18. A detachment from an army was marching in regular column, with 5 men more in depth than in front; but upon the enemy's coming in sight, the front was increased by 845 men; and by this movement the detachment was drawn up in five lines. What was the number of men? Let x represent the number of men in front; then Now, x+845 represents the number in front after it was increased by 845 men. And since the detachment was then drawn up in 5 lines, it had 5 men in depth. Hence, 5(x+845) also represents the whole number of men. x(x+5)=5(x+845). Developing both sides of this Equation, +5=5+4225; from which we shall find x=65, the number in front; hence 65+5— 70, was the number in depth. Then, 65×70=4550 was the whole number of men. 19. A company at a tavern had £8 15s. to pay, but before their bill was settled, two of them went away, when those who remained had 10s. apiece more to pay than before. How many were there in the company at first? Let x represent the number; then since £8 15s. is equal to 175 shillings, 175 x company should have paid; and 175 x-2 represents the number of shillings each one of the original represents the number of shillings each one paid after two had departed, Then, by the conditions of the problem, we shall have 175 x-2 = 175 X +10. Clearing the Equation of fractions, &c., we have 10x2-20x=350; whence x=-7 persons. 20. Some gentlemen made an excursion, and each one took the same sum. Each gentleman had as many servants as there were gentlemen, and the number of dollars which each had was double the whole number of servants; also the whole sum taken was $3456. What was the number of gentlemen? Let x represent the number; then xx, or x2, represents the number of servants; 2r2 represents the number of dollars each gentleman took; and 2x2 ×x, or 2x3, represents the whole sum taken. The Equation will then be 2x33456; whence x=12. 21. Divide the number 20 into two such parts, that the product of the whole number and one of the parts shall be equal to the square of the other. Let x represent the less number; then 20x=(20-x)2. Developing the second member of this Equation, 20x400-40x+x2. -- This Equation will give a 30-10/5, the less number; hence 20—(30—10√/5)=10√/5—10, is the greater. 22. A laborer dug two trenches, one of which was 6 yards longer than the other, for £17 16s., and the digging of each cost as many shillings per yard as there were yards in its length. What was the length of each? Let x represent the number of yards in the first; then xxx, or x2, represents the cost of the first, in shillings; and in shillings. Since £17 16s. is equal to 356 shillings, the Equation will be x2+(x+6)2=356; from which we shall find x= 10 yards, the length of the first; hence 10+6=16 yards, was the length of the second. 23. There are two numbers whose product is 120, and if 2 be added to the less, and 3 subtracted from the greater, the product of the sum and the remainder will also be 120. What are the two numbers? Let x represent the less; then 120 represents the greater. x By the conditions of the problem, the Equation will then be (x+2)× (120 −3)=120. Multiplying together the binomials in the first member, 114x-3x2+240 x From this Equation x will be found equal to 8, the less number; hence 8 B's 24. Two persons lay out some money on speculation. A disposes of his bargain for £11, and gains as much per cent as B lays out. gain is £36, and it appears that A gains 4 times as much per cent as B. What sum did each lay out? Let x represent the number of pounds A laid out; then 11 -X represents his gain on one pound; x 100(11-x) X 100(11-x) X Now 36 dollar; and 100(11-x) Xx =120. represents his gain per cent; and by the problem represents the number of pounds B laid out. 9x 25(11-x)' 2 or , or 9x 36x 100 X 25(11-x) (11-x)' By the conditions of the problem, we shall then have 100(11-x) X represents B's gain on one represents his gain per cent. 36x Clearing this Equation of fractions, 100(11-x)2=144x2. Developing the first member of this Equation, 12100-2200x+100x2=144x2; which will give x= £5, the sum A laid out; hence is the sum B laid out. 100(11-5) 5 25. A set out from C towards D, and traveled 7 miles an hour. After he had gone 32 miles, B set out from D towards C, and went each hour of the whole distance; and after he had traveled as many hours as he went miles in one hour, he met A. Required the distance between the two places. represents the number of hours he traveled; and Ꮖ x2 X , or Let x represent the number of miles between the places; then 19 X 19 X represents the whole distance he traveled. 19 Now, as A traveled 7 miles an hour, and had gone 32 miles before B set out from D, 7x = £120, x 32+7x or 32+ represents the distance A traveled. Since both together traveled the whole distance between the two places, the Equation will be x2 361 x2+11552+133x=361x; from which x will be found equal to 152, or 76 miles. 7x +32+ =x. 19 Clearing this Equation of fractions by multiplying it by the least common multiple of the denominators, which is 361, we have Solution of Two Equations-One or both of the Second or a Higher Degree-Containing Two Unknown Quantities. 1. Find the values of x and y in the Equations From the first Equation we find x= Then 2x2= 10-y 2 2(10-y)2 and " xy= 4 Substituting these values of 22 and xy in the second Equation, 2(10-y)2 10y-y2 +3y2=54. 4 2 Clearing this Equation of fractions, 10y-y2 2 2(10-y)2-20y+2y2+12y2=216. From this Equation we shall find y=4, or — (262); and substituting the first of these values for y in the first Equation, we find x=3; substituting the second value in the same Equation, we find x= 4. 2. Find the values of x and y in the Equations From the first Equation we find x= 9-y; hence Substituting this value of x2 in the second Equation, which will give y=6, or 3; and substituting these values successively, for y in the first Equation, we find x=3, or 6. 3. Find the values of x and У in the Equations Dividing the first Equation by y, we have |