Charles Davies.

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

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62 FIRST 1ESS0N3 IN ALGEBRA.

3. Let — be divided by — . Ans. — ^—.

o 13 60

4 a? 2 4 x

4. Let — — - be divided by 5a?. Ans. — .

7 35

5. Let — — — be divided by — -. .• Ans. .

6 J 3 4a?

6. Let be divided by — . Ans.



a?— 1 J 2 x—l'

5a? . ,. ., _ : 2a 5bx



7. Let — be divided by — . Ans,



3 ; 35' ' 2a



8. Let :• ■ be divided by — ^-. I Ans. — —— .

8cd J Ad 6c 2 x



Q T # x*-b* x 2 +h



a: 2 — 25* + 5 2 * a:— 6 '

.Aft.?. a?-j — -.

h X

10. Divide 6a 2 + — by c 2 -^-?.
5 J 2

60a 2 + 25



.An,?.



11. Divide 18c 2 — a? + 4- by a 2 ——,
b J 5



10c 2 — 5a?-f-5a*



905c 2 — bbx+ba
Ans.



5a 2 b — b 2



12. Divide 20a; 2 -— by a*-^£.
dc 3 * /



20dc 3 fx 2 -8abf
dc*fx 2 -dc*b-hdc4>'



EQUATIONS OF THE FIRST DEGREE. 63



CHAPTER III.

Of Equations of the First Degree.

GO. An Equation is the expression of two equal quanti-
ties with the sign of equality placed between them. Thus,
x—a+b is an equation, in which x is equal to the sum of
a and b.

6 1 . By the definition, every equation is composed of two
parts, separated from each other by the sign == . The part
on the left of the sign, is called the first member ; and the
part on the right, is called the second member. Each mem-
ber may be composed of one or more terms. Thus, in the
equation x=a-\-b, x is the first member, and a-\-b the
second.

62* Every equation maybe regarded as the enunciation,
in algebraic language, of a particular question. Thus, the
equation x-\-x=:30, is the algebraic enunciation of the
following question :



Quest. — 60. What is an equation 7 ! 61. Of how many parts is every
equation composed] How are the parts separated from each other?
What is the part on the left called 1 What is the part on the right called 1
May each member be composed of one or more terms 1 In the equation
x —a-\- b, which is the first member? Which the second ? How many
terms in the first member ? How many in the second 1



64 FIRST LESSONS IN ALGEBRA.

To find a number which being added to itself, shall give a
sum equal to 30.

Were it required to solve this question, we should first
express it in algebraic language, which would give the
equation

x-\-x=30.

By adding x to itself, Ave have

2# = 30.

And by dividing by 2, we obtain
#=15.

Hence we see that the solution of a question by algebra
consists of two distinct parts.

1st. To express algebraically the relation between the known
and unknown quantities.

2nd. To find a value for the unknown quantity, in terms of
those which are known.

This latter part is called the solution of the equation.

The given or known parts of a question, are represented
either by numbers or by the first letters of the alphabet,
a, b, c, &c. The unknown or required parts are repre-
sented by the final letters, x, y, z, &c.

example.

Find a number which, being added to twice itself, the
sum shall be equal to 24.



Quest. — 62. How may you regard every equation 1 What question
does the equation x+a:=30 state] Of how many parts does the solu-
tion of a question by algebra consist 1 Name them. What is the 2nd
part called 1 By what are the known parts of a question represented ?
By what are the unknown parts represented ?



EQUATIONS OF THE FIRST DEGREE. 65

Statement.
Let x represent the number. We shall then have

This is the statement.

Solution.
Having .... as+2a:=24,

we add x+2x,

which gives . . . 3a; =24 ;

and dividing by 3, . x = 8.

63. An equation is said to be verified when the answer
found, being substituted for the unknown quantity, proves
the two members of the equation to be equal to each other.

Thus, in the last equation we found x =8. If we substi-
tute this value for x in the equation

x+2x=2±,

we shall have 8+2x8=8+16=24.

which proves that 8 is the true answer.

64. An equation involving only the first power of the
unknown quantity, is called an equation of the first degree.

Thus, 6a:+3#— 5 = 13,

and ax+bx+c =zd,

are equations of the first degree.

By considering the nature of an equation, we perceive
that it must possess the three following properties :

Qujsst. — 63. When is an equation said to be verified ? 64. When
an equation involves only the first power of the unknown quantity, what
is it called ? What are the three properties of every equation ?

6*



66 FIRST LESSONS IN ALGEBRA.

1st. The two members are composed of quantities of the
same kind : that is, dollars = dollars, pounds = pounds, &c.
2nd. The two members are equal to each other.
3rd. The two members must have the same sign.

65. An axiom is a self-evident truth. We may here
state the following.

1. If equal quantities be added to both members of an equa-
tion, the equality of the members mill not be destroyed.

2. If equal quantities be subtracted from both members of
an equation, the equality will not be destroyed.

3. If both members of an equation be multiplied by the same
number, the equality will not be destroyed.

4. If both members of an equation be divided by the same
number, the equality will not be destroyed.

Transformation of Equations.

66. The transformation of an equation consists in chang-
ing its form without affecting the equality of its members.

The following transformations are of continual use in the
resolution of equations.

First Transformation.

67. When some of the terms of an equation are frac-
tional, to reduce the equation to one in which the terms shall
be entire.

I. Take the equation

2x 3 x



Quest. — 65. What is an axiom 1 Name the four axioms. 66. What
is the transformation of an equation ? 67. What is the first transforma-
tion ? What is the least common multiple of several numbers ? How
do you find the least common multiple ?



EQUATIONS OF THE FIRST DEGREE. 67

First, reduce all the fractions to the same denominator,
by the known rule ; the equation then becomes

4&r 54a? 12a:



72 72 ' 72



and since we can multiply both members by the same num-
ber without destroying the equality, we will multiply them
by 72, which is the same as suppressing the denominator
72, in the fractional terms, and multiplying the entire term
by 72 ; the equation then becomes

48a:-54a?+ 12^=792,

or dividing by 6 8x— - 9 a?-}- 2a? =132.

But this last equation can be obtained in a shorter way, by
finding the least common multiple of the denominators.

The least common multiple of several numbers is the
least number which they will separately divide without a
remainder. When the numbers are small, it may at once
be determined by inspection. The manner of finding the
least common multiple is fully shown in Arithmetic § 87.

Take for example, the last equation

2a? 3 a? ;■

We see that 12 is the least common multiple of the deno-
minators, and if we multiply all the terms of the equation
by 12, and divide by the denominators, we obtain

8a?— 9r+2a? = 132.

the same equation as before found;



68 FIRST LESSONS IN ALGEBRA.

68. Hence, to make the denominators disappear from an
equation, we have the following

RULE.

I. Find the least common multiple of all the denominators.

II. Multiply each of the entire terms by this multiple, and
each of the fractional terms by the quotient of this multiple
divided by the denominator of the term thus multiplied, and
omit the denominators of the fractional terms.

EXAMPLES.

1. Clear the equation of ~r-+~= — 4=3 of its denomi-
nators. Ans. 7x + 5x— 140=105.

2. Clear the equation -^-+-3 — ^= 8 of its denomi-

q y z 1

nators. , Ans. 9x-\-6x-— 2#=432.

<y> (V> /y» <y>

3. Clear the equation —+-5 FT+To^ 20 ofitsde-

nominators. Ans. 18#+12;»— 4#+3tf=720.

4. Clear the equation -r+-^ TT—^ °^ * ts denomi-

nators. Ans. 14a?+10#— 35a?=280.

5. Clear the equation — —+—=15 ofitsdenomi-

4 5 6

nators Ans. 15a:-— 12a? + 10;r =900.



Quest. — 68. Give the rule for clearing an equation of its denomi-
nators.



EQUATIONS OF THE FIRST DEGREE. 69



v k x x

6. Clear the equation — — — +— +— =12 of its de-
nominators. Ans. I8x — l2x+9x + 8x=864:.

a c

7. Clear the equation — —+fz=g.



Ans. ad—bc+bdf=:bdg.



8. In the equation



ax 2c 2 x 4bc 2 x 5a 3 2c 2 _,

r — [Aa= — - — | 3b,

b ab a 3 b 2 a

the least common multiple of the denominators is a 3 b 2 ;
hence clearing the fractions, we obtain

a^bx—2a 2 bc 2 x+Aa"b 2 —Ab 3 c 2 x— 5a 6 +2a 2 b 2 c 2 — 3a 3 R



Second Transformation.

69. When the two members of an equation are entire
polynomials, to transpose certain terms from one member to
the other.

1. Take for example the equation

5a: — 6 = 8 + 2*.

If, in the first place we subtract 2x from both members,
the equality will not be destroyed, and we have

5x—6—2x = 8.

Whence we see that the term 2x, which was additive
in the second member becomes subtractive in the first.



Quest. — 69. What is the second transformation? What do you
understand by transposing a term 1 Give the rule for transposing from
one member to the other.



70 FIRST LESSONS JN ALGEBRA.

In the second place, if we add 6 to both members, the
equality will still exist, and we have

5x—6— 2x+6=zS+6.

Or, since — 6 and +6 destroy each other, we have

5x— 2x=8+6.

Hence the term which was subtractive in the first mem-
ber, passes into the second member with the sign of
addition.

2. Again, take the equation

ax-\-b-=.d— ex.

If we add ex to both members and subtract b from
them, the equation becomes

ax-\-b-\-cx— b=d — cx-\-cx— b.

or reducing ax+cx=d— b.

When a term is taken from one member of an equation
and placed in the other, it is said to be transposed.

Therefore, for the transposition of the terms, we have the
following

RULE.

Any term of an equation may be transposed from one mem-
ber to the other by changing its sign.

TO. We will now apply the preceding principles to the
resolution of equations.

1. Take the equation

4a?— 3=2*+5.



EQUATIONS OP THE FIRST DEGREE. 71

By transposing the terms —3 and 2x, it becomes

4a?— 2#=5+3.

Or, reducing 2x =8.

8
Dividing by 2 x =—=4.

Verification.

If now, 4 be substituted in the place of x in the given
equation

4x— 3 = 2*4-5,

it becomes 4x4—3=2x4+5.

or, 13 = 13.

Hence, the value of x is verified by substituting it for the
unknown quantity in the given equation.
2. For a second example, take the equation

5x 4x 7 13#



12 3 8 6

By making the denominators disappear, we have

10a?-32tf-312=21-52ar,
or, by transposing

10*-32*+52tf=:21 + 312
by reducing 30a? =333

333 111

^-so-^ToT^ 11 ' 1 -

a result which may be verified by substituting it for x in the
given equation.

3. For a third example let us take the equation
(3a— x)(a— b) + 2ax—4b(x+a).



72 FIRST LESSONS IN ALGEBRA*

It is first necessary to perform the multiplications indica-
ted, in order to reduce the two members to two polynomials,
and thus be able to disengage the unknown quantity x, from
the known quantities. Having done that, the equation
becomes,

3a 2 — ax— 3ab+bx-\-2axr=z4bx-\-4ab,
or, by transposing

— -ax+bx-\-2ax — 4bx =z4ab-\-3ab — 3a 2 ,
by reducing ax—3bx =i7ab — 3a 2 .

Or, (Art. 41). (a-3b)x=:7ab-3a 2 .

Dividing both members by a— 3b we find
_ 7ab — 3a 2
a — 3b

Hence, in order to resolve an equation of the first degree,
we have the following general

RULE.

I. If there are any denominators, cause them to disappear,
and perform, in both members, all the algebraic operations
indicated; we thus obtain an equation the two members of
which are entire polynomials.

II. Then transpose all the terms affected with the unknown
quantity into the first member, and all the known terms into
the second member.

III. Reduce to a single term all the terms involving x :
this term will be composed of two factors, one of which will be
x, and the other all the multipliers of x, connected with their
respective signs.

IV. Divide both members by the number or polynomial by
which the unknown quantity is multiplied.

Quest. — 70. What is the first step in resolving an equation of the
first degree % What the second 1 What the third ? What the fourth ?



EQUATIONS OF THE FIRST DEGREE. 73

EXAMPLES.

1. Given 3a?— 2+24 = 31 to find a?. Ans. a?=3.

2. Given a?+18=3a?— 5 to find x. Ans. a?=ll— .

S

3. Given 6— 2a?+10=20 — 3a?— 2 to find a?.

Arcs, a? =2.
i

4. Given a?+—a?+—a?= 11 to find x. Ans. x=z6.

5. Given 2a? — — -a?+l=5a?— 2 to find x. Ans. x=—.

Z 7

6. Given 3ax+— — % — bx— a to find x.

6 — 3a
Ans. xz



"6a-26



7. Given ^— — h~-=20 — ^— — to find x.



Ans. 07=23 - -.
4



8. Given — - — h~rr=4 to find x.



6
An*, a? =3—.
13



^ /-.♦ a? 3a? , 4a? _ . .

9. Given — — +a?=— — 3 to find x.

4 2 8



, ^ ^- 3aa? 2bx M _ _

10. Given ; 4=/ to find a?.



Ans. a?=4.



c<#+4cd
Ans. XZZZ-—Z- — — r— .
3atf— 2fo



74 FIRST LESSONS IN ALGEBRA.

n . Given !^_iipL=4-6 to find ..

56 + 96— 1c
Ans. x= JsS '

x x — 2 a? 13 ~ j

12. Given — —+—=— to find w.

Ans. 07 = 10.

X X X X r f. -.

13. Given r + ~r=f t0 find x -

abed

abedf
AnS ' X = ~bTd - acd+abd - abc'

Note. — What is. the numerical value of a?, when a— I,
5=2, c=3, d=4, &=5, and/=6.

M . G i V en ^-^-^=-12|f to find *

Arcs. 07=14.

15. Given *-i^+±^-=*+l to find *.

Ans. 07 = 6.

16. Given M-4-+4" ^-=2*-43 to find r-

4 o b

Ans. 07=60.



407— 2 3o?— 1 r. j

17. Given 2or — = — - — *o find x.



18 Given 3x+-^j—-oc+a to find oc.



Ans. x=z



Ans. 07=3.
3a+d



6 + & "



x—b a bx bx—a



ax—o a ox ux—u „ ,

19. Given _^— +~=y — - to find x.

3b
Ans. 0? =



3a-26"



EQUATIONS OF THE FIRST DEGREE. 75

20. Find the value of x in the equation

{a+b)(x— b) Aab—b 2 at—bx

3a — —7 2x-\ - .

a — o a-jrb

a*+3tfb + 4a 2 b 2 —6ab 3 +2¥



Ans. #ss-



2b(2a 2 +ab— b 2 )



Of Questions producing Equations of the First Degree
involving but a single unknown quantity.

71. It has already been observed (Art. 62), that the
solution of a question by algebra consists of two distinct
parts :

1st. To express the conditions of the question algebrai-
cally ; and

2d. To disengage the unknown from the known quantities.

We have already explained the manner of finding the
value of the unknown quantity, after the question has been
stated ; and it only remains to point out the best methods
of putting a question in the language of algebra.

This part of the algebraic solution of a question cannot,
like the second, be subjected to any well defined rule.
Sometimes the enunciation of the question furnishes the
equation immediately ; and sometimes it is necessary to dis-
cover, from the enunciation, new conditions from which an
equation may be formed.

Quest. — 71. Into how many parts is the resolution of a question in
algebra divided 1 What is the first step 1 What the second 1 Which
part has already been explained ? Which part is now to be considered ?
Can this part be subjected to exact rules ? Give the general rule fox
stating a question



gfl






TO FIRST LESSONS IN ALGEBRA.

In almost all cases, however, we are enabled to discover
the equation by applying the following

RULE.

Consider the problem solved, and then indicate, by means
of algebraic signs, upon the known and unknown quantities,
the same operations which it would be necessary to perform, in
order to verify the unknown quantity, had it been known.

questions.

1 . To find a number to which if 5 be added, the sum will
be equal to 9.

Denote the number by x.
Then by the conditions

tf + 5 = 9.

This is the statement of the question.
To find the value of x, we transpose 5 to the second
member, which gives

x-9— 5 = 4.

Verification.
4 + 5 = 9.

2. Find a number such, that the sum of one half, one
third, and one fourth of it, augmented by 45, shall be equal
to 448.

Let the required number be denoted by x.

Then, one half of it will be denoted by — .
one third „ „ by -— .



one fourth , „ by — .



3

T



EQUATIONS OF THE FIRST DEGREE 77

And by the conditions,

|-+y+T +45 = 448 -

This is the statement of the question.
To find the value of x, subtract 45 from both members ;
this gives

y+T+T= 403 -

By clearing the terms of their denominators, we obtain

6x+4x+3xz=z4836,

or 13* = 4836.

TT 4836 nrve%

Hence, #=— — =372.

1 o

Verification,

372 372 372

-7^+-^+—- +45 = 186 + 124+93+45=448.

Z o 4

3. What number is that whose third part exceeds its
fourth by 16.

Let the required number be represented by x. Then,

-—-#= the third part.

■—•*= the fourth part.

And by the question

1 1
y*— *=16-

This is the statement. To find the value of x, we clear
the terms of the denominators, which gives

4a?— 3#=192.
and x =192.

7*



78 FIRST LESSONS IN ALGEBRA.



Verification
192 192



64-48=16.



3 4

4. Divide $1000 between A, B and C, so that A shall
have $72 more than B, and C $100 more than A.

Let - x= B's share of the $1000.

Then x-\- 72 = A's share,

and x +172= C's share,

their sum is 3z+244 = $1000.

This is the statement.

By transposing 244 we have

3*=1000-244=756

756
and #=— — =252= B's share.

Hence, x+ 72 =252+ 72 = $324= A's share.
And x+ 172 =252+172 = $424= C's share.

Verification.
252 + 324 + 424 = 1000.

5. Out of a cask of wine which had leaked away a third
part, 21 gallons were afterwards drawn, and the cask being
then guaged, appeared to be half full : how much did it
hold?

Suppose the cask to have held x gallons.

Then, — what leaked away.
o

And — + 21= what had leaked and been drawn.

x 1

Hence, -—+ 21=-— x by the question,
o 2

This is the statement.



EQUATIONS OF THE FIRST DEGREE. 79

To find x, we have

2a+ 126 = 30?,
and — x ±s — 126,

or x = 126,

by changing the signs of both members, which does not
destroy their equality.



Verification.
-^ + 21=42 + 21=63=^



6. A fish was caught whose tail weighed 9lb., his head
weighed as much as his tail and half his body, and his body
weighed as much as his head and tail together ; what was
the weight of the fish 1

Let 2x= the weight of the body.

Then, 9 + a?= weight of the head ;

and since the body weighed as much as both head and tail,

2u? = 9 + 9 + a,

which is the statement. Then,

2x— £=18 and #=18.

Verification.

2#=36/#.= weight of the body.

9+ x —21 lb. = weight of the head.

9ZZ>.= weight of the tail.

Hence, T2lb. = weight of the fish.



80 FIRST LESSONS IN ALGEBRA.

7. The sum of two numbers is 67 and their difference
1 9 : what are the two numbers ?

Let *== the least number.

Then, #+19= the greater.

and by the conditions of the question

2a; + 19 = 67.

This is the statement.

To find #, we first transpose 19, which gives.

2# = 67-19 = 48;

48
hence, #=— = 24, and #+19=43.

Verification.
43+24=67, and 43-24=19.

Another Solution.

Let # represent the greater number :

tnen, #—19 will represent the least,

and, 2#— 19=67, whence 2#=67+19;

86
therefore, #= — =43.

and consequently #—19=43 — 19=24.

General Solution of this Problem.

The sum of two numbers is a, their difference is b.
What are the two numbers ?






EQUATIONS OF THE FIRST DEGREE. 81

Let x be the least number,

x-\-b will represent the greater.
Hence, 2x+b — a, whence 2x=:a—b ;

. r a — b a b

therefore, x=.-— — — — ,

' 2 2 2'

1 1 , 7 a J , 7 a , b

and consequently, x-\-b—— —-+6= - -+—.

2 2 2 2

As the form of these two results is independent of any
value attributed to the letters a and b, it follows that,

Knowing the sum and difference of two numbers, we obtain
the greater by adding the half difference to the half sum, and
the less, by subtracting the half difference from half the sum.

Thus, if the given sum were 237, and the difference 99,

a ♦ ■ 237 . " 237 + 99 336

the greater is —+77-, or =——=168;

.£/'■■ 237 99 138 ^

and the least — — , or — - — =69.

2 2 2

Verification.

168+69=237 and 168-69=99.

8. A person engaged a workman for 48 days. For
each day that he laboured he received 24 cents, and for
each day that he was idle, he paid 12 cents for his board.
At the end of the 48 days, the account was settled, when
the labourer received 504 cents. Required the number of
working days, and the number of days he was idle.



82 FIRST LESSONS IN ALGEBRA.

If these two numbers were known, by multiplying them
respectively by 24 and 12, then subtracting the last product
from the first, the result would be 504. Let us indicate
these operations by means of algebraic signs.

Let x = the number of working days.

48— a? = the number of idle days.
Then, 24 X x = the amount earned,
and 12(48 — x) = the amount paid for his board.
Then, 24a: — 12(48— *)=:504

what he received, which is the statement. Then to find x,
we first multiply by 12, which gives

24*-576 + 12a: = 504.

or, 36tf=504+576 = 1080,

1080 ~ , _ . .

x=z —30 the working days.

whence, 48 — 30 = 18 the idle days.

Verification.

Thirty day's labour, at 24 cents
a day, amounts to 30x24 = 720 cents.

And 18 day's board, at 12 cents

a day, amounts to 18 X 12=216 cents.

The difference is the amount received 504 cents.

General Solution.

This question may be made general, by denoting the
whole number of working and idle days by n.

The amount received for each day he worked by a.
The amount paid for his board, for each idle day, by b.



EQUATIONS OF THE FIRST DEGREE. 83

And the balance due the laborer, or the result of the
account, by c.

As before, let the number of working days be repre-
sented by x.

The number of idle days will be expressed by n—oc.

Hence, what he earns will be expressed by ax.

And the sum to be deducted, on account of his board, by
b(n— x).

The equation of the problem therefore is

ax — b(n — #) — c,

which is the statement. To find x we first multiply by b,
which gives

ax — bn-\-bx=c
or, (a+b)x=zc -\-bn

whence, x= rr~— working days.

a -\-b

c -\-bn an-\-bn — c — bn
and consequently, n— x=n-



a +b " a+b



or n— xz= — = idle days.

a+b J



Let us now suppose w=48, = 24, 5=12, and c=504.
These numbers will give for x the same value as before
found.

9. A person dying leaves half of his property to his wife,
one-sixth to each of two daughters, one-twelfth to a servant,
and the remaining $600 to the poor : what was the amount
of his property ?



84 FIRST LESSONS IN ALGEBRA.

Represent the amount of the property by x.

Then, -^-= what he left to his wife,

— = what he left to one daughter,

and — = — what he left to both daughters,

6 3

— =s what he left to his servant.
12

$600 to the poor.
Then, by the conditions of the question

XXX, n ^ n

T + T+l2+ 600 =*

the amount of the property, which gives x = $7200.

10. A and B play together at cards. A sets down with
$84 and B with $48. Each loses and wins in turn, when
it appears that A has five times as much as B. How much
did A win ?

Let x represent what A won.
Then A rose with $84+ a? dollars,

and B rose with $48— x dollars.

But by the conditions of the question, we have
84+#=5(48-;r),
hence, 84+#=240 — 5x ;

and, 6a: = 156,

consequently, a? =$26 what A won.

Verification.

84+26 = 110 ; 48-26=22 ;

110 = 5(22) = 110



EQUATIONS OF THE FIRST DEGREE. 85

11. A can do a piece of work alone in 10 days, B in 13
days : in what time can they do it if they work together ?

Denote the time by x, and the work to be done by 1.

Then in 1 day A could do — - of the work, and B could

J 10

1 x

do — ; and in x days A could do — of the work, and
13' J 10

B> — : hence, by the conditions of the question
1 o

-+ - 1

10+13 '

which gives 13#+10,z=130 :

130
hence, 23#:=130, #— ~5^| days.

12. A fox, pursued by a greyhound, has a start of 60
leaps. He makes 9 leaps while the greyhound makes but
6 ; but, three leaps of the greyhound are equivalent to 7
of the fox. How many leaps must the greyhound make to
overtake the fox 1

From the enunciation, it is evident that the distance to
be passed over by the greyhound is composed of the 60
leaps which the fox is in advance, plus the distance that the
fox passes over from the moment when the greyhound starts
in pursuit of him. Hence, if we can find the expression for
these two distances, it will be easy to form the equation of


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