1=-1. But this is what is called an identical equation, showing that x is indeterminate, or that any number whatever may be substituted for it. (6.) What sum is that, into how many equal parts is it divided, and what is the amount of each part, wherein The first part The second part The third part =100, and one tenth of the remainder; =200, and one tenth of the then remainder; The fourth part =400, and one tenth of the then remainder; and so on? Suppose the total sum =z. Then, since all the parts are equal, let each =x, by which means the number of parts will be expressed by. This being established, the solution is as follows: The differences in the last column are obtained by subtracting each part from that which follows, and all the portions being equal, the differences should be =0; and as they are expressed exactly alike, it will be sufficient to make one of them equal to nothing, and we have the equation 100-100 =0. Multiplying by 10, we have 1000-x-100=0, or From this, therefore, we know that each part is 900; and taking any one of the equations in the third column, the first for example, it becomes, by substituting the value of x, 900=100+ whence the value of z is obtained; for we 10 900-x=0; consequently x=900. 2-100 have 9000=1000 + z-100, or 9000=900+2; whence z=8100, and consequently=9. Hence the number 8100, and each part =900 and the number of the parts =9. RESOLUTION OF TWO OR MORE EQUATIONS OF THE FIRST DEGREE. 825. It often occurs that we are obliged to introduce two or more unknown quantities into algebraic calculations, and these are represented by the letters x, y, z. If the question is determinate, we arrive at the same number of equations from whence to deduce the unknown quantities. Considering only those equations which contain no powers of an unknown quantity higher than the first, and no products of two or more unknown quantities, it is evident that these equations will have the form az + by + cx=d. The 826. Beginning with two equations, we will endeavour to find from them the values of x and y; and that the case may be considered in a general manner, let the two equations be -I. ax + by = c; and, II. fx + gy=h, in which a, b, c, and f, g, h, are known numbers; it is required from these two equations to obtain the two unknown quantities x and y. most obvious way of proceeding is to determine from both equations the value of one of the unknown quantities, x for example, and to consider the equality of the two values; for then we obtain an equation in which the unknown quantity y appears by itself, and may be determined by the rules we have already given. Knowing y, we have only to substitute its value in one of the quantities that express x. = a 827. According to this rule we obtain from the first equation by, from the second x="-g. Stating these two values equal to one another, a new equation appears, — -ah-agy. ah-fc ag-bf Multiplying by a, the product is c-by-ah-agy; multiplying by f, the product is fc – ƒ by Adding agy, we have fe-fby+agy=ah; subtracting fc, there remains -fby+agy=ah-fe; or (ag-bf)y=ah-fc; lastly, dividing by ag-bf, we have y = 828. In order to substitute this value of y in one of those we have found of x, as in the first, when ac-by, we shall first have -by-ab+bcf; whence c-by=c. - abh+bcf c-by=acg-bcf—abh+bcf __ acg—abh a ag-by ag-af ag-bf ; and dividing by a, x-c-by_cg—bh or ag-bf' 829. To illustrate this method, let it be proposed to find two numbers whose sum may be 15, and difference =7. Let the greater number =x, and the less y; we shall then have, I. x+y=15, and II. x-y=7. The first equation gives x=15-y, and the second r=7+y; whence there results the new equation 15-y=7+y. So that 15=7+2y, 2y=8, and y=4; by which means we find The less number, therefore, is 4, and the greater is 11. x=11. 830. When there are three unknown numbers, and as many equations, as, for example, I. x+y−z=8; II. x+z-y=9; III. y+z−x=10; a value of a is to be deduced from each and from I. we have x=8+z-y; from II., x=9+y−z; and from III. x=y +2-10. Comparing them together, we have the following equations : I. 8+z−y=9+y-z. II. 8+z−y=y+z−10. The first gives 2z-2y=1; the second, 2y=18, or y=9. Substitute this value of y in 22-2y=, and we have 2z-18=1, and 2z=19, so that z=91. We have, therefore, only to determine r, which is found = 8. The letter z thus vanishes in the last equation, and the value of y is immediately found; otherwise we must have had two equations between z and y to have been resolved by the preceding rule. 831. Suppose we had found the three following equations — I. 3x+5y-4z=25. II. 5x-2y+3z=46. Deducing from each the value of r, we have 3 I. x=25—5y++≈. II. x= 46+2y=3x 5 III. 3y+5%-x=62. III. x=3y+5z−62. And comparing these three values together, and the third with the first, we have 3y + 52 25—5y+4z -62= Multiplying by S, 9y+15z-186=25-5y+4z; so that 9y+15z=211 -5y+4z, and 14y+11z=211. Comparing the third with the second, we have 3y+5z -62-46+2046+2y-3% or 46+2y-3z=15y+25z-310, which, reduced, is 356=13y+28z. From these two new equations the value of y may be deduced. |