Page 38 - Elementary Algebra Exercise Book I

ELEMENTARY ALGEBRA EXERCISE BOOK I reAl numBers

1.101 Given ax + by =7, ax + by = 49, ax + by = 133, ax + by = 406 , evaluate
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a + b
2002(x + y) + 2002xy + .
21
Solution: (ax + by)(x + y)= ax + axy + bxy + by =(ax + by ) + (a + b)xy ,
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(ax + by )(x + y)= ax + ax y + bxy + by =(ax + by ) + (ax + by)xy ,
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(ax + by )(x + y)= ax + ax y + bxy + by =(ax + by ) + (ax + by )xy .
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Substitute ax + by =7, ax + by = 49, ax + by = 133, ax + by = 406 into the above
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equalities to obtain
7(x + y) = 49 + (a + b)xy (i)
49(x + y) = 133 + 7xy (ii)
133(x + y) = 406 + 49xy (iii)
(ii) × 7 − (iii) ⇒ x + y =2.5.
(ii) × 19 − (iii) × 7 ⇒ xy = −1.5.
Substitute x + y =2.5, xy = −1.5 into (i): a + b = 21.
a + b 21
Therefore, 2002(x + y) + 2002xy + = 2002 × 2.5 + 2002 × (−1.5) + = 2003.
21 21
,
1.102 If , q, 2p−1 2q−1 are integers, and p> 1,q > 1, find the value of p + q .
p
q
1
Solution 1: If p = q , then 2p − 1 = 2p − 1 =2 − . Since p> 1 is an integer, then 2p−1 =2 −
1
q p p q p
is not an integer, a contradiction to the given problem. Hence, p = q . Without loss of generality,
Let p>q and let 2q − 1 = m (m is a positive integer). Since mp =2q − 1 < 2p − 1 < 2p , then m =1,
p
3
then p =2q − 1, then 2p − 1 = 4q − 3 =4 − . Additionally since 2p − 1 is also a positive
q q q q
integer and q> 1, then q =3, then p =2q − 1= 5, thus p + q =8.
Solution 2: Starting from p>q, let 2p − 1 = m (i), 2q − 1 = n (ii). m, n are both positive integers
q p np +1
and m>n . (ii) is equivalent to q = np+1 , substitute it into (i): 2p − 1= m , thus
2 2
(4 − mn)p = m +2, thus 4 − mn is a positive integer, i.e. mn =1 or mn =2 or mn =3.
Recall that m>n, then we only have two possibilities m =2,n =1 or m =3,n =1. When
m =2,n =1, (i)(ii) lead to p =2,q =3/2 (q is not an integer). When m =3,n =1, (i)(ii)

lead to p =5,q =3, hence p + q =8.

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