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42x-2y+10=28x-21;
... 14x-2y=-31..
9y+12-10x+25=15y;

From (2)

.. 10x+6y=37 .... Eliminating y from (3) and (4), we find that 14

13

= 2.

=

3x

Eliminating x from (3) and (4), we find that
207
= 26

y:

7. 2x-13=3,

4x-y=20.

X

=

10. +1 7

3y+4
5

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NOTE. Sometimes, as in the present instance, the value of the second unknown is more easily found by elimination than by substituting the value of the unknown already found.

2.

y-5

7

1

8.

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4.x - 3

2

(2x-5)=y.

x

9

x=

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=

1

x-y=4.

5. +/-10,

=

1
Xx

=

= 5,

3+y=50.

1

5y=4,

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11. 3x-7y=0,

2 5

7x+3y=7.

3.

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.(1),

(2).

-3x=8.

12. 2-2-0,

=

5 4

.(3).

(4).

1

3x+5y=17.

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SIMULTANEOUS EQUATIONS INVOLVING THREE UNKNOWN QUANTITIES.

161. In order to solve simultaneous equations which contain two unknown quantities we have seen that we must have two equations. Similarly we find that in order to solve simultaneous equations which contain three unknown quantities we must have three equations.

Example 1. Solve 6x+2y-55-13

3x+3y-2z=13
7x+5y-3z=26

subtracting,

RULE. Eliminate one of the unknowns from any pair of the equations, and then eliminate the same unknown from another pair. Two equations involving two unknowns are thus obtained, which may be solved by the rules already given. The remaining unknown is then found by substituting in any one of the given equations.

Choose y as the unknown to be eliminated.
Multiply (1) by 3 and (2) by 2,

x+1 3y-5_x-y

-=

10

18x+6y-15z=39,
6x+6y4z=26;

subtracting,

12x-11z 13.

Again, multiply (1) by 5 and (3) by 2,

subtracting,

and from (1) from (1)

Multiply (4) by 4 and (5) by 3,

30x+10y-252=65,
14x+10y6z=52;

16x-19z 13..

48x-44z=52,

48x-57z=39;

=

13z=13

z=1,

x=2,

y=3.)

8

(1),

(2),

(3)..

..(4).

..(5).

NOTE. After a little practice the student will find that the solution may often be considerably shortened by a suitable combination of the proposed equations. Thus, in the present instance, by adding (1) and (2) and subtracting (3) we obtain 2x-4x=0, or x=2z. Substituting in (1) and (2) we have two easy equations in y and z.

or

Example 2. Solve

From the equation

and from (1) whence x=

1.

we have

Also, from the equation-1+2,

we have

7x-2z=42

And, from the equation +13 =13, we have

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2-1=1/2+1,

3x-y=12

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2y+3z=78 Eliminating z from (2) and (3), we have

3

x+2y+2z=11,

=

7,

2x + y + z
3x+4y+z =14.

21x+4y=282;
12x-4y=48;

Example 3. Consider the equations

5x-3y- z=6 13x-7y+3z=14 7x-4y=8

Multiplying (1) by 3 and adding to (2), we have

28x-16y=32,
7x-4y=8.

=10, y=18. Also by substitution in (2) we obtain z=14.

EXAMPLES XIX. c.

(1).

Thus the combination of equations (1) and (2) leads us to an equation which is identical with (3), and so to find x and y we have but a single equation 7x-4y=8, the solution of which is indeterminate. [Art. 153.]

2.

(2).

In this and similar cases the anomaly arises from the fact that the equations are not independent; in other words, one equation is deducible from the others, and therefore contains no new relation between the unknown quantities which is not already implied in the other equations.

x+3y+4z=14, x+2y+ z= 7, 2x+y+2z= 2.

(3).

(1),

(2),

(3).

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The reciprocals of x and y are

By

16. x+20= +10
2

2

162. DEFINITION. If the product of two quantities be equal to unity, each is said to be the reciprocal of the other. Thus if ab=1, a and b are reciprocals. They are so called because

1

1

a=

=, and b=; and consequently a is related to b exactly as b is related to a.

α

1

and
Ꮳ y

=2z+5
=110-(y+2).

solving the following equations we consider known quantities.

andrespectively, and in

1

and 1

20

y

as the un

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dividing by 7

from (5)

8100

from (5) from (4)

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x

30

X

[ocr errors]

9

1 1
+
2x 4y

y

18

+ =21;

y

46

x

X

18

x

1

=23;

46=23x,

x=2;

y=3.

1

3z

1

==

X By

...

1

x

clearing of fractional coefficients, we obtain

from (1)

3

from (2)

y
15 3 60

from (3)

X

+ =32 У 2 Multiply (4) by 15 and add the result to (6); we have

105 42

+ =77; x y

15 6

+

7

1

4

+ 5y Z

6 3 4
+
x y 2

1

y

6

y

33

x

1

4

-

1

2

=0

11

=0;

2

15

=11;

.. x=3,

y=1,
z=2.)

143

.(1),

(2).

(1),

(2),

(3),

.(4),

.(5),

(6).

..(7);

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