tract from half of the number to be divided, or from half of the 2 Knowing the less part we have the greater by adding to the less the given excess. This remark is sufficient for effecting the solution of the question proposed; but Algebra does more; it furnishes a rule for calculating the greater part without the aid of the less as follows; being the value of this, augmenting it by the excess b, a shows that after having subtracted from the half of b, it is 2 necessary to add to the remainder the whole of b, or two halves of b, which reduces itself to augmenting by the half of b, or by translating this expression we learn, that of the two parts sought the greater is equal to half of the number to be divided plus half of the given excess. In the particular question which I first considered, the number to be divided was 9, the excess of one part above the other 5; in order to resolve it by the rules to which we have just arrived, it will be necessary to perform upon the numbers 9 and 5, the operations indicated upon a and b. 4. I have denoted in the above the less of the two parts by x, and I have deduced from it the greater. If it were required to find directly this last, it should be observed, that representing it by x, the other will be xb, since we pass from the greater to the less by subtracting the excess of the first above the second; the number to be divided will then be expressed by xxb, or by 2x-b, and we have consequently This result makes it evident that 2x exceeds the quantity a by the quantity b, and that consequently 2x= a+b. By taking the half of 2x and of the quantity which is equal to it, we obtain for the value of x a b which gives the same rule as the above for determining the greater of the two parts sought. I will not stop to deduce from it the expression for the smaller. The same relation between the numbers given and the numbers required may be enunciated in many different ways. That which has led to the preceding result is deduced also from the following enunciation : Knowing the sum a of two numbers and their difference b, to find each of those numbers; since, in other words, the number to be divided is the sum of the two numbers sought, and their difference is the excess of the greater above the less. The change in the terms of the enunciation being applied to the rules found above, we have The less of two numbers sought is equal to half of the sum minus half of the difference. The greater is equal to half of the sum plus half of the difference. 5. The following question is similar to the preceding, but a little more complicated. To divide a given number into three such parts, that the excess of the mean above the least may be a given number, and the excess of the greatest above the mean may be another given number. For the sake of distinctness I will first give determinate values to the known numbers. I will suppose that the number to be divided is 230; that the excess of the middle part above the least is 40; and, that of the greatest above the middle one is 60. Denoting the least part by x, the middle one will be the least plus 40, or x + 40, and the greatest will be the middle one plus 60, or x + 40 + 60. Now the three parts taken together must make the number to be divided; whence, x+x+40 + x +40 + 60 = 230. If the given numbers be united in one expression and the unknown ones in another, x is found three times in the result, and for the sake of conciseness we write 3x + 140 = 230. But since it is necessary to add 140 to triple of x to make 230, it follows, that by taking 140 from 230 we have exactly the triple of x, or By adding to 30 the excess 40 of the middle part above the least, we have 70 for the middle part. By adding to 70 the excess 60 of the greatest above the middle part, we have 130 for the greatest. 6. If the known numbers were different from those which I have used in the enunciation, we should still resolve the question by following the steps traced in the preceding article, but we should be obliged to repeat all the reasonings and all the operations, by which we have arrived at the number 30, because there is nothing to show how this number is composed of 230, 40, and 60. To render the solution independent of the particular values of numbers, and to show how the value of the unknown quantity is fixed by means of the known quantities, I will enunciate the problem thus ; To divide a given number a into three such parts, that the excess of the middle one above the least shall be a given number b, and the excess of the greater above the middle one shall be a given number c. Designating as above by x the unknown quantity and making use of the common signs and the symbols a, b, c, which represent the known quantities in the question, the reasoning already given will be repeated. The least part = x, the middle part the greatest = x + b, = x + b + c, and the sum of these three makes the number to be divided; hence, x+x+b+ x + b + c = a. This expression, which is so simple, may be still further abridged; for since it appears that x enters three times into the number to be divided and b twice, instead of x + x + x, I shall write 3x, and instead of + b + b, I shall write + 2b, and it will become 3x + 2b + c = a. From this last expression it is evident, that it is necessary to add to triple the number represented by x, double the number represented by b, and also the number c, in order to make the number a; it follows then, that if from the number a we take double the number b and also the number c, we shall have exactly the triple of x, or that Now a being one third of three times x, we thence conclude that It should be carefully observed, that having assigned no particular value to the numbers represented by a, b, c, the result to which we have come is equally indeterminate as to the value of x; it shews merely what operations it is necessary to perform upon these numbers, when a value is assigned to them, in order thence to deduce the value of the unknown quantity. 2b-c 3 to which x is equal, may be In short, the expression reduced to common language by writing, instead of the letters, the numbers which they represent, and instead of the signs, the kind of operation which they indicate; it will then become, as follows; From the number to be divided, subtract double the excess of the middle part above the least, and also the excess of the greatest above the middle part, and take a third of the remainder. If we apply this rule, we shall determine, by the simple operations of arithmetic, the least part. The number to be divided being for example 230, one excess 40, and the other 60, if we subtract as in the preceding article twice 40, or 80, and 60 from 230, there will remain 90, of which the third part is 30, as we have found already. If the number to be divided were 520, one excess 50 and the other 120, we should subtract twice 50, or 100, and 120 from 520, and there would remain 300, a third of which or 100 would be the smallest part. The others are found by adding 50 to 100, which makes 150, and 120 more to this, which makes 270, so that the parts sought would be 100, 150, 270, and their sum would be 520, as the question requires. It is because the results in algebra are for the most part only an indication of the operations to be performed upon numbers in order to find others, that they are called in general formulas. This question, although more complicated than that of article 1, may still be resolved by ordinary language, as may be seen in the following table, where against each step is placed a translation of it into algebraic characters. PROBLEM. To divide a number into three such parts, that the excess of the middle one above the least shall be a given number, and the excess of the greatest above the middle one shall be another given number. By common language. SOLUTION. The middle part will be the least, plus the excess of the mean above the least. The greatest part will be the middle one, plus the excess of the greatest above the middle one. The three parts will together form the number proposed. Whence the least part, plus the least part, plus the excess of the middle one above the least, plus also the least part, plus the excess of the middle one above the least, plus the excess of the greatest above the middle one, will be equal to the number to be divided. |