Charles Davies.

First lessons in algebra, embracing the elements of the science online

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the problem.

Let x=z the number of leaps made by the greyhound
before he overtakes the fox.

Now, since the fox makes 9 leaps while the greyhound

9 3

makes but 6, the fox will make — - or — - leaps while

o &

8



86 FIRST LESSONS IN ALGEBRA.

the greyhound makes 1 ; and, therefore, while the greyhound

3
makes x leaps, the fox will make —x leaps.

Z

Hence, the distance which the greyhound must pass over
3

will be expressed by 60-{ x leaps of the fox.

Z

It might be supposed, that in order to obtain the equation,

3
it would be sufficient to place x equal to 60 -j — —x; but

z

in doing so, a manifest error would be committed ; for the

leaps of the greyhound are greater than those of the fox, and

we should then equate numbers of different denominations ;

that is, numbers referring to different units. Hence it is

necessary to express the leaps of the fox by means of those

of the greyhound, or reciprocally. Now, according to the

enunciation, 3 leaps of the greyhound are equivalent to 7

leaps of the fox, then 1 leap of the greyhound is equiva-

7
lent to — leaps of the fox, and consequently x leaps of

7x
the greyhound are equivalent to -— ■ of the fox.

o

Hence, we have the equation

7* ™ , 3
T =60+-* ;

making the denominators disappear

14* =360 +9,?,

whence 5* =360 and x =72.

Therefore the greyhound will make 72 leaps to overtake
the fox, and during this time the fox will make

3

72 x— or 108.
z



I



UNIVERSITY J

or J

EQUATIONS OF THE FIRST DEGREE. 87

Verification*
The 72 leaps of the greyhound are equivalent to

lilies

leaps of the fox. And

60+108 = 168,

the leaps which the fox made from the beginning.

13. A father leaves his property, amounting to $2520, to
four sons, A, B, C, and D. C is to have $360, B as much
as C and D together, and A twice as much as B less
$1000 : how much does A, B, and D receive ?

Ans. A $760, B $880, D $520.

14. An estate of $7500 is to be divided between a widow,
two sons, and three daughters, so that each son shall receive
twice as much as each daughter, and the widow herself
$500 more than all the children : what was her share, and
what the share of each child ?

• Widow's share $4000.
Ans. 1 Each son's $1000.

' Each daughter's $500.

15. A company of 180 persons consists of men, women,
and children. The men are 8 more in number than the
women, and the children 20 more than the men and women
together : how many of each sort in the company ?

Ans. 44 men, 36 women, 100 children.

16. A father divides $2000 among five sons, so that each
elder should receive $40 more than his next younger bro-
ther : what is the share of the youngest? Ans. $320.

17. A purse of $2850 is to be divided among three per-
sons, A, B, and C. A's share is to be to B's as 6 to 11.



i



88 FIRST LESSONS IN ALGEBRA.

and C is to have $300 more than A and B together : what
is each one's share ? ■

Ans. A's $450, B's $825, C's $1575.

18. Two pedestrians start from the same point; the first
steps twice as far as the second, but the second makes 5
steps while the first makes but one. At the end of a certain
time they are 300 feet apart. Now, allowing each of the
longer paces to be 3 feet, how far will each have travelled?

Ans. 1st, 200 feet ; 2nd, 500.

19. Two carpenters, 24 journeymen, and 8 apprentices,
received at the end of a certain time $144. The carpen-
ters received $1 per day, each journeyman half a dollar,
and each apprentice 25 cents : how many days were they
employed? Ans. 9 days.

20. A capitalist receives a yearly income of $2940 : four-
fifths of his money bears an interest of 4 per cent, and the
remainder at 5 per cent : how much has he at interest ?

Ans. 70000.

21. A cistern containing 60 gallons of water has three
unequal cocks for discharging it ; the largest will empty it
in one hoar, the second in two hours, and the third in three :
in what time will the cistern be emptied if tliey all run
together? Ans. 32^ T min.

22. In a certain orchard -J are apple trees, \ peach trees,
J plum trees, 120 cherry trees, and 80 pear trees : how
many trees in the orchard ? Ans. 2400.

23. A farmer being asked how many sheep he had,
answered that he had them in five fields; in the 1st he
had 1, in the 2nd -|, in the 3rd J, and in the 4th A, and in
the 5th 450 : how many had he ? Ans. 1200.

24. My horse and saddle together are worth $132, and
the horse is worth ten times as much as the saddle : what
is the value of the horse ? Ans. 120



EQUATIONS OF THE FIRST DEGREE. 89

25. The rent of an estate is this year 8 per cent greater
than it was last. This year it is $1890 : what was it last
year? Ans. $1750„

26. What number is that from which, if 5 be subtracted,
^ of the remainder will be 40 | Ans. 65 %

27. A post is 1 in the mud, J in the water, and 10 feet
above the water : what is the whole length of the post 1

Ans. 24 feet.

28. After paying \ and \ of my money, I had 66 guineas
left in my purse : how many guineas were in it at first ?

Ans. 120.

29. A person was desirous of giving 3 pence apiece to
some beggars, but found he had not money enough in his
pocket by 8 pence : he therefore gave them each 2 pence
and had 3 pence remaining : required the number of beggars.

Ans. 11.

30. A person in play lost i of his money, and then won
3 shillings ; after which he lost -J- of what he then had ;
and this done, found that he had but 12 shillings remaining :
what had he at first 1 Ans. 20s.

31. Two persons, A and B, lay out equal sums of money
in trade ; A gains $126, and B loses $87, and A.'s money
is now double of B's : what did each lay out ?

Ans. $300.

32. A person goes to a tavern with a certain sum of mo-
ney in his pocket, where he spends 2 shillings ; he then
borrows as much money as he had left, and going to another
tavern, he there spends 2 shillings also ; then borro wing-
again as much money as was left, he went to a third tavern,
where likewise he spent two shillings and borrowed as
much as he had left ; and again spending 2 shillings at a
fourth tavern, he then had nothing remaining. What had
he at first ? Ans. 3s. 9d.

8*



90 FIRST LESSONS IN ALGEBRA.



Of Equations of the First Degree involving two or
more unknown quantities.

72. Although several of the questions hitherto resolved
contained in their enunciation more than one unknown quan-
tity, we have resolved them all by employing but one sym-
bol. The reason of this is, that we have been able, from
the conditions of the enunciation, to express easily the other
unknown quantities by means of this symbol ; but we are
unable to do this in all problems containing more than one
unknown quantity.

To ascertain how problems of this kind are resolved, let
us take some of those which have been resolved by means
of one unknown quantity.

1. Given the sum of two numbers equal to 36 and their
difference equal to 12, to find the numbers.

Let x= the greater, and y= the less number.

Then, by the 1st condition #+y=36,

and by the 2nd, x— y=l2.

By adding (Art. 65, Ax. 1), .... 2*=48.

By subtracting (Art. 65, Ax. 2), . . . 2y=24.

Each of these equations contains but one unknown quantity.

48
From the first we obtain a?=— =24.

Z

24

And from the second ?/=— = 12.

2

Verification.

x+y=36 gives 24+12 = 36.
a— y=12 „ 24-12 = 12.



EQUATIONS OF THE FIRST DEGREE. 91

General Solution.
Let x = the greater, and y the less number.

Then by the conditions #+y=#,

and x—yz=.b.

By adding, (Art. 65, Ax. 1), . . . 2x=a+b.

By subtracting, (Art. 65, Ax. 2), . . 2y = a—b>

Each of these equations contains but one unknown quantity,

a+b
From the first we obtain x= — — .

z

a — b
And from the second y=— - — .

y 2

Verification.

a-\-b a — b 2a . a+b a—b 2b
— — — a: and — = — =b.

2^22' 2 22

For a second example, let us also take a problem that has
been already solved.

2, A person engaged a workman for 48 days. For each
day that he labored he was to receive 24 cents, and for each
day that he was idle he was to pay 12 cents for his board.
At the end of the 48 days the account was settled, when
the laborer received 504 cents. Required the number of
working days, and the number of days he was idle.

Let x= the number of working days,

y= the number of idle days.
Then, 24x = what he earned,
and 12y= what he paid for his board.

Then, by the conditions of the question, we have

x-\-y =48,
and 24a: — 12y = 504.

This is the statement of the question.



92 FIRST LESSONS IN ALGEBRA.

It has already been shown (Art. 65, Ax. 3), that the two
members of an equation can be multiplied by the same num-
ber, without destroying the equality. Let, then, the first
equation be multiplied by 24, the coefficient of x in the
second : we shall then have

24a?+24y=1152,

2 4a— 12y = 504.

And by subtracting, 36y= 648,

648 ,o

and V = -36-= 18 '

Substituting this value of y in the equation

24a?— 12y = 504, we have 24a?— 216 = 504,

which gives

720
24*:=: 504 + 2 16 = 720, and x= — —=30.

24

Verification.

x+ y— 48 gives 30+18= 48,

24a:— 12y=504 gives 24x30 — 12x18 = 504.

Elimination.

73. The method which has just been explained of com-
bining two equations, involving two unknown quantities, and
deducing therefrom a single equation involving but one, is
called elimination.



Quest. — 73. What is elimination 1 How many methods of elimina-
tion are there 1 Give the rule for elimination by addition and subtrac-
tion ! What is the first step 1 What the second ? What the third 1



EQUATIONS OF THE FIRST DEGREE. 93

There are three principal methods of elimination :

1st. By addition and subtraction.

2d. By substitution.

3d. By comparison.
We will consider these methods separately.

Elimination by Addition and Subtraction.

1. Take the two equations

3x— 2y=7
&*+2y=48.

If we add these two equations, member to member, we
obtain

Ux=z55.
which gives, by dividing by 11

x— 5 :

and substituting this value in either of the given equations,
we find

y=4.

2. Again, take the equations

8x-\-2y=z48
3x+2y=23.

If we subtract the 2nd equation from the first, we obtain
5x=25,
which gives, by dividing by 5

x=z5 :
and by substituting this value, we find

y=4.



94 FIRST LESSONS IN ALGEBRA.

3. Take the two equations

5#+7y = 43.
ll#+9y=69.

If, in these equations, one of the unknown quantities was
affected with the same coefficient, we might, by a simple
subtraction, form a new equation which would contain but
one unknown quantity.

Now, if both members of the first equation be multiplied
by 9, the coefficient of y in the second, and the two mem-
bers of the second by 7, the coefficient of y in the first, we
will obtain

45#+63y = 387,
77*+63y = 483.

Subtracting, then, the first of these equations from the
second, there results

32a? =96, whence x =3.

Again, if we multiply both members of the first equation
by 11, the coefficient of x in the second, and both members
of the second by 5, the coefficient of x in the first, we will
form the two equations

55#+77y = 473,
55#+45y=345.

Subtracting, then, the second of these two equations from
the first, there results

32y = 128, whence y=4.
Therefore x =3 and y=4, are the values of x and y.

Verification.
5*+7y=43 gives 5x3 + 7x4 = 15+28=43;
lla?+95=69 „ 11x3 + 9x4=33 + 36 = 69.



EQUATIONS OF THE FIRST DEGREE. 95

The method of elimination just explained, is called the
method by addition and subtraction.

To eliminate by this method we have the following

RULE.

I. See which of the unknown quantities you will eliminate.

II. Make the coefficient of this unknown quantity the same
in both equations, either by multiplication or division.

III. If the signs of the like terms are the same in both
equations, subtract one equation from the other ; but if the
signs are unlike, add them.

EXAMPLES.

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

3x — y=3,

y-\-2x — 7.

Ans. x=z2, y=3

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

4x-7y=z -22,
5x+2y=37.

Ans. x=5, y=z6.

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

2x+6y = 42,
8x—6y=z 3.

Ans. x =4^, y=z5^.

7. Find the values of .t and y in the equations

8x — 9y=l.
6x—3y=z4x.

Ans. *==£, yssft



96 FIRST LESSONS IN ALGEBRA.

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

14*— 15y=12,
7*+ 8y=r37.

Ans. x=3, y=2.

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

1 j_ X A






— *+— y= 6 i-



3^2*

Jtis*. a?=6, y=9.

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

x— y~— 2.

A;z5. a?=14, y=16.

11. Says A to B, you give me $40 of your money, and
I shall then have 5 times as much as you will have left.
Now they both had $120 : how much had each 1

Ans. Each had $60.

12. A Father says to his son, " twenty years ago, my
age was four times yours ; now, it is just double :" what were
their ages ? . ( Father's 60 years.

< Son's 30 years.

13. A Father divides his property between his two sons.
At the end of the first year the elder had spent one quarter
of his, and the younger had made $1000, and their property
was then equal. After this the elder spends $500 and the
younger makes $2000, when it appears the younger has
just double the elder : what had each from the father ?

Elder $4000.



1 Younger $2000.



EQUATIONS OF THE FIRST DEGREE. 97

14. If John give Charles 15 apples, they will have the
same number; but if Charles give 15 to John, John will
have 15 times as many wanting 10 as Charles will have
left. How many had each 1 \ i John 50.

( Charles 20.

15. Two clerks, A and B, have salaries which are to-
gether equal to $900. A spends -^ per year of what he
receives, and B adds as much to his as A spends. At the
end of the year they have equal sums : what was the salary

of each? , < A's=:500.

Ans. <

( B's=400.



7 Elimination by Substitution.

7 4. Let us again take the equations

5a:+7y:=43,
lla:+9y=i69.

Find the value of x in the first equation, which gives



Substitute this value of x in the second equation, and we
have

llX 43 J 7y +9y=69,

or, 473 — 77y+45y = 345,

or, — 32y= — 128.

Hence, y=4,

43-28 ■
and, a?= — - — =3.

5

9



98 FIRST LESSONS IN ALGEBRA.

This method is called the method by substitution: we
have for it the following

RULE.

Find the value of one of the unknown quantities in either
of the equations, and substitute this value for the same unknown
quantity in the other equation : there will thus arise a new
equation with but one unknown quantity.

Remark. — This method of elimination is used to great
advantage when the coefficient of either of the unknown
quantities is unity.

EXAMPLES.

1. Find, by the last method, the values of x and y in the
equations

3x — y — l and 3y— 2#=4

Ans. x=l, y=^2.

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

by — 4x=— 22 and 3y + 4a:=:38.

Ans. # — 8, y=2.

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

x+8y = 18 and y — 3x=— 29.

Ans. x=zlO, y=l.

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

2
5x—y=13 and 8x-\-—y=29.

Ans. x =3 J, y—\\.

Quest. — 74. Give the rule for elimination by substitution ? When is
it desirable to use this method !



EQUATIONS OF THE FIRST DEGREE. $9

5. Find the values of x and y from the equations

10a?— —=69 and lOy— — =49.
5 y 7

Ans. g=s7f y = 5.

6. Find the values of x and y from the equations

*+T*-f= 10 and T*fe?*

Arcs. a?=8, y=10.

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

T-y +5 = 3 ' -+f^i7i.

Ans. #=15, y=14.
8 Find the values of x and y in the equations

Arcs, a? =3 J, y=4.
9. Find the values of x and y from the equations

4 h6 = 5 and tt:=0.

8 4^ 12 16

-y-~ A/i5. #=12, y=16.

10. Find the values of x and y from the equations

-|-|-l = -9 and 5*-g=29.

A/i5. #=6, y=7.

11. Two misers A and B sit down to count over their
money. They both have $20000, and B has three times as
much as A : how much has each 1

A . . $5000.
$15000.



Ans. \ B *



FIRST LESSONS IN ALGEBRA.

A person has two purses. If he puts $7 into the
first purse, it is worth three times as much as the second :
but if he puts $7 into the second it becomes worth five
times as much as the first : what is the value of each purse ?

Ans. 1st, $2: 2nd, $3.

13. Two numbers have the following properties : if the
first be multiplied by 6 the product will be equal to the
second multiplied by 5 ; and one subtracted from the first
leaves the same remainder as 2 subtracted from the second :
what are the numbers 1 Ans. 5 and 6.

14. Find two numbers with the following properties :
the first increased by 2 to be 3^ times greater than the
second : and the second increased by 4 to be half the first :
what are the numbers ? Ans. 24 and 8.

15. A father says to his son, " twelve years ago I was
twice as old as you are now : four times your age, at that
time, plus twelve years, will express my age twelve years
hence :" what were their ages ? . i Father 72 years.

7U ' I Son 30 "

Elimination by Comparison.

75. Take the same equations

5a+7y=43,
U*+9y=69.

Finding the value of x in the first equation, we have

43-7y

*=— 5 l

and finding the value of x in the second, we obtain
- 69— 9y

*- ii y



EQUATIONS OF THE FIRST DEGREE. 101

Let these two values of x be placed equal to each other,
and we have

43— 7y _ 69 — 9y
5 ~ 11 '

Or, 473-77y=:34-5-45y ;

Or, — 32y= — 128.

Hence, y^^.

A A 69 ~ 36 Q

And, x= — — = 3.

This method of elimination is called the method by
c )mparison, for which we have the following

RULE.

I. Find the value of the same unknown quantity in each
e^ nation.

II. Place these values equal to each other; and a new
equation will arise with but one unknown quantity.

EXAMPLES.

1. Find, by the last rule, the values of x and y in the
equations

3z+|-+6=42 and y~=U%.

Ans. #=11, y = 15.



Quest. — 75. Give the rule for elimination by comparison 1 What is
the first step 1 What the second 1

9*



102 FIRST LESSONS IN ALGEBRA.

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

f-f+5=6 and \ +4=^+6.

Ans. #=28, y=20.

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

~ =1 and 3y—xz=6.

10 4 8 y

Ans. x=9, y—5.

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

y— 3 = —x+5 and ^-^-=y— 3|.

^4tt,y. a;— 2, y=9.

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

V — x x • , x , y

^Ln^. # = 16, y = 7.

6. Find the values of x and y from the equations

y+x y—x 2y

Z h- =% — ;r> and x+y = l6.

2^2 3 ' y

^4^5. # = 10, y=6.

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

2x—3y 2 y— 1

Ans. a?=l, y=3.

8. Find the values of and y in the equations

2y+3*=:y+43, y-~=y-y.

4/w. a;s=10, y=13.



EQUATIONS OF THE FIRST DEGREE. 103

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

4y— X -^-z=zx+l8, and 27— y = #+y+4.

Ans. #=9, y=7.

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

Aw*. #=10, y=20.

76. Having explained the principal methods of elimina-
tion, we shall add a few examples which may be solved by
either ; and often indeed, it may be advantageous to use
them all even in the same question.



GENERAL EXAMPLES.

1. Given 2a:+3y=16, and 3#— 2y=ll to find the
values of x and y. Ans. a? =5, y=2.

_ -,. 2x , 3y 9 , 3« 2y 61 m Z t .

2. Gl ven -+JL = _ and T +-f = M to find the

values of x and y. Ans. x= — , y=z — .

9 o



3. Given — +7y=99, and -^-+7a?=51, to find the
values of x and y. Ans. x~7 9 y=:14.

4. Given

to find the values of x and y. Ans. #=60, y=40.



104 FIRST LESSONS IN ALGEBRA.



x QUESTIONS.

1. What fraction is that, to the numerator of which, if 1

be added, its value will be — , but if one be added to its

o

denominator, its value will be — ?



Let the fraction be represented by
Then, by the question



x+l 1^*1

=— and — -—-= - -.

y 3 y+1 4

Whence 3#+3=y and 4ac=yH-l.

Therefore, by subtracting,

x— 3 = 1 or a?=4.
Hence, 12 + 3=y;

therefore, y=15.

2. A market-woman bought a certain number of eggs at
2 for a penny, and as many others, at 3 for a penny ; and
having sold them again altogether, at the rate of 5 for 2d,
found that she had lost 4d : how many eggs had she 1

Let 2x = the whole number of eggs.

Then x= the number of eggs of each sort.

Then will ~7r Xz= the cost of the first sort,

and ~^-oc=z the cost of the second sort.

But 5 : 2 : : 2a? : — the amount for whict the egg

D

were sold.



EQUATIONS OF THE FIRST DEGREE. 105

Hence, by the question

11 4x

Therefore, 15*+ 10a?— 24*:=120,

or a?:=120j

the number of eggs of each sort,

3. A person possessed a capital of 30,000 dollars, for
which he drew a certain interest ; but he owed the sum of
20,000 dollars, for which he paid a certain interest. The
interest that he received exceeded that which he paid by
800 dollars. Another person possessed 35,000 dollars, for
which he received interest at the second of the above rates ;
but he owed 24,000 dollars., for which he paid interest at
the first of the above rates. The interest that he received
exceeded that which he paid by 310 dollars. Required the
two rates of interest

Let x and y denote the two rates of interest ; that is, the
interest of $100 for the given time.

To obtain the interest of $30,000 at the first rate, denoted
by x, we form the proportion

100 : x :: 30,000 : : S °f°° X or 300*,
100

And for the interest $20,000, the rate being y,

100 : y : : 20,000 : : ^l or 200y.

But from the enunciation, the difference between these
two interests is equal to 800 dollars.

We have, then, for the first equation of the problem,

300*— 200y=800.



106 FIRST LESSONS IN ALGEBRA.

By writing algebraically the second condition of the pro-
blem, we obtain the other equation,

350y — 240#:=310.

Both members of the first equation being divisible by 1 00,
and those of the second by 10, we may put the following,
in place of them :

3x—2y = 8, 35y — 24 a: =31.

To eliminate a?, multiply the first equation by 8, and then
add it to the second ; there results

1 9y = 9 5, whenc e y = 5.

Substituting for y, in the first equation, its value, this
equation becomes

3x— 10 = 8, whence x—6.
Therefore, the first rate is 6 per cent, and the second 5.

Verification.

$30,000, placed at 6 per cent, gives 300 x 6 = $ 1 800.
$20,000, „ 5 „ „ 200x5 = $1000.

And we have 1800 — 1000 = 800.

The second condition can be verified in the same manner.

4. What two numbers are those, whose difference is 7,
and sum 33 ? Ans. 13 and 20.

5. To divide the number 75 into two such parts, that
three times the greater may exceed seven times the less
by 15. Ans. 54 and 21.

6. In a mixture of wine and cider, 1 of the whole plus
25 gallons was wine, and ^ part minus 5 gallons was cider :
how many gallons were there of each ?

Ans. 85 of wine, and 35 of cider.



EQUATIONS OF THE FIRST DEGREE. 107

7. A bill of .£120 was paid in guineas and moidores, and
the number of pieces of both sorts that were used was just
100. If the guinea be estimated at 21s, and the moidore
at 27s, how many were there of each ? Ans. 50 of each.

8. Two travellers set out at the same time from London
and York, whose distance apart is 150 miles. One of them
goes 8 miles a day, and the other 7 : in what time will they
meet? Ans. In 10 days.

9. At a certain election, 375 persons voted for two can-
didates, and the candidate chosen had a majority of 91 :
how many voted for each ?

Ans. 233 for one, and 142 for the other.

10. A person has two horses, and a saddle worth j£50.
Now, if the saddle be put on the back of the first horse, it
will make his value double that of the second ; but if it be
put on the back of the second, it will make his value triple
that of the first. What is the value of each horse ?

Ans. One .£30, and the other £40.

1 1 . The hour and minute hands of a clock are exactly
together at 12 o'clock : when are they next together ?

Ans. Ihr. b^min.

12. A man and his wife usually drank out a cask of beer
in 12 days ; but when the man was from home, it lasted
the woman 30 days : how many days would the man alone
be in drinking it 1 Ans. 20 days.

13. If 32 pounds of sea-water contain 1 pound of salt,
how much fresh water must be added to these 32 pounds,
in order that the quantity of salt contained in 32 pounds of
the new mixture shall be reduced to 2 ounces, or ± of a
pound? Ans. 2241b.

14. A person who possessed 100,000 dollars, placed the
greater part of it out at 5 per cent interest, and the other


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Online LibraryCharles DaviesFirst lessons in algebra, embracing the elements of the science → online text (page 5 of 12)