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63. Particular answer to 61st $; particular answer to 62d 1}.

SECTION XXVIII.

2=

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+5.

ART. 84.

From the 1st to the 5th inclusive, performed.

6. This question as enunciated, if x represents the number, gives

7 x 52

+5.

10 From this equation we find

- 350. Changing the sign of x in the original equation, we have

5 x

+5; or, by transposition, 10

7 5x 7 x

10 To rectify the question, therefore, we enunciate it thus : What number is that of which exceeds To of it by 5 ?

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?

2 8. If

,
Y

2
Y
-2

Y
From these equations we find

x = -3, and y=- 13. Changing the original equations, in a manner similar to that pursued in the solution of the 5th question, we have

+ 2 and =; from which the modifica.

Y tion to be made in the enunciation is manifest.

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,-2 = }; and

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y + 2

9. Let x = the greater, and y = the less; then, X-Y = 20, and 6 x

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 question 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

52-3y=24, and 7%-5 y = 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.

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32

+

+49.

Art. 85.
1. Let a represent his money in cents. Then

5%

+ 40 = 3 12

4 This equation gives 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. 2. Let x =

A's age; then, x + 10 = B's, x+ 20 = C's, and - 34 = D's. We have, therefore, 1+10

2x+40 52 — 170
+
+

10.

= 2x

32

This equation gives

12 2 — 12 x or 0.2 = 60 - 60; and

2 = . Hence, the value of x is indeterminate; consequently, their ages cannot be ascertained from the conditions.

3. If x represent the number killed to-day, and y the number killed yesterday, we have

y

+5, and y= -5. 2

2 These equations become, by multiplication and transposition,

3x-2y=30, and 3x – 2y=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 — 3x= 20, or 0.2= 20, and x = 4. Or, by subtracting and canceling the terms containing x, we have

2y-2y=20, or 0.y=20, and y=%. 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.

SECTION XXX.

5. 36.
6. 93.
7th and 8th performed.

Art. 86.

1. 28. 2. 3. 3. 78.

4. 29. ART. 87.

1. 37. 2. 49.

3. 19.
4. 351.

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1. A's 21,

B's 27 years.

2. A 10, B 12 miles per

hour. A went 100, B 120 miles. 3. 5 yards and 6 yards. 4. 10 rows; 12 trees in a row ; 960 bushels. 5. 30 miles. 6. 50-662 + inches.

7. 12.649 + rods. 8. 43.818— rods. 9. 16. 10. Greater 60; less 15. 11. Length 30, breadth 25, height 20 feet. 12. 37 yards in each; prices 42s. and 32s. per yard. 13. 30 persons; $150 each. 14. Greater 57; less 45. 15. $150.

SECTION XXXIII.

Art. 95.

1. By the 2d statement, 9 yards. Art. 97.

1st and 2d performed.
3. 15 rods.
4. 9 rods;

per

rod. 5. Greater 90; less 85. 6. Length 100, breadth 60 rods. 7. 10 sheep; 13 calves. 8. 40 hours; 4 miles

per

hour. 9. $100. 10. 80 feet. 11. Coffee $0:10, tea $0-50 per pound. 12. 4 rods. 13. 225. 14. 5 sons. 15. 12 fowls. 16. A's stock $100. 17. 2 rods. 18. 10 gallons.

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