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MISCELLANEOUS EXAMPLES. 1. Given 5x+3434x +36; to find the value of x. By transposition, 53-4x=36–34. Rule lst. .. by collecting the terms, 22.

2. Given 4ax-5b=3d8 +2c; to find the value of x. By transposition, 4ax_-3dx=2c +56. Rule 1st. By collecting, (40—3d)x=2c+56.

2c+56 .. by division, x= Rule 2d.

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3. Given

3 By multiplying all the terms by 6, the least common multiple of the denominators, it becomes 21-10+ 3x=72 -2x +- 20. Rule 3d.

And by transposition, 2x + 3x + 2x=72+20+10. Rule Ist.

Hence, by collecting the terms, it becomes 7x= 102. i by division, by Rule 2d,

x=144. 68. SCHOLIUM 1. If the relation between , and the known quantities be not given in the form of an equation, but of a proportion, it may be changed into the form of an equation, by making the first term divided by the second =the third divided by the fourth, -see definitions ;-or by making the product of the extremes=that of the means. For when a : 6::c: d, by definition, 6 =,, and multiplying both sides by bd, it becomes ad=bc; or the product of the extremes is equal to that of the means.

5x+4 18-X 4. Given :: 7:4; to find the value of x.

126-72 By multiplying extremes and means, 10x+8= Multiplying by 4 it becomes 40x+32=126–78. :. by transposition, 47x=94. Rule 1st. And hence,

x=2. Rule 2d. 69. ScHoLIUM 2. When an equation appears under the form of the equality of two fractions, it may frequently be solved with much elegance, by making the sum of the numerator and denominator on the first side, divided by their difference, equal to the sum of the numerator and denominator on the second, divided by their difference. If only one side be a fraction, it may be reduced to the above form by writing 1 for a denominator on the integral side. The above principle may be demonstrated thus :

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Let i = (1.)

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(2.)
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Also g—1=:-). Ax. 2.
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(3.) And by dividing (2) by

a+b c+d (3) we obtain

the rule given above.

Ja+ sax EXAMPLE. Given

; to find x. Ja-dax

1ta By the Scholium,

2a
2/a><

lta
VA

1-a

1+2ata? Or by squaring,

1—2a+amo

1—2ata? By inverting,

1+2a+a?.

1-2ata? Reducing the first side, 1

1+2a+a

1--2ataa Transposing,

1+2a+a2

4a Reducing the first side,

(1+a)"
4a?

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EXERCISES.

1. Given -6+25=36; to find x.
2. Given 3-5=23_X; to find x.
3. Given 73-3=5x + 13; to find X.
4. Given 3x+5=10x-16; to find x.
5. Given 2x +11=7x-14; to find s.
6. Given 15x—24=20+£x; to find x.
7. Given x+=4x—17; to find X.

Ans. x=17.
Ans. x=7.
Ans. x=8.
Ans. X=3.
Ans. x=5.
Ans. x=3.

Ans. x=6.

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9. Given +*+ő tö+

=46; to find X. Ans.x=48. 10. Given + - =}; to find x.

Ans. x=. 11. Given *+++*?=14+*; to findt. Ans.x=13.

5x+1 12. Given

=16- ; to find X. Ans. x=7.

4

x+10 a 13. Given um

+4;

to find x. Ans.x=5.

x+1
2

+

2+2
3

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2214. Given at

-12- ; to find X. Ans. x=5. 2 1 6

ab

Ans. x= 15. Given ät=c; to find x. 16. Given ax+69=bx+a'; to find x.

Ans. x=a+b.

a+2c 17. Given bx+2x-a=3x +2c; to findx. Ans.de

ata 618. Given 3x—; +c==

; to find s. 3

3a+ab_363 Ans. x=

8ab+ 3abc-3b

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PROBLEMS IN SIMPLE EQUATIONS.

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EXAMPLE 1. What number is that to which, if its half and its fifth part be added, the sum will be 34 ?

Let & represent the number sought; then its half will be and its fifth part will be . Therefore we will have by 23 the question,

x+3+5= 34, and multiplying by 10; 10x +5x + 2x=340, Rule 3d; hence by collecting the terms, 17x=340,

x= 20, by Rule 2d. 2. What number is that whose third part exceeds its seventh part by 4?

Let x represent the number sought; then its third part will be, and its seventh part. Therefore by the question,

37

4,

ge multiplying both sides by 21, 7x43x=84, Rule 3d; hence collecting the terms,

4x=84, .: x=21, Rule 2d.

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3. A cistern can be filled by two pipes in 12 hours, and by the first of them alone in 20 hours: in what time would it be filled by the second alone?

Put x for the time required:
Then would be the quantity supplied by the second

12 in 12 hours; and would be the quantity supplied by

20 the first in 12 hours. But in twelve hours the two running together filled the

12

12 cistern,

-+ =l, and multiplied by 20x, 240+ 12x=20x, Rule 3d; by transposition, &c.

240= 8x, Rule 1st;

30= X, Rule 2d. 4. How much rye, at four shillings and sixpence a bushel, must be mixed with 50 bushels of wheat, at six shillings a bushel, that the mixture may be worth five shillings a bushel ?

Let x be the number of bushels of rye.
Then 9x= the price of the rye in sixpences.

600= the price of the wheat in sixpences. (50+x)10= the price of the mixture.

9x+600=500+ 10x by the question.

Hence 100=x by transposition. 70. ScHolium. The transferring of problems into algebraic equations will be facilitated by studying carefully the following remarks:—1st, If the sum of two numbers be given, and one of the numbers be represented by x, then the

other will be the sum minus X. 2d, If the difference of two numbers be given, and the less be represented by x, the other will be x plus the given difference; and if the greater be represented by x, the other will be x minus the difference. 3d, If the product of two numbers be given, and one of them be represented by x, the other will be the product divided by x. 4th, If the quotient of two numbers be given, and one of them be represented by x, the other will be the quotient multiplied by x. 5th, If the sum of two numbers be represented by s, and the less number by X, then the difference of the numbers will be s-2x; and if the greater be represented by x, their difference will be (2x-8).

19. What number is that which being increased by its half and its third part, the sum will be 154?

Ans. 84.

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20. What number is that to which, if its third and fourth parts be added, the sum will exceed its sixth part by 17?

Ans. 12. 21. At a certain election; 311 persons voted, and the successful candidate had a majority of 79: how many voted for each ?

Ans. 195 and 116. 22. What number is that from which, if 50 be subtracted, the remainder will be equal to its half, together with its fourth and sixth parts ?

Ans. 600. 23. A husband, on the day of his marriage, was thrice as old as his wife, but after they had lived together 15 years, he was only twice as old: what were their ages on the marriage day?

Ans. Husband's, 45; wife's, 15. 24. It is required to divide L.300 between A, B, and C, so that A may have twice as much as B, and C as much as A and B together: what was the share of each?

Ans. A's, L.100; B's, L.50; and C's, L.150. 25. A cistern can be filled with water by one pipe in 12 hours, and by another in 8: in what time would it be filled by both running together?

Ans. 41 hours. 26. Two pieces of cloth, which together measured 40 yards, were of equal value, and the one sold at 3s., and the other at 7s. a-yard: how many yards were of each?

Ans. 12 yards at 7s., and 28 at 3s. 27. A has three times as much money as B, and if B give him L.50, A will have four times as much as B: find the money of each. Ans. A's, L.750; B's, L.250.

28. After 34 gallons had been drawn from one of two equal casks, and 80 from the other, there remained thrice as much in the first as in the second: what did each contain when full?

Ans. 103 gallons. 29. A person paid a bill of L.100 with half-guineas and crowns, using in all 202 pieces: how many pieces were there of each sort? Ans. 180 half-guineas, 22 crowns.

30. There is a cistern which can be supplied with water from three different pipes; from the first it can be filled in 8 hours, from the second in 16 hours, and from the third in 10 hours: in what time will it be filled if the three pipes be all set running at the same time?

Ans. 3 hours 2815 minutes. 31. Solve the above question generally on the supposition that the first pipe can fill the cistern in a hours, the second in b, and the third in c.

abc Ans.

ab+ac+bc

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