Page 60 - Elementary Algebra Exercise Book I

ELEMENTARY ALGEBRA EXERCISE BOOK I equAtions

2
x
When y = 101, then (i) implies 102 = 100 ⇒ x = .
lg 102
2
2
We can verify that x − 10 = 1 =1 and x − 10 = 1 ,y = 101 are the two solutions of the original
=
=
,y
lg 102
lg 2
system of equations.
2.34 If a, b, c are nonzero real numbers, solve the system of equations
xy
= c,
ay + bx
yz
= a,
bz + cy
zx
= b.
az + cx

Solution:

xy
= c
ay + bx
yz
= a
bz + cy
zx
= b
az + cx

⇔

ay + bx 1
=
xy c
bz + cy 1
=
yz a
az + cx 1
=
zx b

⇔
a b 1
+ = (i)
x y c
b c 1
+ = (ii)
y z a
a c 1
+ = (iii)
x z b

a b c 1 1 1 1
(i)+(ii)+(iii): + + = + + (iv). Then (iv)-(ii), (iv)-(iii), (iv)-(i):
x y z 2 a b c
2
2
2
= 2a bc ,y = 2b ca ,z = 2c ab .
ab+ac−bc bc+ab−ac ca+bc−ab

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