3y - 4

from (1)

4.x - 3 Example 2. Solve 3.x

.(1), 7

2
1
(2.x – 5)=y ....

(2). 5 3 Clear of fractions. Thus

42x – 2y +10=28x – 21;
.. 14.x – 2y=-31.

(3). From (2)

9y + 12 – 10x + 25=15y;

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

14

13 Eliminating x from (3) and (4), we find that

207 y=

26 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.

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.

RULE. Eliminate one of the unknowns from any pair of the equations, and then eliminate the same unknown from another pair. I'wo 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. Example 1. Solve 62 +2x - 55=13

(1), 3x + 3y – 2z=13

(2), 7x + 5y - 3z=26

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

18x+6y - 152=39,

6x + 6y – 4z=26; subtracting, 12.x – 112=13..

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

30x + 10y – 252=65,

14x+10y – 6=52; subtracting, 16x – 192=13..

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

48x – 44z=52,

48x – 572=39; subtracting,

132=13

z=1,) and from (1)

x=2, from (1)

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 – 4z=0, or x=22. Substituting in (1) and (2) we have two easy equations in y and 2.

y=3.)

we have

we have

From the equation

1=
2
3x – Y=12

(1).
-1= +2,
2
7x – 22=42

(2). y And, from the equation =13,

2 we have 2y + 3z=78

(3). Eliminating z from (2) and (3), we have

21x+4y=282; and from (1)

12x – 4y=48; whence x=10, y=18. Also by substitution in (2) we obtain z=14.

+

(2),

or

Example 3. Consider the equations
5x – 3y - 2=6

(1),
13.x – 7y + 3z=14
7. – 4y=8

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

28x – 16y=32,

7x – 4y=8. 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.]

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.

EXAMPLES XIX. C.

1.

2.

x+2y + 2z=11, 2x+y +z 7, 3x+4y+z = 14.

x+3y + 4z=14, x+2y + 2= 7, 2.3+ y +22= 2.

a=

a

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
and
7

;

and consequently a is related to b exactly as b is related to a.

1

1
The reciprocals of x and y are and respectively, and in

y
1
and

1 solving the following equations we consider

as the unknown quantities.

(1),

(2).

9 Example 1. Solve

= 1

y
10 6
+ = 7

ข Multiply (1) by 2 and (2) by 3; thus

16 18

:2,
2 y
30 18

+

=21;

=23;

(1);

(2),

(3),

y

46 adding,

2 nultiplying up,

46=23x,

x=2; and by substituting in (1),

Y=3.

1 1 1 1 Example 2. Solve +

2.x 4y 3z 4

1 1

Зу 1 1 4 2

+ = 2

5y Z 15 clearing of fractional coefficients, we obtain

6 3 4
+

3.
у z

3 1 from (2)

=0. C Y

15 3 60 from (3)

32 y

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

105 42

=77;

Y

15 6 dividing by 7

+ 11

y

18 6 from (5)

-0;
y
33

=11;

from (1)

.(4),

..(5),

+

(6).

+

....(7);