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From which we find y=4; and substituting this value for y, in either of the equations (1), or (2), we find x = 1.

Such equations are called simultaneous.

Simultaneous equations are those in which the values of the unknown quantities are the same in both.

We have seen that it requires two simultaneous equations, containing two unknown quantities, to determine the values of the unknown quantities. In the same way, it could be shown that it would require three equations containing three unknown quantities, four equations containing four unknown quantities, and so on, to determine the values of the unknown quantities. In general, there must be as many equations as there are unknown quantities. The equations necessary to determine any number of unknown quantities, constitute a group of simultaneous equations.

Such equations are solved by successive elimination.

ELIMINATION.

84. Elimination is the operation of combining two equations so as to get rid of one of the unknown quantities which enter them.

There are three principal methods of elimination: by addition, or subtraction; by substitution; and by comparison.

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Multiplying both members of (1) by 4, and of (2) by 6, (axiom 3°), we have, 28x + 24y = 80

(3) 54.x 247

(4)

.

.

.

= 84

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.

Adding (3) and (4), member to member, (axiom 1°), we have,

82x = 164.

Here, y has been eliminated by addition.

Again, multiplying both members of (1) by 9, and of (2) by 17, 63x + 547

(5) 63x – 28y = 98

(6)

= 180

.

Subtracting (6) from (5), member from member, we have,

82y

= 82.

Here, x has been eliminated by subtraction. In the same manner, an unknown quantity may be eliminated from any two simultaneous equations; hence, the

RULE.

Prepare the equations, so that the coefficients of the quantity to be eliminated shall be numerically equal in both; if the signs are unlike, add the equations, member to member; if alike, subtract them, member from member,

In preparing the above equations, we multiplied both members of each, by the coefficient of the quantity to be eliminated in the other. They may be prepared in other ways. A better way, in most cases, is to find the least common multiple of the coefficients

of the quantity to be eliminated; then multiply both members of each equation by the quotient of this least common multiple by the coefficient of the quantity to be eliminated in that equation. In the first case considered, the least common multiple of 4 and 6 is 12; we might have multiplied both members of (1) by *, or 2, and of (2) by us, or 3, giving,

14x + 12y = 40
27.0 — 12y

= 42. Whence, by addition,

41x

= 82.

Here, y has been eliminated as before, but we have a simpler equation.

2o. Elimination by Substitution.

86. Take the same equations as before:

700 + 6y = 20
9x 4y

(1) (2)

= 14

Finding, from (1), the value of y in terms of 2,

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Substituting this value of y, in (2), we have,

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Here, y has been eliminated by substitution. In the same manner, we may eliminate an unknown quantity between any two simultaneous equations; hence, the

RULE.

Find from one of the equations the value of the quantity to be eliminated, in terms of the

other quantities; substitute this value for that quantity in the other equation.

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Finding the values of x in terms of y, from each of equations (1) and (2), we have,

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Placing these two values equal to each other, we have,

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Here, x has been eliminated by comparison. In the same manner, an unknown quantity may be eliminated between any two simultaneous equations; hence, the

RULE.

Find from each equation the value of the quantity to be eliminated; place these values equal to each other.

Of the three rules given, any one can be used, as may be most convenient. As a general thing, that is employed which gives rise to the simplest equations.

SOLUTION OF GROUPS OF SIMULTANEOUS EQUATIONS.

88. Take the group of three equations:

.

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3x + 4y
2z = 10

(1)
5x
2y + 3z = 16

(2) 4.x + 2y + 22 = 22

(3) Combining (1) and (2), also (1) and (3), eliminating % in each case, we have the new group, 19x + 8y = 62

(4) 7 + 6y = 32

(5) Combining (4) and (5), eliminating y, we have the single equation,

29x = 58; :... X = 2.

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Substituting this value of x in (5), we bave,

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Substituting these values of x and y in (1), we have,

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In the same manner, any group of simultaneous equations may be solved; hence, the

RULE.

1. Combine one equation of the group with each of the others, eliminating one unknown quantity : there will result a new group containing one equation less than the original group.

II. Combine one equation of this group with each of the others, eliminating a second unknown

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