tinuing n years, it will be worth a+ (??) a + ( 39 ) 2a + ( ?} ) 3 a + ( 39 ) * a . . . . + (29)”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(3)+1a, the value of which second term is easily calculated by logarithms. SOLUTION OF PROBLEMS. α 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: women, go to a tavern. The men spend 24 shillings, and the man, it appears, has spent 1 shilling more than each woman. men and the number of the women? Let the number of the men = x; Twenty persons, men and women as much; but each What was the number of 20 — x. 24 I Now, these x men having spent 24 shillings, each man's share must be shillings. Again, the 20-r women having also spent 24 shillings, the share of each woman is 20 shillings. 24 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 =8, which value will correspond to the equation, for 2-1, 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 +a=b, 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 raa+a+1. 820. In the equation a-8a=20-6a, we find x=20-6a+ 8a, or x=20 + 2a. And in 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 ar=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, 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 d_b+c the same form ax=d-b+c and x=' Hence, a If 2x+5=17, we have 2x=12 and x=6. d-b+c 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. bc ax ax a When we have such an equation as =c, 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=' ax Let fr-4-1, we have 3r=. and 2x=15; whence r=, or 7. bd+bc a In the case of two or more terms containing the letter z 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+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 + }x =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 = 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=ab+c' d 823. Second. If there be terms containing a 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, x must be first subtracted from both sides, which gives 2x+2=10; whence 2x=8, and x = 4. Let x+4=20-x, it is evident that 2x+4=20, and thence 2x=16, and x=8. Let x+8=32-3.x, we have 4x+8=32, then 4x=24, and x=6. Let 15-x=20-2x, we have then 15+x=20, and x=5. Let 1+x=5−x, we have 1 + }x=5, and r=4; 3x=8; and, lastly, x=}=2. 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 -8-12, we must first add 8, and we have =20; then, multiplying by z, we have 100=20x, and dividing by 20, x=5, 100 100 5x+3 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 x 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+, 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.2=L.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 32-6 Now,+5 Subtracting 30, there remains 2=Jr. 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 25-x H =49. 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 241. (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 =x; the double = 2x. 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 x, so that x-1=x-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 fourth part Suppose the total sum =z. =100, and one tenth of the remainder ; 200, and one tenth of the then remainder; =300, and one tenth of the then remainder; =400, and one tenth of the then remainder; and so on? 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 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—2—100 ̧ =0. Multiplying by 10, we have 1000-x-100=0, or 900-x=0; consequently x=900. 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 z-100 have 9000=1000+z-100, or 9000=900+z; 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. 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. The most obvious way of proceeding is to determine from both equations the value of one of the unknown quantities, a 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. 827. According to this rule we obtain from the first equation by, from the second Stating these two values equal to one another, a new equation appears,— x= h-gy f a ah-fc ag-bf Multiplying by a, the product is c-by-ah-agy; multiplying by f, the product is fc-fby -ah-agy. Adding agy, we have fc-fby+agy=ah; subtracting fc, there remains -fby+agy=ah-fe; or (ag-bf) y=ah-fe; 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 x= c-by we shall first have -by= abhbef; whence c-by=c. ag-bf c-by-acg-bcf-abh+bcf __ acg—abh ag-bf a ag-af c-by_cg-bh a ay=bf' •—abh+bf 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 x=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 x is to be deduced from each and from I. we have x=8+2-y; from II., x=9+y-2; and from III. x=y +2-10. Comparing them together, we have the following equations : I. 8+2y=9+y―z II. 8+2-y=y+z-10. The first gives 2z-2y=1; the second, 2y=18, or y=9. Substitute this value of y in 2z-2y=, and we have 2z-18=1, and 2z=19, so that z=9 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 And comparing these three values together, and the third with the first, we have 3y+52 25-5y+4z -62= 4. Multiplying by S, 9y+15z—186=25-5y+4z; so that 9y+15z=211 3 ♦-5y+4z, and 14y+11z=211. Comparing the third with the second, we have 3y+5z 46+2y-Sz or 46+2y-Sz=15y+ 25z−310, which, reduced, is $56=13y+ 28z. From these two new equations the value of y may be deduced. 211-11z $56-28% = These two values form the new equation which becomes 2743-143z =4984-392z, or 249z=2241; whence z=9. This value being substituted in one of the two equations of y and z, we find y=8, and by a similar substitution in one of the three values of x, x=7. 832. If more than three unknown quantities are to be determined, and as many equations to be resolved, the same manner must be pursued; but the calculations are often tedious; and it is to be observed that in each particular case means may be resorted to for facilitating the resolution. These consist in introducing, besides the principal unknown quantities, some new one, arbitrarily assumed; such, for instance, as the sum of all the rest. But practice only can teach this; and the architect is in this, and remaining pages of this chapter, as much informed on the subject as his practice is likely ever to require. RESOLUTION OF PURE QUADRATIC EQUATIONS. 835. If an equation contains the square or second power of the unknown quantity without any of the higher powers, it is said to be of the second degree. An equation containing the third power of the unknown quantity belongs to cubic equations, and its resolution requires particular rules. There are, therefore, only three kinds of terms in an equa tion of the second degree. I. The terms which do not contain the unknown quantity at all, or which contain only known numbers. II. The terms wherein only the first power of the unknown quantity is found. III. The terms which contain the square, or second power of the unknown quantity. Thus, a signifying an unknown quantity, and the letters a, b, c, d, &c. being known numbers, the terms of the first kind will have the form a, those of the second kind will have the form br, and those of the third kind will have the form cxx. 834. It has been already seen that two or more terms of the same kind may be united together and considered as a single term; thus the formula arr—brx + cxx may be con sidered as a single term if thus represented (a−b+c)xx; since, in fact, (a−b+c) is a known quantity. When such terms are found on both sides the sign =, we have seen they may be brought to one side, and then reduced to a single term. For example, in the equation 2xx-3x+4=5xx— 8x+11; Lastly, subtracting 11, the remainder is 3xx=5x—7. 835. All the terms may also be brought to one side of the sign leaving only O on the other. Thus, the above equation, remembering to change the signs, will assume this form, Sxx-5x+7=0. Hence, the following general formula represents all equations of the second degree axx + bx + c = 0, wherein the sign ± is read plus or minus, and shows that the terms to which it is prefixed may be positive or negative. To this formula all quadratic equations may be reduced. Suppose, for instance, the equation ax+b ex+f The fractions must be first destroyed. Multiplying for this purpose by ca+d, we have ax+b=cezz+¢fr+edz+fd ; then, by gx + h, we have agxx + bgx + ahx + bh: =cexx + cfx + edx + fd, an equation of the second degree, and one which may be reduced to the three following terms, which are transposed by arrangement in the usual manner : 0=agxx + bgx + bh This equation is, perhaps, more clearly exhibited in the following form: 0=(ag―ce)xx + (bg+ah — cf — ed)x + bh—fd. 836. Equations of the second degree are called complete when the three kinds of terms are found in them, and their resolution is more difficult; on which account we shall first consider those in which one of the terms is wanting. If the term rx be not found in the equation, it is not a quadratic, but belongs to those whereof we have already treated. If the term containing known numbers only were wanting, the equation would have the form arr + bx=0, which, being divisible by x, may be reduced to ax + b=0, which, also, is a |