Page 83 - Elementary Algebra Exercise Book I

ELEMENTARY ALGEBRA EXERCISE BOOK I equAtions

Obviously (iv)(v)(iv) should have the same sign on the right hand side. Hence, the original
2 2
2 2
2 2
2 2
2 2
2 2
2 2
2 2
2 2
system has two solutions: ( a b +a c −b c , a b +b c −a c , a c +b c −a b ),
2abc 2abc 2abc
2 2
2 2
2 2
2 2
2 2
2 2
2 2
2 2
2 2
a b +a c −b c a b +b c −a c a c +b c −a b
(− ,− ,− ).
2abc 2abc 2abc
Solution 2: (i)× (ii)× (iii): (x + y) (x + z) (y + z) = a b c ⇒ (x + y)(x + z)(y + z) = ±abc
2
2
2
2 2 2
bc ac ab
(iv). (iv)/(i),(iv)/(ii),(iv)/(iii) ⇒ y + z = ± ,x + z = ± ,x + y = ± . The right hand side
a b c
should have the same sign, thus
bc
y + z =
a
ac
x + z =
b
ab
x + y =
c
or
bc
y + z = −
a
ac
x + z = −
b
ab
x + y = −
c
They lead to the two solutions same as Solution 1.
√ √
2.90 Nonnegative real numbers x,y,z satisfy 4 5x+9y+4z − 68 × 2 5x+9y+4z + 256 = 0 .

Find the maximum and minimum values of x + y + z .
√
2
Solution: Let 2 5x+9y+4z = t , then t − 68t + 256 = 0 ⇒ (t − 4)(t − 64) = 0 ⇒ t =4 or
t = 64.
√ √
2
When t =4, 2 5x+9y+4z =4 =2 ⇒ 5x +9y +4z =2 ⇒ 5x +9y +4z =4.
√ √
6
When t = 64, 2 5x+9y+4z = 64 = 2 ⇒ 5x +9y +4z =6 ⇒ 5x +9y +4z = 36.
Since x,y,z are nonnegative real numbers, 4(x + y + z) ≤ 5x +9y +4z ≤ 9(x + y + z).
When 5x +9y +4z = 36, x + y + z ≤ 9, thus x + y + z has the maximum value 9, which
can be obtained when x = y =0,z =9.

When 5x +9y +4z =4, x + y + z ≥ 4/9, thus x + y + z has the minimum value 4/9,
which can be obtained when x = z =0,y =4/9.

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