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SIMULTANEOUS EQUATIONS. 71. When two or more unknown quantities are to be determined, there must always be as many independent equations as there are unknown quantities; and since the values of these unknown quantities must be the same in all the equations, the values are said to subsist simultaneously, and the equations are called simultaneous equations.

CASE I. To determine two unknown quantities from two independent equations.

72. RULE I. Make the coefficients of one of the unknown quantities the same in both equations, then by adding or subtracting these equations, there will result an equation containing only the other unknown, whose value may be found by the previous rules.

Note 1. The equations are to be subtracted when the quantity whose coefficient is rendered the same in both equations, has the same sign in each, and added when it has opposite signs.

Note 2. The coefficients of either of the unknown quantities may always be rendered the same in both equations, by multiplying the first equation by the coefficient of the unknown quantity, which is to be made to disappear in the second equation, and the second equation by the coefficient of the same letter in the first. By this means the coefficients of that quantity will evidently be the same in both, for it will be the product of its two coefficients in the original equations.

73. Rule II. Find a value of one of the unknown quantities in terms of the other from each of the equations, and make these values equal to each other, which will give an equation containing only the other unknown, from which its value can be found by the previous rules.

74. RULE III. Find a value of one of the unknown quantities in terms of the other from one of the equations, and substitute this value instead of it in the other, from which there will again result an equation containing only one unknown quantity, which may be solved as before.

75. RULE IV. Multiply one of the equations by a con-' ditional quantity n, then add this product and the other equation together: let n fulfil the condition of making the coefficient of one of the unknown quantities 0, then the equation will only contain the other unknown ; determine the value of n that fulfils the above condition, and substitute this value instead of it in the resulting equation, and the value of the other will be determined.

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Note 3. All the above operations can be performed first in relation to one of the unknown quantities, and then in relation to the other, which will give independent values of each; or the value of one of the unknown quantities being found, its value can be substituted instead of it in either of the given equations, from which the value of the other can be determined.

EXAMPLE. Given 3x+5y=35, and 72—4y=19; to find the values of x and y.

By Rule Ist.
1 3x + 5y= 35.

2 7x - 4y= 19. (1) x 7 3 21x+35y=245. (2) x 3 4 21x-12y= 57. (3)–(4) 5

47y=188. (5):47 6

is y=

4. (1) x 4

7 12.0 + 20y=140. (2) x 5 8 35x—20y= 95. (7) +(8) 9 473 =235. (9):47 10

5.

By Rule 20.
From (1.) by transposition and division, x=

3

19+4y And from (2.)

7 19+4y

Ax. Ist.

7
245–35y=57+12y by mult. by 21.

188=474 by transposition.

.4=y by dividing by 47.
And substituting this value of y in (1.) we obtain

3x+20=35,
Hence 3x=15,
And x=5, as before.
By Rule 3d.

35-54 It has already been found from (1), that x=

3 stituting this value, instead of x in (2) it becomes 7(55)—4y=19

. 245–354—12y=57, by mult. by 3.

188=474, by transposition.

:. 4=y. In the same manner may x be found from either equation, by substituting a value of y found from the other equation.

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35—5y

3

; suby=10.

By Rule 4th. Multiplying (1) by n, and adding (2), we obtain (3n+7)x+(ön-4)y=35n+19, which, if the coefficient of y be 0, becomes (3n+7)x=35n+19; and since the coefficient of y is =0, 5n—4=0; hence n= =* ; substituting this value of n in the equation (3n+7)x=35n + 19, it becomes

9 =47. Hence 47x=235.

And ..x=5. Next, let n be such as to render the coefficient of x=0, then the equation will become (5n-4)y=35n+19; and

7 since 3n+7=0:: n=-5, substituting this value instead of n in the equation, and changing the signs, it becomes

15 y=623 Hence 47y=188.

And .. y=4. 1. Given

3x +2y=561 to find the values of x and y. 2x+3y=54

Ans. x=12, 2. Given

5x7y= 81 to find the values of x and y. 4x— y=34)

Ans. x=10, y=6. 3. Given *x+3y=7 to find the values of x and y. +y=9

Ans. x=9, y=8. 4. Given J 3x+ ļy=38 to find the values of x and y. x+2y=12

Ans. x=12, y=4. 5. Given (x-y:=20 to find the values of x and y. x +y =10

Ans. x=6, y=4.

to find the values of x and y. x-y=d) Ans. x=}(s+d), y={(s—d). 3x+2y +2x=16 to find the values of x

"

6 7. Given

and 2x-3 2x+7y

y. =4

Ans. x=6, y=3.

to find the values of x and y. 8. Given ax+by=0

ch'-cb ca'c'a a'x+by=d Ans. x=

Fab'ab3Y=ba' - ab The above equations may all be solved, by substituting in the answer to the (8) the proper values of a, b, c, a', u', and d', with their proper signs; only (7) would require to be reduced to the proper form before the substitution can be made.

CASE II. 76. To determine the values of three or more unknown quantities from as many independent equations as there are unknown quantities.

6. Given S x+y=8

+

9

11

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Rule I. Multiply each of the equations by such multipliers as will make the resulting coefficient of one of the unknown quantities the same in all the equations, then by adding or subtracting these equations, a new set of equations, one less in number, and containing one unknown quantity less than before, will be obtained; with which proceed as before.

77. RULE II. Find a value of one of the unknown quantities, from each of the equations, in terms of the other unknowns; then equate these values, and a new set of equations will be obtained, containing one unknown quantity less than formerly; with which proceed as before.

78. RULE III. Find a value of one of the unknown quantities from one equation, and substitute this value, instead of it, in the others, which will evidently give one equation less than formerly, containing all the unknown quantities, except that whose value was found, with which equations proceed as before.

79. RULE IV. Multiply all the equations except one by some conditional multipliers, as m, n, p, &c., then add all the multiplied equations and the unmultiplied one together, and a new equation will be obtained, in which if we make all the coefficients of the unknown quantities except one equal to 0, a set of equations among the conditional multipliers will be obtained, which will determine their values, and these values substituted in the resulting equation will give the value of the other unknown quantity whose coefficient was not considered to be 0.

The above rules, as well as those given in Case I. are all evidently true from the axioms.

EXAMPLE,

30—2y+62=22) to find the values of x, y, Given 2x+7y82=24

and z. (4x + y + z=30 Multiplying the first equation by 4, the second by 6, and the third by 3, they become by Rule 1.

1 12x–8y +24z=88.
2 12x +42y—482=144.

3 12c +3y +3=90. (2)–(1)

4 50y-722=56. (2)–(3)

5

397–512=54. (4): 2

6

257-36=28. (5): 3

7 137–172=18. (6) x 13

8 3257-468z=364. (7) X 25

325y-4252–150,

(9)–(8)

10 432=86. (10): 43

11 .:.2=2. By subst. in (6) 12 257–72=28. Transposition 13 25y=100. (13):25

14 ..y=4. By subst. in (1) 15 12.–32 +48=88. Transposition 16 12x=72. (16): 12

17 1.x=6. The solution by Rules 2d and 3d are left as exercises for the pupils. The solution by Rule 4th is as follows: multiplying the first of the given equations by m, the second by n, and adding these products to the 3d of the given equations, we obtain the following: (3m+2n+4)x -(2m-7n-1)y+(6m-8n+1)=22m +241 +30.

Let now m and n be such as to make the coefficient of y and 2 each equal to 0, and the following equations are obtained, in which m and n are the unknown quantities.

1 2m-n-1=0.

2 6m—8n+1=0.
(1) x3 3 6m-21n-3=0.
(2)-(3) 4 13n+4=0.
Transp. 5 13n=-4.
(5):13

n=-is. In the same manner, or by substitution, we obtain m= 15

and since the coefficients of y and 2 are each=0, 26 have (3m +2n+4)x=22m +24n + 30, in which, by substituting the values of mand n, there results, 45 8

330 96
+4)=
=-

+30,
13

26 13 hence 127x=934

43x=258, by mult. by 26.

x=6. In the same manner, if the coefficients of x and z be equated to 0, we can find the value of y; or if the coefficients of x and y be equated to 0, we can find the value of z.

(x+y=z=10 to find the values of x, y, 1. Given

X-Y +2= 6 y+2x= 2 Ans. x=8, y=6, 2=4. x + y + z=35

to find the values of x, 2. Given

x+2y+32=66

3x+3y + 4z=13) Ans. x=12, y=15, z=8. 3. Given

to find the values of x, y, and z. +z=26

Ans. x=16, y=12, z=10.

we

and %.

y, and z.

x+y=28

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y+z=22

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