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SIMPLE EQUATIONS. Art. 1. Simple equations, or equations of the first degree, may involve one, two, three, or more than three unknown quantities.

If a simple equation involve only one unknown quantity, the equation admits of only one value of that quantity.

The solution of all equations in general will be found to depend on the axioms stated on the eighth page of Section IV.

In the application of the second and third axioms in the solution of equations, it will be seen that if an equal quantity be added to or subtracted from each side of an equation, the process is the same as simply transposing or removing the quantity from one side to the other side of the equation with its algebraical sign changed; as the addition of ta to each side of the equation mxa =b, is mx—ata =b+a; and is equivalent to the removal of -a from one side to ta on the other, as mx=b+a. And the subtraction of ta from each side of the equation mx+a=b, is muta-b=b-a, and is equivalent to the removal of ta from one side to -a on the other, as mx = b-a. Hence the second and third axioms are united in the general rule, that any quantity may be transposed from one side to the other side of an equation by changing its sign. By the application of this rule the terms involving the unknown and known quantities can be separated and placed together, the former on one side and the latter on the other side of the equation; and when the equation is reduced to the form mx =c, by the seventh axiom the unknown quantity is determined to

с be x=-

m And by this axiom an equation may be simplified if every term of it contain the same common factor which does not involve any unknown quantity.

Also by the sixth axiom an equation may be cleared from fractions by multiplying each side of the equation by the denominators of the fractions, or by their least common multiple, as the equation *+ * = o becomes bx + ax = abc by multiplying each side by a

c a and b, the denominators of the fractions.

And further, the signs of all the terms of an equation may be changed, as the process is equivalent to multiplying the terms on each side by -1.

It is also obvious that if the same quantity with the same sign be found on both sides of an equation, it may be removed from each side.

If a term involving the unknown quantity in an equation be

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affected with a fractional index, by means of the eleventh axiom the fractional index can be reduced to an integral index, and the separation of the known and unknown quantities become possible, as if (mx+b)}=c, by raising each side of the equation to the second power, it becomes mx+b=co, and the known quantity b can be separated from

6 x the unknown quantity.

2. To solve the general equations a,x+by=0aqx+by = 0,, involving two unknown quantities, or to find the values of x and y which satisfy both equations.

When there are given two independent equations of the first degree which simultaneously involve two unknown quantities x and y, it is obvious that, in order to determine the value of one of them, thero must first be deduced from the two given equations an equation which shall contain only one of the unknown quantities.

There are three methods by which the values of x and y may be determined from these equations, a,x+6y=C1, 42x+b.y = Cz.

First method: By finding the value of one of the unknown quantities in terms of the other and known quantities, from one equation, and substituting it in the other equation. Let the value of x be found from the first equation ay.c+b,y=ci.

61-by By transposing by, a,x=67-by, and x = then substituting

dy this value of x in the second equation, ayt+b.y=cy, it becomes ag(c-by)

+bay=, clearing the equation from fractions, Qy 220—a,b,y +a,bay = 2,c2, and by transposition (a,bz-a,b )y = 2,0;—0,0,, * &,C2 - A,C,

and the value of x is found by substituting this abe-abi' value of y in either of the two given equations.

с —ъ,у с by c. 2 ас1,0
obry Q a, α, α, α,β, ,β,

a, abıazbi
b26, -6,C

a, b, - a,b, And these values of x and y, when substituted in the given equations, atisfy both equations by reducing each to an identity.

a,(6,0-6,ca), bacı-acı)
Thus a,x+by=

+
a,b,-azbi a, b, a,b,
a,b,c, -a,b,c,ta,b,c,- a,b,

a, b, a,b,
(a,be-a,b)c

a,b2-aeb And a +by =

az(b2c, —6,02), b2(4,22—2,c)

+
a,b,ca,b, a, b,- a,b,
azb,c—agb,c +a,b,c,a,b,

cybe-azbi

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and .. y=

.. 2=

or =

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C=

(a,b,- a,b,c,

a, b, - azbi Second method: By equating the values of the same unknown quantity from each equation. From the first equation, a,x+b,y=0, x=9–6,4,

a From the second, agx+b2y = (2, x=

C2-b,y

а. Next equating these values of x,

0,-by C.-by
an

A2
Clearing the equation from fractions, azcı a,b,y = 2,62 -- a,bay,
and (a,b, - a,b)y=0,0,– Azę, by transposition,

.. Y=

ac,- A,C,

a, b, -0,6
and the value of x can be found as before,

b2c, – 6,02

a, b, - a,b, Third method : By equalising the coefficients of one of the unknown quantities in each equation, and then eliminating it by addition or subtraction of the equations.

In the given equations a,x+by=(1, 2,x+by=c, if a, ay, the coefficients of x, be prime to each other, multiply the first equation by Qg, and the second equation by an,

then a, a,x+a,by=aqcı, and aja,x+ab.y = 2,02. Next add or subtract these equations according as the signs of the coefficients of x are either one positive and the other negative, or both positive or both negative, and the resulting equation is

(a,b, - a,b)y = 22c; – 2,62,

5,6,- 6201

as before.

azb, - ab It

may be remarked that the signs of the terms in the values of 2 and y appear with signs contrary to the terms in the values found by the other methods. This circumstance arises from subtracting the second equation from the first. If the first had been subtracted from the second, the signs would not have been changed.

The values of x and y are of the same value in both instances, as a fraction remains unchanged in value when the signs of every term of the numerator and denominator are changed.

Also, if the known quantities involved in the equations be positive or negative integers or fractions, the values of x and y in general will be either integers or fractions positive or negative.*

aze, – AC2, and x=

.. Y=

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a,b, - ab'

The values of u and y in two simple equations whon reduced to the form

m

or

m

m

6, and that of the other

Let the value of z assume the form ,, then from ==

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m

9

It is possible, however, that the relation of the constants in the equations to each other may be such that the values of x and y may

0

0 assume the forms of

0 0 If the values of one of the unknown quantities assume the form of 0 o 0

6. b, ca

a, b, - azbi
baf1 - 6.c, = 0, and a, b, - a,b, = 0,
:: 6,0 = b;ca, and "_b.

cb2
also a,b= a,b,, and a b Hence 947
a =

be

с а
a, b,

Cg
Next let the value of y assume the form then from y=

a,C2 - ,

a, b, - a,b a, b, a,b, = 0, and a co-2,4 =m, .. a,bera,bi, and ; also acq=mtaze, and a 9.

a. а.с. с. But from the former ", _9, @ 9, which is impossible,

+-, а,

C2 ас, с. if m have a definite value, therefore the given equations are incompatible, and do not admit of any definite values of x and y.

If, however, the values of x and y both assume the form of 0 then ac,-0.9, = 0, a,b-a,b, = 0, and 6.6, -6,6, = 0, a, b, & ab C if a

a,
dz
6, а, с. boca

a, b, c, and a, = naz, b, = nba, a = ne,, which shews that the co-efficients of one equation are respectively equimultiples of the corresponding coefficients of the other. Hence it follows that one of the equations may be deduced from the other, and the two equations do not express two independent conditions.*

m

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+

m

C2

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En, then

=

=

a ,x+b, y=cq, azx+b2y=ca, can be found by substituting the particular for the general values of the constants in the values of x and y, as, for example, To find the values of x and y in the equations 5x + 8y = 31, 2x – 3y=0. Generally x=

6,6, -6,C,, a,€, –a,C,

a, b, - a,b a, b, -a,b,
Ilere a, =5, 6, 8, C, = 31; 0,=2, 6,= -3, c, = 0.
-93-0 -93

+3,
-15-16 -31

0-62 62 y=

-15-16 -31 ..3 and 2 are the values of x and y in the equations 5x+8y=31, 2x – 3y=0.

The subject of indeterminate equations arises from the fact of some problems involving more unknown quantities than independent conditions to determine them.

+2.

And in this case there is only one independent equation involving two unknown quantities. Hence the equation is indeterminate, and admits of an indefinite number of solutions, as it is evident that for every value of x in the equation there will always be some corresponding value of y.

If there be three equations involving two unknown quantities, and if one of the equations can be deduced from either of the other two, then these three equations become only two independent equations involving two unknown quantities.

If one of the equations be deducible from either or both of the other two, the three equations can be reduced to one equation having two unknown quantities, and is therefore indeterminate.

If, however, all the three equations are independent of each other, they may express conditions incompatible with each other, or be contradictory. But if the equations be taken two and two at a time, the three sets may admit each of different solutions with definite values of the unknown quantities.

3. To solve the general equations involving three unknown quantities, 0;&+by+z=d, (1), 2g.+by+0,8 = d, (2), Qgx+by+cox=d(3); or to find the values of x, y, & which will satisfy the equations.

Here any one of the three unknown quantities can be eliminated between any two of the three equations. First, eliminate x between the first and second equations.

Multiply (1) by ag, then a, a,&+azby +2,0,2 = aydı,

Multiply (2) by ang ają,& tabyta czł = a,da,
.. (azb, –a,b2)y+(a,0,-a,caz=azd, -a;d, by subtraction,

or py+qz=r, putting p, q, r for the coefficients. Next, eliminate x between the first and third equations.

Multiply (1) by as, then a azt+azby +230,2 = aydı,

The particular problems of Diophantus are characterised by their simplicity and neatness of solution, and the peculiar skill of the artifices he employs, and they afford considerable scope for exercising the ingenuity and sharpening the invention of the student.

Bachet and Fermat, in the seventeenth century, made some additions to the knowledge of this subject as left by Diophantus. Bachet was the first who gave a complete method of solving all indeterminate problems of the first degree. Fermat devised some methods for the solution of indeterminate problems above the first degree. Diophantus solved a problem which is equivalent to that of finding a so that a2+1 shall be a square number. Fermat extended it to that of finding a so that ax? + b shall be a square number, a and b being some constant numbers.

In the second part of the third English edition of Euler's Algebra, the student will find the subject of indeterminate problems very amply treated, containing some generalisations of the Diophantine problems involving higher powers than the first of the unknown quantities.

In the additions of M. de la Grange at the end of the volume will be found the method of applying continued fractions to the solution of indeterminate problems.

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