HIGHER ALGEBRA.
CHAPTER
I.
RATIO.
Ratio is the relation which one quantity
Definition.
bears to another of the same kind, the comparison being made by
considering what multiple, part, or parts, one quantity is of the
1.
other.
•
The
and
A
A
B
is usually written A
to
The quantities
B.
are called the terms of the ratio.
The first term is
ratio of
B
:
called the antecedent, the second
To
2.
by
-^
,
B
;
find
what multiple or part A
hence the ratio
and we
term the consequent.
A
B may
:
we
divide A
be measured by the fraction
is
shall usually find it convenient to
of B,
adopt this notation.
In order to compare two quantities they must be expressed in
terms of the same unit. Thus the ratio of £2 to 15s. is measured
2x20
....
by the traction —^—
or
8
-
.
A
ratio expresses the number of times that one quantity contains another, and therefore every ratio is an abstract quantity.
Note.
3.
Since by the laws of fractions,
a
b
=
ma
mJ'
follows that the ratio a
b is equal to the ratio ma
mb
that is, the value of a ratio remains unaltered if the antecedent
and the consequent are multiplied or divided by the same quantity.
it
:
H. H. A.
:
1
;
HIGHER ALGEBRA.
2
or more ratios may be compared by reducing their
Thus suppose
equivalent fractions to a common denominator.
Two
4.
bx
x
and — = =— hence
h
by
by
y
the ratio a
b is greater than, equal to, or less than the ratio
x
y according as ay is greater than, equal to, or less than bx.
_
a
_
and x
o
:
y are two
J
:
a
a
JNow - = ~V
xt
x-
ratios.
,
i
:
3
:
:
The
5.
two fractions can be expressed as a
ratio of
C
Ch
of
two
Thus the
integers.
°
—
ratio
ratio
—
a
:
b
is
measured by the
a
—
fraction
,
or =—
:
and
therefore equivalent
is
to
the
ratio
be
c
d
ad
be.
:
or both, of the terms of a ratio be a surd
quantity, then no two integers can be found which will exactly
measure their ratio. Thus the ratio J'2 1 cannot be exactly
expressed by any two integers.
If either,
6.
:
If the ratio of any two quantities can be
Definition.
expressed exactly by the ratio of two integers, the quantities
otherwise, they are said to be
are said to be commensurable
incommensurable.
7.
;
Although we cannot find two integers which will exactly
measure the ratio of two incommensurable quantities, we can
always find two integers whose ratio differs from that required
by as small a quantity as we please.
J5 =
V
J5
>
—
mm(>
2-236068...
Thus
-.
4
,
,
.
and therefore
™
A1I ,
-559017...
=
4
559017
so that the difference
559018
,
and <
-jooOOOO
between the
;
559017
ratios
:
1000000 and
than -000001. By carrying the decimals further, a
closer approximation may be arrived at.
J5
:
4
is less
Ratios are compounded by multiplying toDefinition.
gether the fractions which denote them ; or by multiplying together the antecedents for a new antecedent, and the consequents
for a new consequent.
8.
Example.
Find the
ratio
2a
compounded
:
Sb, Q>ab
:
of the three ratios
5c 2 c
,
:
a
KATIO.
m
The
.
,
,.
required ratio =
.
2a
Gab
-x
x 6b
a
be
c
--„
1
_4a
~ DC
'
Definition. When the ratio a b is compounded with
2
2
and is called the duplicate ratio
itself the resulting ratio is a
b
3
3
Similarly a
b is called the triplicate ratio of a
of a b.
b.
9.
:
:
,
:
:
Also a 2
b
:
2"
is
Examples.
:
called the subduplicate ratio of
a
:
b.
The duplicate ratio of 2a 3b is 4a 2 96-.
The subduplicate ratio of 49 25 is 7 5.
The triplicate ratio of 2x 1 is 8a; 3 1.
(1)
:
(2)
:
:
(3)
:
A
Definition.
:
:
said to he a ratio of greater
inequality, of less inequality, or of equality, according as the
antecedent is greater than, less than, or equal to the consequent.
10.
A
ratio
is
of greater inequality is diminished, and a ratio of
inequality is increased, by adding the same quantity to both
11.
less
ratio
terms.
its
a ,
,,
Let T be the
..
,
ratio,
and
,
,
let
6
adding x to both
its
Now
a+x
=
be the new ratio formed by
J
b + x
terms.
erms.
a
a+x
ax — bx
b
b
+x
b(b+x)
x(a — b)
~b(b + x)
}
and a - b is positive or negative according as a
less than b.
a+x
a
y
ence it a > b,
>
T
^
o
b + x
H.
is
greater or
/,
;
«
j
d.it
a <b,
a
7-
<
b
x
a
~t~
b
+ x
;
which proves the proposition.
can be proved that a ratio of greater inequality
increased, and a ratio of less inequality is diminished, by taking
Similarly
is
the
it
same quantity from both
its terms.
two or more ratios are equal many useful propositions may be proved by introducing a single symbol to
12.
When
denote each of the equal ratios.
1—2
;
HIGHER ALGEBRA.
4
The proof of the following important theorem
the method of procedure.
a
b
//
c
e
d
f
will illustrate
'
1
.
,
.
.
each of
J these ratios
where
r,
p, q,
/pan
r-
n
+ qc + re +
j\pb n + qd n + rtsn +
11
= —
-
—,
b
d
7
a—
bk,
^
pan =pbn
~>—
.
. .
,
)
/
= dk,
c
qc"
—&
j
•
• •
')
J
= qd"k
e
n
=fk, ...;
re
,
n
= rf"k",...
n
n
pa" + gc" + re +
'''
.
quantities ivhatever.
ace
= —
Ijet
whence
.
(
n are any
then
.\ n
n
_pb"k" + qd"k + rf"k +
...
H
pb + qd"+rf+...
n
fb +
'
qcl
n
+r/i +...
= k";
i
'pa"
e
2)b"
a
c
+ re" + .\ n _
= k = ^ = -,=
d
b
+ qd" + ?'/" + .../
+
qc"
.
,
.
giving different values to p, q, r, n many particular cases
of this general proposition may be deduced ; or they may be
proved independently by using the same method. For instance,
By
a
_c
e
b-~d'f-'"
each of these ratios
b+d +f+
a result of such frequent utility that the following verbal equivalent should be noticed
When a series of fractions are equal,
each of them is equal to the sum of all the numerators divided by the
sum of all the denominators.
:
Example
(I
1.
If b
= C- =€>
d
,
shew that
J
+ 2c 2 e - Sae 2/ _ ace
~b 4 + 2^/-36/3 ~bdf
az b
Let
_ _^._A,,
«-£-£-X;.
6
then
a = bk,
rf
c
= dk, e =fk
;
RATIO.
*
W
aa6+2c»g-3qgy _
""
k 4 + 2tl-f - Bbf3
'*
5
fc
+2d?fk* - 3bf 3 k 3
4
+ 2r/-/ - 36/8
a
...
c
e
ace
=
bdf'
Example
=f=
a
If
2.
#+a
it
z
y
c
=r=-=
A;
,
prove that
,
y2 + &2
y+b
*2 + a2
x
Let a
c
b
2
+ c2 _
z+c
3
so that x = «£,
s a + a3
„
then
(
.c
+ y +2 2 + (a + & + c) 2
a; + ?/ + 2 + a + &+c
)
= 6/c, = ch
?/
2;
—ah + a
= aW+a* =
(k*
+a
x*+a*
ar
+a
y
a
?/
+y
ga +e»_
•
+&
z
+c
(ife
a
'
+ l)a
—
L__
'
Jc+1
-
a:
;
+l)o
+1
(&
2
;
+ l)&
(fc
£+1
/c
+ l)c
&+1
2
Jfc 2 + l)(a + 6 + c)
fc+1
8
Jfc
(a+6+c) 8 +(a+ 6+c) a
&(a + & + c) + a +
+c
+ he) % + (a + b + c) 2
(ka + kb + kc)+a + b + c
(x+y+z)*+(a+ b + cf
_ (lea
_
6
+
kb
x+y+z+a+b+c
13.
an equation
If
quantities,
we may
is
homogeneous with respect
for these quantities substitute in the equation
any others proportional to them.
lx
is
homogeneous in
portional to
x, y, %
x
11
a
£
Put h = — =
3
instance, the equation
Let
a, j3,
y be three quantities pro-
respectively.
z
= -
,
so that
x-
ak,
y=
/3k, z
= yk
;
y
Ia f3k
4
+ ma(3 2 yk* + n^y'k4 =
3
is,
For
2 2
y + mxifz + ny z —
x, y, z.
3
then
that
75
to certain
7a /?
2 2
+ ma/3 2 y + nj3 y =
0,
;
an equation of the same form as the original
a, /?, y in the places of x, y, z respectively.
one, but with
—
.
.
HIGHER ALGEBRA.
6
The following theorem
14.
.
If
.
y*
.
y~
,
.— ,....
,
nominators are
n
unequal fractions, of which the de-
be
r-
n
3
2
1
Q
i
important.
is
of the same sign, then
all
+ a8 + a 3 +
b +b 2 +b 3 +
a,
...
+ an
•'•
+bn
l
lies
in magnitude between the greatest
Suppose that
all
and
least
of them.
the denominators are positive.
and denote
least fraction,
the fraction
a
— /c
a
1
>k
—
ko r
a> kb
.-.
:
=*
be the
then
;
.'.a r
i
'
b
y-
by k
it
Let
b
«
>
'
i
l
a2
b
> k;
by
so on;
addition,
a ,+«2
+ an >
+ «3 +
(
+b + K +
b.+b 29 + b.+
3
+b n
we may prove
^
is tlie
'
br
that
+ an
+*.+*
+K
l
.
+
<
at
V
greatest of the given fractions.
In like manner the theorem may be proved when
denominators are negative.
15.
'>
ar
.
a + a2 + a 3 +
6
+K) k
,
+ au
1
Similarly
bl
a + a 2 + a3 +
l
where
;
a
and
.*.
a 2 > kb 2
.-.
The ready application
all
the
of the general principle involved
Art. 12 is of such great value in all branches of mathematics,
that the student should be able to use it with some freedom in
any particular case that may arise, without necessarily introducing
an auxiliary symbol.
in
Example
If
-
b
prove that
—=
+ c-a
X
1.
x+y+z
a+b
+c
V
c
+ a-b
z
=
,
a
+ b-c
= *&+*)+? (*+*)+ * (*+V)
2(ax + by + cz)
=
;
;
RATIO.
t
Each
i
,i
i=
ofe the given efi actions
•
_
sum °f numerators
—
__
sum of denominators
x+y+z
'
a
"
+b+c
(
''
Again, if we multiply both numerator and denominator of the three
given fractions by y + z, z + x, x + y respectively,
+ z)
=
j
\
(y + z)(b + c-a)
?(« + *>
{l
each fractions
-
(z
+ x)
(c
'(* + *
-
__.
+ a-b)
(x + y)
)
(a+b-e)
sum of numerators
sum of denominators
x
=
.'.
from
(1)
and
a
prove that
If
2.
_x
(y
+ z)+y
+ b + c~
(2).
+ x)+z (x + y)
(ax + by + cz)
2
(z
m(nc + la-mb)
l(mb + nc-la)
m
I
x(by + cz-ax)
We have
+ y (z + x ) +z {x + y)
2ax + 2by + 2cz
+z)
(2),
x+y+z
Example
(y
y
(cz
y
I
m
mb + nc —
nc
la
(la
+ mb - nc)
'
n
+ ax-by)
x
n
z(ax + by -cz)
z
+ la-mb
n
la + mb —
nc
v
z
-+-
=m
n
'"2/a"
= two
similar expressions
ny + mz _lz + nx
b
a
y,
_ mx + ly
c
Multiply the first of these fractions above and below by
and the third by z then
.r,
the second by
;
nxy + mxz
ax
_
+ nxy _ mxz + lyz
Jyz
by
cz
= _2lyz
by
+ cz- ax
= two
m
I
x
(by
+ cz -ax)
similar expressions
y
(cz
+ ax-by)
z
n
(ax + by-cz)'
:
HIGHER ALGEBRA.
8
16.
If
we have two
quantities in the first
equations containing three
degree, such as
a x+b y+
l
a2 x +
we cannot
c
z=Q
l
l
b 2y
+ c2 z =
solve these completely
unknown
(1),
(2),
but by writing them in the
;
form
we
can,
II
X
by regarding - and - as the unknowns, solve in the
z
z
ordinary
way and
x
%
or,
obtain
- b2c
"
afi 2 - a 2 b
b c2
l
y
i
'
z
__
"
l
cxa
2
afi 2
- c2 a
-a
2
l
>
'
b
l
more symmetrically,
b c2
x
- b2c
x
y
cla
x
2
-
c a,
2
afi 2
-
a_p x
'
,(3).
when we have two equations of the type
and (2) we may always by the above formula
It thus appears that
represented by (1)
write down the ratios x y z in terms of the coefficients of the
equations by the following rule
:
:
Write down the coefficients
those of y; and repeat these as
of x, y, z in order, beginning with
in the diagram.
Multiply the coefficients across in the way indicated by the
arrows, remembering that in forming the products any one
obtained by descending is positive, and any one obtained by
ascending is negative.
The three results
h i cz- h fv
are proportional to
This
is
c
x
a2-
c
2
an a
A- a b
2
>
x, y, z respectively.
called the
Rule of Cross Multiplication,
RATIO.
Example
Find the
1.
x
ratios of
7x=4y + Qz
By
down
from the equations
3z = 12x + Uy.
y
:
t
:
z
we have 7x - Ay - 8-2 = 0,
12x + lly-Sz = 0.
transposition
"Write
9
the coeilicients, thus
-4
-8
-3
11
-4
7
12
11,
whence we obtain the products
(-8)xl2-(-3)x7,
(-4)x(-3)-llx(-8),
or
-75,
100,
x
=
-
is,
4
Example
2.
Eliminate
y
-*-
-3
(2)
and
+ ^ 1 + c 12 =
a^ + ^y + c^^O
(2),
Ogaj+fegy+c^^O
(3).
Example
3.
(1),
?/
*__
" Vs
_
C 2«i
denoting each of these ratios by
and dividing out by A-, we obtain
(Va -
This relation
.
5
by cross multiplication,
(3),
k> C 3
Oj
= z?
from the equations
x, y, z
a 1 a;
From
z
y
x
,.
that
125;
100 ~ ^75~"125'
•'*
x,
7 x 11 - 12 x (-4),
63ca)
+ &i
k,
(^'"3
y
~ C 3«2
„
j*
«2 6 3
.
~ ll ih
'
by multiplying up, substituting in
- c 3 a a) + ('i
(«
A-
" A-)
= °-
called the eliminant of the given equations.
is
Solve the equations
ax + by + cz =
(1),
=
hex + cay + abz = (b - c)
x+ y+
From
(1)
and
(2),
.-.
Substituting in
z
(2),
(c-a) (a-b)
(3).
by cross multiplication,
x
z
—y
b-c
= ^— =
c-a
x = k (b-
c),
— k, suppose
a-bT
y — k (c - a), z — k(a- b).
:
(3),
k {bc(b-c) + ca
(c
-
a)
+ ab (a -
k{-{b-c)(c-
b)}
a) {a - &)
.-.
^ln'nce
(1),
\
={b- c)
-
a) {a
= (b-e) [e - a)
fcss-lj
x = c -b, y — a-r,
(c
z
= b - a.
{a
-
b),
-
b)
;
HIGHER ALGEBRA.
10
z
=
1,
equations (1) and (2) become
ax+
bxy
+
c
= 0,
v+
h 2y
+
c2
we put
If in Art. 16
17.
x
and
(3)
t
=°
>
becomes
x
-
b x c2
y
cx a
b 2c l
a
l
b2
-a b
a
- c2a
*
-a b
afi 2
l
- a2 b
aj> %
i
2
'
]
l
Hence any two simultaneous equations involving two unknowns in the first degree may be solved by the rule of cross
multiplication.
Example.
By
5x-3y -1 = 0, x + 2y = 12.
Solve
= 0,
5x - 3y - 1
transposition,
x + 2y -12 = 0;
x
36 + 2
*'•
=
x=
whence
1
y
- 1 + 60 ~
38
10 + 3
59
=
is' y lS'
EXAMPLES.
I.
Find the ratio compounded of
1.
and the duplicate
(1)
the ratio 2a
(2)
the subduplicate ratio of 64
(3)
the duplicate ratio of
2.
If
#+7
:
36,
:
2a
-j-
:
9,
/6a?
:
-
M
-,
2 (# + 14) in the duplicate ratio of 5
make
What number must
it
equal to
y=3
5.
If x
6.
If 15 (2a-2
:
1
:
:
4,
3
:
ratio
:
8,
:
Sax
56.
:
2by.
find x.
12 so that the greater
be added to each term of the ratio 5
\
find the ratio of
- y 2 ) = *7xy,
ab.
:
ratio 27
and the
Find two numbers in the ratio of 7
exceeds the less by 275.
4.
ratio of 9b 2
and the
3.
to
;
7x-4y
find the ratio of
x
:
:
3x+y.
y.
:
37
—
"
RATIO.
7
?=£ = «
If
2rt 4 &
2
+ 3a 2 e 2 -5eV = "_
_^__^__
prove that
= =
6ca
-7
If v
8.
11
prove that
,
is
-j
a
l
equal
t<»
y
a
9.
If
shew that
10.
If
y
_
r+p-q p + q-r
q + r-p
(q - r) x + (r - p) y + (p - q) z = 0.
——
x-z
—
==-
=-
-
find the ratios of
,
y
z
Khew
#
if
2 (*+?+*)
tliat
12.
a+o + c
1<J.
-
+a
—„-
If
prove
1
16.
If
3
„ t
.r-fa 2
c
+ 6 z3 + c 3 _(.y + + *) 3 + (q + &+c)
t
+ 2 + c- — (>c+y + \« /_
_\s
y* + b
5) + (a + o + c)y
3
3
-
3
?/
-
;,,
.,
2
2
,
.,
i
i
26 +
2c-a
=
2c
+ 2a"-
.
_
— 2A-+2y-g
=
6
j.
•
,
J
2a + 26-c
+y 2 + ^2 = («.v+^ + ^) 2
x a=y b = z c.
I (my + rut - Ix) = m (nz + Ix - my) = n (Ix + my - nz\
y+z-x = z+x-y x+y-z
=
n
(a 2 +6 2 + c2 )
(.i-
2
)
:
:
,
:
——
-
-j
m
I
Shew
-
•
that the eliminant of
ax + cy +
bz
= Q,
a3
is
17.
o
2y + 2g-.v _
— 2g + 2.-c-y
II
shew that
15.
6c + <?a-|-a6
a
BheW that
14.
z.
:
i'=^ = 2-,
.tfS
shew
'y
_ (6+o).r+(C +«)y+(« ± i)i
If
tliat
:
y+ z = z+ v == r+ ^
pb + qc pc + qa pa + qb'
'
ii
x
Eliminate
ctx
x, y, z
= 0, bx + </y + c; = 0,
+ & 3 + c3 -3«6c = 0.
cx + by + az
from the equations
+ hy + (/z = 0, hx + by-\-fz = 0, gjc+fy+C2=0.
:
HIGHER ALGEBRA.
12
18.
x = cy + bz, y = az+cx z=bx + ay,
If
}
X
z
=
=
j—*
y
1 - a
\-b
\—c
19.
2
II
shew that
n
•>
i l
1
9
L
Given that a(y + z)=x, b(z + x)=y, c(x+y)=z,
prove that
bc
+ ca + ab + 2abc = l.
Solve the following equations
3x-4y + 7z =
20.
:
z,
3x-2y+17z =
0,
x* + 3f + 2zs = l67.
tyz + 3sa?=4an/,
2tys - Sac = 4ry,
22.
x+y=
21.
0,
2x-y-2z = 0,
to?-f+£=l8.
3x 2 - 2y 2 + oz 2 = 0,
23.
5.0-4^ + 73 = 6.
+-^L^
.--*
If
I
Ja+Jb
m
+
Jb+Jo
—— =
shew that
- 3y 2 -I5z 2 = 0,
7a*
ff+2y+32=19.
24.
•
(b - c)
(a-b)(c-\/ab)
+
(a -
*„<>,
n
</c+V«
=
?==be)
V
'
=(c - a)
(b
-
\J ac)
Solve the equations
ax + by + cz = 0,
bcx + cay + abz = 0,
xyz + abc (a3x + b 3y + &z) = 0.
25.
a.-£+&y
26.
+ C2=a 2# + & 2y +
6' 2
2==0,
x + y + z + (b-c)(c-a) (a-b) = 0.
27.
a(y+x)=x,
If
b(z + x)=y, c(x+y)=z,
X
—
7— = —
=
—
a -be) b(l-ca) c(l-ab)
2
prove
that
1
28.
If
?/
-
,
2
*
s2
{I
ax + ky+gz = 0, kx + by + fz^0, gx+fy + cz = 0,
prove that
^
x2
bc-f 2
(2)
(be
-f
y
2
z2
ca-g 2
2
)
{ea -
g
ab-h 2
2
)
(ab
- h 2 ) = (fg - eh)
(gk - af) (A/- bg).