Then, 592, by by co-ef. of x, (8) 60y = 600, 4y8, by subtracting, as before. And, y = 2 shillings, the gain of the wife and son. (12) This problem could be also solved by finding the value of x or y in each of the equations, and then take each of the answers to form the side of a new equation, from which we could find the value of the remaining quantity, as we have before shown; but the most simple mode appears to be by multiplying each equation by a number, for to make a quantity in both identical. 5.-A courier went from Taunton to Somerton, and back again, in 4 hours, and found that he went at the rate of 16 miles an hour, but returned at the rate of 8 miles an hour. We demand the distance between the two places? Let x be the time going, y the time coming back, It is evident that the rate of returning is only half that of going. Wherefore 2x = y, that is, the rate of going is double that of returning. But the time taken in going and returning is 4 hours, therefore + y is equal to 4. Now we have the equation as follows. Then we have and 2x = y, x + y = 4, 3x4, by Subtraction, x = 1}, the time taken going to Somerton. Again, taking the same equations, 2x Y, x + y = 4, We have, 2x + 2y = 8, and, y = by multiplying the second line by 2, in order to make the first terms identical. Then we have, 3y=8, by subtracting 1st line from 3rd. 23, the time taken in returning to Taunton. Now, by multiplying the value of y by the rate o returning, or the value of x by the rate of going, we have the distance required, thus: 2 × 8=213, distance between Taunton and Somerton. 1 × 16 = 211. دو 6.—A father, being asked the age of his son, replied: if from double the age he now is, you take three times his age at three shall have his actual age. years old, you Ans...The child was 9 years old. 7.—A hare is 50 leaps in advance of a grey-hound, and takes 4 leaps to the grey-hound's 3; but 2 of the dog's leaps are equal to 3 of the hare's. We demand how many leaps must the grey-hound take, to catch the hare? Ans...300. 8.- Given, x + x − x = 1, to find x? .x= 9.- Given, 12 + x = 2 + √x, to find x? 10.- Given, x + y = 8, Ans... 6 x2 - y2 = d, Equations having three or more unknown quantities may be similarly solved; by first eliminating the same quantity in two of the equations, and then eliminate the same quantity in the next, and either of the others; proceeding similarly as in equations with one or two unknown quantities. EQUATIONS OF THE SECOND DEGREE. The equations of which we have treated involve only the first power of the unknown quantity - but if we propose the following question - To find a number such, that being multiplied by its quintuple, the product shall be equal to 125-we have a pure equation of the second degree or power. Therefore if we designate this number by x, we find x × 5x = 5x2 = 125. Then by dividing by 5, and extracting the square root of both sides of the equation, we have x = 5. We must bear in mind that the square root of Algebraic quantities admits of two signs, id est, plus or minus. An example will suffice. Thus (+5)2 + 251 according to the rule for = or (-5)2: +25 Multiplication. Therefore it is evident that the square root of 25 is either + or 5. The next form is that of adfected or compound quantities, that is, involving both the first and second power of the unknown quantities, the remainder being known quantities. Thus x2 + px = q is the general formula; x2 and x being the unknown terms; p and զ the known. Now it is manifest by the 4 Prop. B. 2, that if a line or number be divided into two parts, the sum of their squares shall be less than the square of the sum, by twice the product of the parts-therefore either of the following Rules is required to complete the square of each member of a quadratic equation before its root can be extracted. RULE I. Add the square of half the co-efficient of the lowest power of a to both members of the equation, and the squares shall be complete; but if the highest power of have a co-efficient, reduce it to unity, by dividing the whole equation by it, and then proceed as above. Multiply the whole equation by four times the coefficient of the highest power of x; then add to both members of the product, the square of the original coefficient of the lowest power of x, and the squares shall be complete. This is called the Hindoo method, and is sometimes more suitable than the first, because the co-efficient of the highest power of x has not to be removed before adding what completes the squares. 1.- Given, 2x2 + 8x 2070, to find x? It becomes, 16x2 + 64x 160= 560, by × by 8, then, 16x2 + 64x + 64784, by + the (8)2 to both, &c. and, 4x + 8 = 28. by Evolution. Ans...x=9. 2. To find a number such, that if you multiply it by 8, the product shall be equal to the square of the same number, having 12 added to it? It is necessary to remember that terms involving the unknown quantity must be made to occupy one side of the equation, before proceeding to find its value; and also, that if we square or extract the root of one side of an equation, we must proceed similarly with the other, in order to keep up the equality between them. END OF THIRD BOOK. NUMERICAL PROOF OF EUCLID'S SECOND BOOK. Let numbers be placed on each given line, according to its divisions, named in the enunciation. 20 Prop. 1....10.10 10. Now, proceeding to prove the Proposition as directed by the enunciation, we have, 20 x 30 = 600. Again, 10 x 20 + 10 × 20 + 10 × 20 = 600. Therefore, each side being equal to 600, it is evident that the rectangle contained by the two right lines (supposed to contain 20 and 30 units respectively) is equal to the rectangles contained by the undivided line (20) and the several parts of the divided line. We shall similarly prove the other Propositions. Prop. 2....10. 20 (30)2 = 900. And 30 x 10 + 30 x 20 = 900. Prop 3....10. 20 30 x 10 = 300. And (10)2 + 20 × 10 = 300. (10)2 + (20)2 + 2 x 10 x 20 = 900. 10 × 20 + (5)3 =225. Again, (15)2 = 225. |