Letters used for unknown Quantities.

CHAPTER III.

EQUATIONS OF THE FIRST DEGREE.

SECTION I.

Putting Problems into Equations.

101. The first step in the algebraic solution of a problem is the expressing of its conditions in algebraic language; this is called putting the problem into equations.

102. No rule can be given for putting questions into equations, which is universally applicable. The following rule, can, however, be used in most cases, and problems, in which it will not succeed, must be considered as exercises for the ingenuity.

Represent the required quantities by letters of the alphabet. Perform or indicate upon these letters the same operations which it is necessary to perform upon their values, when obtained, in order to verify them.

It is usual to represent the unknown quantities by the last letters of the alphabet, as v, w, x, y, z.

103. EXAMP¡.ES.

The following problems are to be put into equations. 1. A person had a certain sum of money before him From this he first took away the third part, and put in its

Examples of putting Questions into Equations.

stead $50; a short time after, from the sum thus increased he took away the fourth part, and put again in its stead He then counted his money, and found $ 120.

$70.

What was the original sum?

Method of putting into equations. Let

≈ the original sum expressed in dollars.

After taking away the third part and putting in its stead $50, there remains two thirds of the original sum increased by $ 50, or

x+50.

If from this sum is taken a fourth part, there remains three fourths; to which is to be added $70, giving

(3x+50)+70 = { x+1071;

which is found to be equal to $120. We have, therefore, for the required equation,

2. A merchant adds yearly to his capital one third of it, but takes from it at the end of each year $1000 for his expenses. At the end of the third year, after deducting the last $1000, he finds himself in possession of twice the sum he had at first. How much did he possess originally?

Ans. If x = the original capital in dollars, the required equation is

64x 4111 =2x.

3. A courier, who goes 31 miles every 5 hours, is sent from a certain place; when he was gone 8 hours, another was sent after him at the rate of 22 miles every 3 hours. How soon will the second overtake the first?

Solution. If x = the required number of hours, the number of hours which the first courier is on the road is ≈ +8;

Examples of putting Questions into Equations.

and the distance which he goes is obtained from the proportion

5: x+8=311: distance gone by 1st courier,

whence, by art. 88,

distance gone by 1st courier

=

63(x+8).

The distance gone by the second courier is obtained from the proportion

whence

3:x= 22 distance gone by 2d courier ;

distance gone by 2d courier 15 x.

But as both couriers go the same distance, the required equation is

13(x+8)= 15 x.

4. A courier went from this place, n days ago, at the rate of a miles a day. Another has just started, in pursuit of him, at the rate of b miles a day. In how many days will the second courier overtake the first?

Ans. If x the required number of days, the required equation is

bx = a(x + n).

5. A regiment marches from the place A, on the road to B, at the rate of 7 leagues every 2 days; 8 days after, another regiment marches from B, on the road to A, at the rate of 31 leagues every 6 days. If the distance between A and B is 80 leagues, in how many days after the departure of the first regiment will the two regiments meet? Ans. If x the required number of days, the required equation is

7x+3⁄41 (x −8)=80.

Examples of putting Questions into Equations.

6. A hostile corps has set out two days ago from a certain place, and goes 27 miles daily. Another corps wishes to march in pursuit of it from the same place, and so quickly that it may reach the other in 6 days. How many miles must it march daily to accomplish it?

Ans. If x = the required number of miles, the required equation is

6x=216.

7. From two different sized orifices of a reservoir, the water runs with unequal velocities. We know that the orifices are in size as 5: 13, and the velocities of the fluid are as 8:7; we know farther, that in a certain time there issued from the one 561 cubic feet more than there did from the other. How much water, then, did each orifice discharge in this space of time?

Solution. Let x = Let the quantity discharged by the first orifice.

As the size of the second orifice is ths of that of the first, the water discharged from the second orifice, if it flowed at the same rate, would be

13 x.

But as the water flows from the second orifice with a velocity ths of that which it should have to discharge 3 z in the given time, its actual discharge must be

8. A dog pursues a hare. When the dog started, the hare had made 50 paces before him. The hare takes 6 paces to the dog's five; and 9 of the hare's paces are equal to 7 of the dog's. How many paces can the hare take before the dog catches her?

Examples of putting Questions into Equations.

Ans. If x= the required number of paces, the required

equation is

9. A work is to be printed, so that each page may contain a certain number of lines, and each line a certain number of letters. If we wished each page to contain 3 lines more, and each line 4 letters more, then there would be 224 letters more on each page; but if we wished to have 2 lines less in a page, and 3 letters less in each line, then each page would contain 145 letters less. How many lines are there in each page? and how many letters in each line? Solution. Let

x=

y

and we shall have

the number of lines in a page,

the number of letters in a line,

x y = the number of letters in a page.

But if there were 3 lines more in a page, and 4 letters more in a line, the number of letters in a page would be

(x+3) (y +4)=xy+4x+3 y + 12,

which exceeds the required number of letters in a page by 4x+3y+12;

whence we have for one of the required equations

4x+3y+12= 224 ;

and, in the same way, the condition, that 2 lines less in page and 3 letters less in a line make 145 letters less in a page, gives the equation

or

x y — (x — 2) (y — 3) = 145 ;

3x+2y-6= 145.

10. Three soldiers, in a battle, make $ 96 booty, which they wish to share equally. In order to do this, A, who made most, gives B and C as much as they already had; in