Page 50 - Elementary Algebra Exercise Book I

ELEMENTARY ALGEBRA EXERCISE BOOK I equAtions

2 EQUATIONS

5 8
2.1 Given the equation x − b = x + 142, find the smallest positive integer b such that
2 5
the solution x is a positive integer.

5 8 9 9
Solution: x − b = x + 142 ⇒ b = x − 142. Since b is a positive integer, then x
2 5 10 10
should be a positive integer and greater than 142. Thus x should be a multiple of 10. To

minimize b , x = 160, then b = 9 × 160 − 142 = 2, that is, the smallest positive integer
10
b is 2.

2.2 Solve x − a − b + x − b − c + x − c − a =3 .
c a b
Solution 1: The equation implies a, b, c =0. Multiply abc on both sides of the equation:
(x − a − b)ab +(x − b − c)bc +(x − c − a)ca =3abc ⇔ (ab + bc + ca)x =3abc
(x − a − b)ab +(x − b − c)bc +(x − c − a)ca =3abc ⇔ (ab + bc + ca)x =3abc + +
ab(a + b)+ bc(b + c)+ ca(c + a) ⇔ (ab + bc + ca)x =(a + b + c)(ab + bc + ca) + ca). When
ab(a + b)+ bc(b + c)+ ca(c + a) ⇔ (ab + bc + ca)x =(a + b + c)(ab + bc
ab + bc + ca =0 , x = a + b + c ; When ab + bc + ca =0, x can be any real number.

Solution 2:
x − a − b − b x − b − c b − c x − c − a c − a x − a − b a − b x − b − c b − c x − c − a c − a
x − a
x −
x −
x −
x −
x −
−1+ −1+ x − c − a
+ + x − b − c
x − a − b + + x − c − a =3 −1+ −1+ x − b − c − −
=3 ⇔ ⇔ x − a − b
c c + a a + b b =3 ⇔ c c −1+ a a −1+ b b −
b
b
c
c
a
x − (a + b
x − (a + b + c) + c) x − (a + b + c) + c) x − (a + b + c) + c) a 1 1
x − (a + b
x − (a + b
=0 ⇔ [x−(a+b+c)]( + +
1=
1= 0 ⇔ 0 ⇔ x − (a + b + c) + + x − (a + b + c) =0 ⇔ [x−(a+b+c)]( 1
+ + x − (a + b + c)
a a +
1= 0 ⇔ c c + a a + b b =0 ⇔ [x−(a+b+c)](
a
ab + bc
1 1 1 1 c ab + bc + ca + ca b a
=0 =0
+ )= 0 ⇔ [x − (a + b
1
1
+ )= 0 ⇔ [x − (a + b + c)] + c)] ab + bc + ca When ab + bc + ca =0, x = a + b + c ;
c c
b b + )= 0 ⇔ [x − (a + b + c)] abc abc =0.
b c abc
When ab + bc + ca =0, x can be any real number.
2.3 Find the condition for a such that the equation |ax − 2y − 3| + |5x +9| =0 has the
solution (x, y) where x, y have the same sign.
Solution: 9 9 < 0,y =
|ax − 2y − 3| + |5x +9| =0 ⇒ ax − 2y − 3=0, 5x +9 = 0 ⇒ x = −
|ax − 2y − 3| + |5x +9| =0 ⇒ ax − 2y − 3=0, 5x +9 = 0 ⇒ x = − < 0,y =
5
ax 3 9 3 5
ax 3 − 9 = − 3 a −
5
− 2 = − 2 a − 10 . Since x, y have the same sign, y < 0 ⇒ a> − .
2
2 2 10 2 3
2.4 Find all positive integer solutions of the equation 123x + 57y = 531.
6 − 3x
Solution: 123x + 57y = 531 ⇔ 41x + 19y + 177 ⇔ y =9 − 2x + . Thus
19
x =2,y =5 is a specific solution, then all positive integer solutions should have the
form x =2 − 19t, y = 5 + 41t, where t is an integer.
5 5 2 2
5
<t<
41 41 19 19
2 − 19t> 0,t + 41t> 0 ⇒−2 − 19t> 0,t + 41t> 0 ⇒− <t< , thus the only integer t =0. Hence, the
original equation only has one positive integer solution x =2,y =5.
Download free eBooks at bookboon.com
50