Algebra for High Schools and Colleges

PROBLEMS

In Pure Equations and Affected Quadratics containing but One Unknown Quantity.

1. Find two numbers such that their product shall be 750, and the quotient of the greater divided by the less, 3.

Let x represent the greater of the two numbers;

750

then will represent the less; and the Equation will be

750 x2

or =3.

х 750

From this equation we shall find x=50, or -50.

Each of these

values will satisfy the Equation of the problem; but only the positive one can be taken to answer the conditions of the problem itself, in which the required numbers are understood to be positive, as in the problems of common Arithmetic. Ans. 50, and 15. .

2. Find a number such that if and of it be multiplied together, and the product divided by 3, the quotient will be 2983. Ans. 224.

3. A mercer bought a piece of silk for £16 4s.; and the number of shillings that he paid per yard, was to the number of yards, as 4 to How many yards did he buy? and what was the price per yard?

Let a represent the number of shillings he paid per yard;

then 49::x: the number of yards.

9x
,
4

But without forming a Proportion, the number of yards is readily known to be of the price per yard.

Ans. 27 yards, at 12s. per yard.

4. Find two numbers which shall be to each other as 2 to 3, and the sum of whose squares shall be 208. Ans. 8 and 12.

5. A person bought a quantity of cloth for $120; and if he had bought 6 yards more for the same sum, the price per yard would have been $1 less. What was the number of yards? and the price per yard? Ans. 24 yards, at $5 per yard.

6. Divide the number 20 into two such parts that the squares of these parts may be in the proportion of 4 to 9. Ans. 8, and 12.

7. A merchant bought a quantity of flour for $100, which he sold again at $5 per barrel, and in so doing gained as much as each barrel cost him. What was the number of barrels ? Ans. 20.

8. Divide the number 800 into two such parts that the less divided by the greater, may be to the greater divided by the less, as 9 to 25. Let x represent the less number:-we shall then have the Proportion

800- х

800-x

:9:25;

which will be converted into an Equation by putting the product of the two extremes equal to the product of the two means.

Ans. 300, and 500 9. Two fields which differ in quantity by 10 acres, were each sold for $2800, and one of them was valued at $5 an acre more than the other. What was the number of acres in each? Ans. 70, and 80.

10. A and B started together on a journey of 150 miles. A traveled 3 miles an hour more than B, and completed the journey 8 hours before him. At what rate did each travel per hour?

Ans. 9, and 6 miles. 11. A man traveled 105 miles, and then found that if he had gone 2 miles less per hour, he would have been 6 hours longer on his jourAt what rate did he travel per hour? Ans. 7 miles.

ney.

12. A person has two pieces of silk which together contain 14 yards. Each piece is worth as many shillings per yard as there are yards in the piece, and their whole values are in the proportion of 9 to 16; how many yards are there in each piece? Ans. 6, and 8 yards. 13. A merchant sold a piece of linen for $39, and in so doing gained as much per cent. as it cost him. What was the cost of the linen?

Ans. $30.

14. A grazier bought as many sheep as cost him $100. After reserving 5 of the number, he sold the remainder for $135, and gained $1 a head on them; how many sheep did he buy?

Ans. 50.

15. Find two numbers which shall be in the proportion of 7 to 9, and have the difference of their squares equal to 128.

Ans. 14, and 18.

16. An officer would arrange 2400 men in a solid body, so that each rank may exceed each file by 43 men. How many must be placed in rank and file?

Ans. 75, and 32.

17. Two partners gained £18 by trade. A's money was employed in the business 12 months; and B's, which was £30, 16 months. A received for his capital and gain £26; what was the amount of his capital?

Let x represent A's capital; then 26-x will be his gain; and since the gain is in the compound ratio of the capital and the time it was employed, we have

12x+16 × 30: 12x: 18: 26-x.

The first ratio in this proportion may be simplified by dividing the antecedent and the consequent by 12, (158). Ans. £20.

18. A detachment from an army was marching in regular column, with 5 men more in depth than in front; but upon the enemy's coming in sight, the front was increased by 845 men; and by this movement the detachment was drawn up in five lines. What was the number of men? Ans. 4550.

19. A company at a tavern had £8 15s. to pay, but before their bill was settled, two of them went away, when those who remained had 10s. apiece more to pay than before. How many were there in the company at first?

Ans. 7.

20. Some gentlemen made an excursion, and each one took the same sum. Each gentleman had as many servants as there were gentlemen, and the number of dollars which each had was double the whole number of servants; also the whole sum taken with them was $3456. What was the number of gentlemen? Ans. 12. 21. Divide the number 20 into two such parts, that the product of the whole number and one of the parts shall be equal to the square of the other. Ans. 105-10, and 30-10/5. 22. A laborer dug two trenches, one of which was 6 yards longer than the other, for £17 16s., and the digging of each cost as many shillings per yard as there were yards in its length. What was the length of each? Ans. 10, and 16 yards.

23. There are two numbers whose product is 120, and if 2 be added to the less, and 3 subtracted from the greater, the product of the sum and the remainder will also be 120. What are the two numbers? Ans. 8 and 15.

24. Two persons lay out some money on speculation. A disposes of his bargain for £11, and gains as much per cent. as B lays out; B's gain is £36, and it appears that A gains 4 times as much per cent. as B. What sum did each lay out? Ans. A £5, B £120.

25. A set out from C towards D, and traveled 7 miles an hour. After he had gone 32 miles, B set out from D towards C, and went each hour of the whole distance; and after he had traveled as many hours as he went miles in one hour, he met A. Required the distance between the two places. Ans. 152, or 76 miles.

SOLUTION OF Two EQUATIONS-ONE OR BOTH OF THE SECOND OR A HIGHER DEGREE-CONTAINING TWO UNKNOWN QUANTITIES.

(264) For the solution of two Equations, containing two unknown quantities, the method which naturally occurs is,

1. By elimination between the given equations to derive a new equation containing but one of the unknown quantities, and thence to find the value of that quantity.

2. By substituting this quantity for its symbol in one of the equations containing the other unknown quantity, to determine thence the value of that quantity.

There are, however, some facilitating expedients to be applied, in certain cases, to equations of the second and higher degrees: these will be exemplified as we proceed.

But the solution of two Equations—one or both of the second or a higher degree-containing two unknown quantities, may be impossible by the method of quadratics,-from the impossibility of deriving from them a new equation containing but one unknown quantity, which will admit of a quadratic solution

EXAMPLES

1. Find the values of x and y in the equations
2x+y=10, and 2x2-xy+3y2=54.

By substituting these values in the second eqnation, we find

2(100—20y+y2), 10y—y2

+3y2=54.

The value of y may be found from this equation; and by substi tuting the value of y for y in the first equation, the value of x may readily be determined.

Observe that the left hand fraction in the last equation may be reduced to lower terms; and the solution of the equation be thus somewhat simplified.

Ans. x=3, or; y=4, or -4.

2. Find the values of x and y in the equations.

x+y=9, and x2+y2=45.

Ans. x 3, or 6; y=6, or 3

Whenever x and y may be interchanged with each other, without changing the form of the given equations-as in the preceding example-the two values of one of these letters may be taken, in reverse order, for the two values of the other.

3. Find the values of x and y in the equations xy=28, and x2+y2=65.

Ans. x=7, or ±4; y=±4, or ±7.

4. Find the values of x and y in the equations
x+4y=14, and 4x-2y+y?=11.