Page 15 - Elementary Algebra Exercise Book I

ELEMENTARY ALGEBRA EXERCISE BOOK I reAl numBers

2
2
2
2
2
2
Solution: 5x − 6xy +2y +2x − 2y +3 = (x − y + 1) + (2x − y) +2 . Since (x − y + 1) ≥ 0, (2x − y) ≥ 0

(x − y + 1) ≥ 0, (2x − y) ≥ 0, the minimum value of 5x − 6xy +2y +2x − 2y +3 is 2, and the
2
2
2
2
associated values of x, y are x =1,y =2 (solved from x − y +1 = 0, 2x − y =0).
1.34 Factoring x + x + x +2.
4
2
3
3
2
4
2
2
Solution: Let x +x +x +2 = (x +ax+1)(x +bx+2) = x +(a+b)x +(ab+3)x +(2a+b)x+2,
2
3
4
3
2
4
then equaling the corresponding coefficients to obtain a + b =1, ab +3 = 1, 2a + b =0 ⇒ a = −1,b =2 ⇒ x + x + x +2 =
4
3
2
a + b =1, ab +3 = 1, 2a + b =0 ⇒ a = −1,b =2 ⇒ x + x + x +2 =
2
2
(x − x + 1)(x +2x + 2)
a + b =1, ab +3 = 1, 2a + b =0 ⇒ a = −1,b =2 ⇒ x + x + x +2 = (x − x + 1)(x +2x + 2).
2
2
3
2
4

2
2
(x − x + 1)(x +2x + 2)
1.35 Let a, b, c are lengths of three sides of a triangle, and they satisfy a − 16b − c +6ab + 10bc =0 ,
2
2
2
show a + c =2b .
2
2
2
2
2
2
2
2
2
Proof: a −16b −c +6ab+10bc = a +6ab+9b −25b +10bc−c =(a+3b) −(5b−c) =
2
2
2
2
2
2
2
2
2
a −16b −c +6ab+10bc = a +6ab+9b −25b +10bc−c =(a+3b) −(5b−c) =
(a +3b − 5b + c)(a +3b +5b − c) =(a − 2b + c)(a +8b − c) =0
(a +3b − 5b + c)(a +3b +5b − c) =(a − 2b + c)(a +8b − c) =0. Since a, b, c represent
lengths of three sides of a triangle, a +8b − c> 0, thus a − 2b + c =0 ⇒ a + c =2b .
1 1 3
1.36 x, y are prime numbers, x = yz , + = , find the value of x +5y +2z .
x y z
1 1 3
Solution: + = ⇒ yz + xz =3xy . Since x = yz, we have x + xz =3xy . Since x =0,
x y z
we have 1+ z =3y. Since y, z are prime numbers, the only possibility is y =2,z =5,x =2 × 5 = 10 .

Hence, x +5y +2z = 10 + 5 × 2+2 × 5 = 30.
a + b b + c c + a
1.37 Given = = where a, b, c are distinct, show 8a +9b +5c =0.
a − b 2(b − c) 3(c − a)
a + b b + c c + a
Proof: Let = = = k , then a + b = k(a − b) (i), b + c =2k(b − c)
a − b 2(b − c) 3(c − a)
(ii), c + a =3k(c − a) (iii). (i) × 6 + (ii) × 3 + (iii) × 2 ⇒ 8a +9b +5c =0.
y
1.38 The positive integers x,y,z satisfy x,y,z = 11, x,y,z + = 14, and x + y = z , find a
x
+
z z
positive integer for x + y if possible.
z
x + y x + y
Solution: Since is a positive integer, we can let = k where k is a positive
z z
x
y
+
integer, then x + y = k (i). The sum of x,y,z = 11 and x,y,z + = 14 leads to x,y,z + x+y = 25
+y
z z z
(ii). Substitute (i) into (ii): kz+k = 25 ⇒ k = 25 . Therefore, z =4 or 24. However when
z+1
x + y
z = 24, k =1 which violates x + y = z . Hence, z =4,k =5 , then =5.
z
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