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CHAPTER X.

PROBLEMS LEADING TO SIMPLE EQUATIONS.

77. The principles of the last chapter may now be employed to solve various problems.

The method of procedure is as follows:

Represent the unknown quantity by a symbol, as x, and express in symbolical language the conditions of the question ; we thus obtain a simple equation which can be solved by the methods already given in Chapter viii.

Example 1. Find two numbers whose sum is 28, and whose difference is 4.

Let x be the smaller number, then x+4 is the greater.
Their sum is x + (x+4), which is to be equal to 28.
Hence

x + x +4=28;

2x=24;

.: x=12, and

x +4=16, so that the numbers are 12 and 16.

The beginner is advised to test his solution by proving that it satisfies the conditions of the question.

Example 2. Divide 60 into two parts, so that three times the greater may exceed 100 by as much as 8 times the less falls short of 200.

Let x be the greater part, then 60 – x is the less.
Three times the greater part is 3x, and its excess over 100 is

3x – 100.
Eight times the less is 8 (60 – x), and its defect from 200 is

200 – 8 (60 – x).
Whence the symbolical statement of the question is

3x - 100=200 – 8 (60 – x);

3.x - 100=200 – 480 +8x, 480 -- 100 - 200=8x - 3x,

5x=180;

:. x=36, the greater part, and

60 – x=24, the less.

Example 3. Divide $47 between A, B, C, so that A may have $10 more than B, and B $8 more than C.

Suppose that C has x dollars; then B has x+8 dollars, and A has x+8+10 dollars. Hence

x+(2+8)+(x+8+10)=47;
2+3+8+*+8+10=47,

3x=21;

.. X=7; so that C has $7, B $15, A $25.

Example 4. A person spent $112.80 in buying geese and ducks ; if each goose cost 14 dimes, and each duck 6 dimes, and if the total number of birds bought was 108, how many of each did he buy?

In questions of this kind it is of essential importance to have all quantities expressed in the same denomination; in the present instance it will be convenient to express the money in dimes.

Let x be the number of geese, then 108– is the number of ducks.

Since each goose costs 14 dimes, x geese cost 14x dimes.

And since each duck costs 6 dimes, 108 — ducks cost 6(108—2) dimes. Therefore the amount spent is

14x + 6(108 — ) dimes, but the question states that the amount is also $112.80, that is, 1128 dimes. Hence

14x+6(108 — 2)=1128; dividing by 2, 7x+324 3x = 564,

4x=240,

x=60, the number of geese, and

108 — X=48, the number of ducks. Example 5. A is twice as old as B; ten years ago he was four times as old; what are their present ages ?

Let B's age be x years, then A's age is 2x years.

Ten years ago their ages were respectively, x— 10 and 2x – 10 years; thus we have

2x-10=4(3— 10);
2x-10=4x — 40,

2x=30;

... x = =15, so that B is 15 years old, A 30 years.

Note. In the above examples the unknown quantity x represents a number of dollars, ducks, years, &c.; and the student must be careful to avoid beginning a solution with a supposition of the kind, “let x= A's share” or “let x=the ducks,” or any statement so vague and inexact.

EXAMPLES X. 1. One number exceeds another by 5, and their sum is 29; find them.

2. The difference between two numbers is 8; if 2 be added to the greater the result will be three times the smaller: find the numbers,

3. Find a number such that its excess over 50 may be greater by 11 than its defect from 89.

4. A man walks 10 miles, then travels a certain distance by train, and then twice as far by coach. If the whole journey is 70 miles, how far does he travel by train?

5. What two numbers are those whose sum is 58, and difference 28?

6. If 288 be added to a certain number, the result will be equal to three times the excess of the number over 12: find the number.

7. Twenty-three times a certain number is as much above 14 as 16 is above seven times the number: find it.

8. Divide 105 into two parts, one of which diminished by 20 shall be equal to the other diminished by 15.

9. Find three consecutive numbers whose sum shall equal 84.

10. The sum of two numbers is 8, and one of them with 22 added to it is five times the other: find the numbers.

11. Find two numbers differing by 10 whose sum is equal to twice their difference.

12. A and B begin to play each with $60. If they play till A's money is double B's, what does A win?

13. Find a number such that if 5, 15, and 35 are added to it, the product of the first and third results may be equal to the square of the second.

14. The difference between the squares of two consecutive numbers is 121: find the numbers.

15. The difference of two numbers is 3, and the difference of their squares is 27: find the numbers.

16. Divide $380 between A, B, and C, so that B may have $30 more than A, and C may have $20 more than B.

17. A sum of $7 is made up of 46 coins which are either quarters or dimes : how many are there of each?

18. If silk costs five times as much as linen, and I spend $48 in buying 22 yards of silk and 50 yards of linen ; find the cost of each per yard.

19. A father is four times as old as his son: in 24 years he will only be twice as old; find their ages.

20. A is 25 years older than B, and A's age is as much above 20 as B's is below 85 : find their ages.

21. A's age is six times B's, and fifteen years hence A will be three times as old as B: find their ages.

22. A sum of $16 was paid in dollars, half-dollars, and dimes. The number of half-dollars used was four times the number of dollars and twice the number of dimes; how many were there of each ?

23. The sum of the ages of A and B is 30 years, and five years hence A will be three times as old as B: find their present ages.

24. I purchase 127 bushels of grain. If the number of bushels of wheat be double that of the corn, and seven more than five times the number of bushels of corn equals the number of bushels of oats, find the number of bushels of each.

25. The length of a room exceeds its breadth by 3 feet; if the length had been increased by 3 feet, and the breadth diminished by 2 feet, the area would not have been altered: find the dimensions.

26. The length of a room exceeds its breadth by 8 feet; if each had been increased by 2 feet, the area would have been increased by 60 square feet: find the original dimensions of the room.

CHAPTER XI.

RESOLUTION INTO FACTORS.

78. DEFINITION. When an algebraical expression is the product of two or more expressions each of these latter quantities is called a factor of it, and the determination of these quantities is called the resolution of the expression into its factors.

In this chapter we shall explain the principal rules by which the resolution of expressions into their component factors may be effected.

WHEN EACH OF THE TERMS IS DIVISIBLE BY A COMMON

FACTOR. 79. The expression may be simplified by dividing each term separately by this factor, and enclosing the* quotient within brackets; the common factor being placed outside as a coefficient.

Example 1. The terms of the expression 3a2—6ab have a common factor 3a;

... 3a2–6ab=3a(a-26). Example 2. 5a2bx3 – 15abx2 – 2003x2=5bx2(a2x — 30–462).

EXAMPLES XI. a.

Resolve into factors :

1. Q: — ax.

2. 23 - 22. 3. 2a-2a2. 4. a2-ab2. 5. 7p2 +p.

6, 8x — 2x2. 7. 5ax-5a3x2. 8. 3x2+005. 9. 202 + xy. 10. cô –co. 11. 5x – 25x2y. 12. 15+25x2. 13. 16x+64x2y. 14. 1592-225a4. 15. 54–81x. 16. 10x3 — 25x4y. 17, 3x3 — 32+x. 18. 6x3 +234 +4x5.

19. 23 — 22y + xy2. 20. 3a4 – 3a’b+62262.

21. 2x2y3 6x2y2 + 2xy3. 22. 6x3 - 9xły +12xy2.

23. 5x5 — 10a_x3 – 15a3x. 24. 7a-7a3 +14a4.

25. 38a3x5 +57a4x2.

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