63. Particular answer to 61st ; particular answer to 62d 1. SECTION XXVIII. ART. 84. From the 1st to the 5th inclusive, performed. 6. This question as enunciated, if x represents the number, Changing the sign of x in the original equation, we have To rectify the question, therefore, we enunciate it thus: 7. This question gives the number of years 40; and the equation being modified accordingly, we find that the question should have been : How many years after marriage was his age to hers as 7 to 6? Changing the original equations, in a manner similar to that pursued in the solution of the 5th question, we have tion to be made in the enunciation is manifest. 9. Let the greater, and y = the less; then, x-y=20, and 6x -3y=96. These equations give x 12, and y = - 8. Changing the sign of y in the original equations, we have x+y=20, and 6x+3y=96; and the ques tion should be changed accordingly. 10. Let x and y represent the number of gallons, which flow through the cocks A and B, respectively, in a minute. Then, 5x+3y=24, and 7x+5y = 32. These equations give x 6 gallons, and y-2 gallons. Changing the sign of y in the original equations, we have 5x-3y=24, and 7x-5y=32. Hence it appears that water flows out, instead of flowing in, through the cock B. 11. A's $5000; B's $3000; C's $2000. Hence, either C is in debt $2000; or he possesses $2000, and a certain number of times his estate is subtracted, in each case, instead of being added. SECTION XXIX. ART. 85. 1. Let x represent his money in cents. Then 3x-3x=9; that is, (3-3)x or 0.x=9; therefore, x= Hence, the question is impossible, or the value of x is infinite. A's age; then, x+10= B's, x+20= C's, and We have, therefore, 2. Let x 34 D's. 2x+40 5x–170 2 + + 3 6 This equation gives 12x-12x or 0.x 60-60; and Hence, the value of x is indeterminate; consequently, their ages cannot be ascertained from the conditions. 3. If z represent the number killed to-day, and y the number killed yesterday, we have 3 x These equations become, by multiplication and transposition, 2y=30, and 3 x — ·2 y = 10, in which the first members are identical, while the second members are different. Hence, the two conditions are incompatible with each other. By subtracting and canceling the terms containing y, we have -3x=20, or 0. x 20, and x = 20. 3 x Or, by subtracting and canceling the terms containing x, we have 2y-2y=20, or 0. y = 20, and y = 20. The problem, therefore, is impossible; or the values of x and y are infinite. Other forms for the answers may be obtained by different methods of elimination; but they are all essentially the same, because they are all infinite. |