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244. Having fully explained what concerns the elimination of unknown quantities in simple equations, and also illustrated the characters by which it may be known, if the proposed equations be determinate, indeterminate, or impossible; we may now proceed to the resolution of examples in determinate equations of the first degree: the pràctical rules that are necessary for this purpose, shall be pointed out in the two following sections.

§ II. RESOLUTION OF SIMPLE EQUATIONS,

Involving two unknown Quantities.

245. When there are two independent simple equations, involving two unknown quantities, the value of each of them may be found by any of the following practical rules, which are easily deduced from the Articles in the preceding Section.

RULE I.

246. Multiply the first equation by the coefficient of one of the unknown quantities in the second equation, and the second equation by the coefficient of the same unknown quantity in the first. If the signs of the term involving the unknown quantity be alike in both, subtract one equation from the other; if unlike, add them together, and an equation arises in which only one unknown quantity is found.

Having obtained the value of the unknown quan. tity from this equation, the other may be determined by substituting in either equation the value of the quantity found, and thus reducing the equation to one which contains only the other unknown quantity.

Or. Multiply or divide the given equations by such numbers, or quantities, as will make the term that contains one of the unknown quantities the same in each equation, and then proceed as before.

S 2x+3y=23, Ex. 1. Given

to find the values

5x-2y=10, S of wand y. Multiply the 1st equ" by 5, then 10x+15y=115;

2nd by 2, then 10x – 4y= 20;

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23–3y

by subtraction, i9y=95,

95 by division,

Y=

19' Now, from the first of the preceding equations

23-15 we shall have =

F(since y=5) 2

2 8

-4. 4

The values of sand y might be found in a similar manner, thus : Mult, the 1st equation by 2, then 4x+6y=46 ; 2nd

by 3, then 15x-by=30;

.

.. by addition, 19x=76,

76 by division, w= 4.

19 Now, from the first of the preceding equations, 23- 2x

15 we shall have y=

= 3

3 :5.

=(since w=4)23-8

3)

40+ 9y=35,1 Ex. 2. Given

to find the

6+12y=48, S values of x and y. Muit. the 1st equation by 6, then 24x+-54y=210; 2nd

4, 24x +48y=192 ;

.

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.. by subtraction, 6y=18,

18 by division, y= 3.

6 Now, from the first of the preceding equations, 35-97

35-9X3 we shall have x= =(since y=3) 4

4 35-27

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The values of x and y may be found thus; у

3 Mult, the 1st equation by 3, then 12.x+27y=105; 2nd

12x+24y= 96 ;

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9

3

... by subtraction, 3y=9,
by division, y= -3.

35-27 8
And ..x

4 4 : The numbers 3 and 2, by which we multiplied the given equations, are found thus ; :: The product of two numbers or quantities, divided by their greatest common measure, will give their least common multiple, (Art. 146). 6 X4

=12 the least common multiple,

12 Then 33, the number by which the first equa4

12 tion is multiplied ; and -2, the number by which

6 the second equation is multiplied.

By proceeding in a similar manner with other equations, the final equation will be always reduced to its lowest terms. Ex. 3. Given $5x+4y=58,

to find the va

3x + 7y=67, lues of x and

y. Mult. the 2nd equation by 5, then 15x+35y=335; Ist.

3, 15x+12y=174;

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and y=

;7ے

30

.. by subtraction, 23y=161,

161

23 whence, 5.x=58-4y=58--28=30,

and

-6.

5 If the second equation had been 'multiplied by 4, and subtracted from the first when multiplied by 7, an equation would have arisen involving only , the value of which might be determined, and thence, by substitution, the value of y. Ex. 4. Given

to find the 57-6y=-10, values of x and y. Mult, the 1st equation by 3,

1830 — 6y= 42; but 5x-by=-10;

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=}

.. by subtraction, '13x=52, and a=t. 5x+10 20+10

=30=5. 6

6

whence y=

or

247. These values being substituted in the place of x and y in each of the equations, shall render both members identically equal, or, what is the same thing, each of the equations will reduce to 0=0.

Thus, by substituting 4 for ®, and 5 for y, in the above equations, they become 6 X 4-2 X 53 -14,

1. 14;? 5 X 4-6X55--10; S

-10=10. Therefore, by transposition,

14-14=0, or 0=0; and -10+10=0, or 0=0. Since (Art. 56) 14-14=1, and 10-10=0.

If these conditions do not take place, it is evident that there must be an error in the calculation : therefore, the student, whenever he has any doubt respecting the answer, should always make similar substitutions.

.
11x3y=

to find the va41-7y=

: lues of x and

y. Mult the 1stequation by 7, then 77x+ 217=700, 2nd

3,

12x -217= 12;

Ex. 5. Given { 18+g=100;}

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... by addition, 89x=712,

712 by division, =

and .. =8; whence 3y=100--11x=100-11X8=100-88

898

12;

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