36. Divide the number 240 into two parts, such that 7 times the first shall equal 5 times the second. Ans. 100 and 140. 37. In a garrison of 2400 men, there are 3 times as many cavalry as artillery, and twice as many infantry as artillery and cavalry together. How many are there of each kind ? Ans. 200 artillery, 600 cavalry, and 1600 infantry. 38. Divide 21000 dollars between A, B, C, and D, so that A's part shall be of D's, B's part of C's, and C's part of D's. How many. dollars will each receive? Ans. A, 3200; B, 4800; C, 6000; D, 7000. 39. A capital was put out at 61 per cent. for one year, when the capital and interest together amounted to 1917 dollars. How many dollars were there in the capital? Ans.: 1800. 40. A boatman rows with the tide 42 miles in 3 hours. In returning, the tide is but as strong, and it takes 102 hours to row the same distance. At what rate per hour did the tide run in each case. Ans. 6 and 4 miles. 41. A cistern can be filled by two cocks; the first would fill it in 70 minutes, and the second in 80 min. utes. In how many minutes would they both fill it together? Ans. 374 AQUATIONS OF THE FIRST DEGREE, CONTAINING MORE THAN ONE UNKNOWN QUANTITY. 89. If we have a single equation, containing two unknown quantities, as 2x + 3y = 14, we may find the value of one of them in terms of the other, as follows: 14 3y (1.) If now, we Now, if the value of y is unknown, that of 2 will also be unknown. Hence, from this equation alone, the value of 2 cannot be determined. have a second equation, 3x + 2y = 11, we may, in like manner, find the value of x in terms If now, the values of x and y are the same in the two equations, we shall have the second members of (1) and (2) equal to each other; giving the equation, From which we find y = 4; and substituting this ralue for y, in either of the equations (1), or (2), we lind 2 =1 Such equations are called simultaneous. Hence, SIMULTANEOUS EQUATIONS are those in which the ralues of the unknown quantities are the same in both. We have seen that it requires two simultaneous equa tions, containing two unknown quantities, to determine the values of the unknown quantities. In the same way, it could be shown that it would require three equations containing three unknown quantities, four equations con. taining four unknown quantities, and so on, to determine the values of the unknown quantities. In general, there must be as many cquations as there are unknown quantities. The equations necessary to determine any number of unknown quantities, constitute a group of simultaneous equations. Such equations are solved by successive eliminations. ELIMINATION, 90. ELIMINATION is the operation of combining two equations in such a manner as to get rid of one of the unknown quantities which enter them. There are three principal methods of elimination, by Addition or Subtraction; by Substitution ; and by Com. parison. . Multiplying both members of (1) by 4, and of (2) by 6 (Axiom 3), 28x + 24y = 80 (3.) 54x 24y (4.) Adding (3) and (4), member to member Axiom 1), = 84 . 82X = 164. Here, y has been eliminated by addition. Again, multiplying both members of (1) by 9, and of (2) by 7, 63x + 549 ( 5.) (6.) Subtracting (6) from (5), member from member, (Axiom 2), 82y = 82. Here, x has been eliminated by subtraction. In the same way, an unknown quantity may be eliminated from any two simultaneous equations. Hence, the RU LE. other ways. Prepare the equations, so that the coefficients of the quantity to be eliminated shall be numerically equal in both; if the signs are unlike, add the equations, member to member ; if alike, subtract them, member from member. In preparing the equations above, we muitiplied both members of each, by the coefficient of the quantity to be eliminated in the other. They may be prepared in A better way, in most cases, is to find the least common multiple of the coefficients of the quantity to be eliminated; then multiply both members of each equation by the quotient of this least common multiple by the coefficient of the quantity to be eliminated in that equation. In the first case considered, the least common multiple of 4 and 6 is 12; we might have 12 multiplied both members of (1) by or 2, and of 6 12 (2) by or 3, giving, 4 14x + 12y 27x – 12y = 42. = 40 Whence, by addition, 412 = 82. Here, y has been eliminated as before, but we have a simpler equation. Finding, from (1), the value of y in terms of the Substituting this value for y, in (2), Here, y has been eliminated by substitution. In the same way, may eliminate an unknown quantity between any two equations. Hence, the we RULE. Find from one of the equations the value of the quantity to be eliminated; substitute this value for thar quantity in the other equation. |