2.8 Solution of simultaneous equations 57
EXAMPLE 2.33Solve
(1) x 1 + 1 y 1 = 13
(2) 2x 1 + 12 y 1 = 16
In this case, doubling equation (1)gives equation (2)and there is effectively only
one independent equation; both equations represent the same line. The equations are
said to be linearly dependentand it is only possible to obtain a partial solution; both
equations givex 1 = 131 − 1 yfor all values of y. We will return to the general problem of
solving systems of linear equations in Chapters 17 and 20.
0 Exercise 69
EXAMPLE 2.34Three linear equations
(1) x 1 + 1 y 1 + 1 z 1 = 13
(2) 2x 1 + 13 y 1 + 14 z 1 = 112
(3) x 1 − 1 y 1 − 12 z 1 = 1 − 5
To solve, first eliminate xfrom equations (2)and (3)by subtracting 21 × 1 (1)from (2)
and (1)from (3):
(1) x 1 + y 1 + 1 z 1 = 13
(2′) y 1 + 12 z 1 = 16
(3′) − 2 y 1 − 13 z 1 = 1 − 8
Now eliminate yfrom (3′)by adding 21 × 1 (2′)to(3′):
(1) x 1 + 1 y 1 + 1 z 1 = 13
(2′) y 1 + 12 z 1 = 16
(3′′) z 1 = 14
The equations can now be solved in reverse order:(3′′)isz 1 = 14 , then(2′)isy 1 + 181 = 16
so thaty 1 = 1 − 2 , and (1)isx 1 − 121 + 141 = 13 so thatx 1 = 11.
The method used in this example is a general systematic method for solving any
number of simultaneous linear equations. It is discussed further in Chapter 20.
0 Exercise 70
EXAMPLE 2.35A line and an ellipse
(1) 3 x 1 + 14 y 1 = 14
(2) 2x
21 + 13 xy 1 + 12 y
21 = 116
Equation (1)can be solved for yin terms of xand the result substituted in equation (2)
of the ellipse (see Section 19.5 on quadratic forms). Thus, from (1),y 1 = 111 − 13 x 24 , and
(2) becomes
x
21 − 1161 = 1 (x 1 + 1 4)(x 1 − 1 4) 1 = 10