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? Let the number of the men = x ; That of the women then will be = 20-x. Now, these x men having spent 24 shillings, each man's share must be # shillings. Again, the 20-x women having also spent 24 shillings, the share of each woman is *: shillings. 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 women; consequently # - 1 = #.From this last equation we have to find the value of x. We shall hereafter see that r=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. r Resolution of simiPLE EQUATIONs, or of THE FIRST DEGREE. 816. When the number sought, or unknown quantity represented by r, 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 ar=b+a. 819. So, if the equation has the form x-a=aa + 1, by adding we have r=aa + a + 1. 820. In the equation a -8a=20–6a, we find x = 20-6a+8a, or x=20+2a. And in * + 6a=2O + 3a, we have r = 20 + 3a–6a, or x = 20-3a. (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-r. Adding r, 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. These parts, then, make up the number 1600; we have therefore – 3r + 400 = 1600; 3x = 1200; and x =400. The third part, therefore, is 400, the second 500, and the greatest 700. (S.) 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. Adding }x, we have 32=30+}r. Subtracting 30, there remains 2=}r. Multiplying by 6, we have r=12. Wherefore the lesser part = 12, the greater = 20. (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-r. The latter, divided by the former, ought to give the quotient 49; therefore 25-r ar =49. Multiplying by z we have 25–2 =49x. Adding r, 25=50r. Dividing by 50, z=y. 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 = x ; the double = 2x. Subtracting 1, the remainder is 2x-1; doubling this, we have 4x–2. Subtracting 2, the remainder is 4 r–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 = 100, and one tenth of the remainder; The second part =200, and one tenth of the then remainder; The third part = 300, 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 =2. 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–£=0. Multiplying by 10, we have 1000-x -100=0, or 900-r=o; consequently r=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 have 9000=1000 + 2 - 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 oit 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 r, 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 + cr=d. 826. Beginning with two equations, we will endeavour to find from them the values of z and y; and that the case may be considered in a general manner, let the two equations be –I. ax + by =c; and, II. fr + gy=h, in which a, b, c, and fg, 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 r. 827. According to this rule we obtain from the first equation *=''", from the second h-gy =4984-3922, or 24.92=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 r, 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. 833. 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 equation 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, x 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 car. 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–brr=crr may be considered as a single term if thus represented (a–b 4 c):rr; 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-33 + 4 = 5.x.r-8x + 1 1 ; We first subtract 2xx, and the remainder is – 3.x + 4 = 5xx -8a -i- 11. Then adding 8x, we have 5.x + 4 = 3rr + 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 0 on the other. Thus, the above equation, remembering to change the signs, will assume this form, 3xx–5x +7=0. Hence, the following general formula represents all equations of the second degree— arx + ba + 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 ar+b_ext/ cr-i-d"gr-i-h" The fractions must be first destroyed. Multiplying for this purpose by cri d, we have *r 4-b= *::::::::A; then, by gr: h, we have agra + bgr + ahx + bh=cerz + cf.r Hedrifa, 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: — O= agrx + bgx + bh —cer r + ahr-fa – cf.r -ed.r. This equation is, perhaps, more clearly exhibited in the following form : – 0=(ag-ce)rr + (bg + ah-cf-ed)r + bh-fa. 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 ra 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 + br=0, which, being divisible by r, may be reduced to ax + b =0, which, also, is a simple equation, not belonging to the present class. |