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Question 77334: Q.1) If a, b, c, are real numbers such that a2+2b=6, b2+4c= -7 and c2+6a= -13, then the value of a2+b2+c2 is equal to :
(A) 14 (B) 21 (C) 28 (D) 35
Q.2) If x, y and z are positive integers with xy= 24, xz= 48 and yz=72, then x+y+z is:
(A) 18 (B) 19 (C) 20 (D) 22
Answer by Edwin McCravy(20055)   (Show Source):
You can put this solution on YOUR website! Q.1) If a, b, c, are real numbers such that a²+2b=6, b²+4c= -7 and c²+6a= -13,
then the value of a²+b²+c² is equal to :
a²+2b=6
b²+4c= -7
c²+6a= -13
I don't believe this one has a solution in real numbers. It doesn't
according to my graphing calculator. Are you sure you copied it right?
Maybe you didn't mean a² by a2. Did you?
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Q.2) If x, y and z are positive integers with xy= 24, xz= 48 and yz=72,
then x+y+z is:
(A) 18 (B) 19 (C) 20 (D) 22
This one is straight-forward:
xy = 24
xz = 48
Divide equals by equals
xy 24
---- = ----
xz 48
1 1
xy 24
---- = ----
xz 48
1 2
y 1
--- = ---
z 2
Cross-multiply
z = 2y
Substitute that in
yz = 72
y(2y) = 72
2y² = 72
y² = 36
y = 6
Substitute that in
yz = 72
6z = 7
z = 12
Substitute y = 6 in
xy = 24
x(6) = 24
6x = 24
x = 4
x = 4, y = 6, z = 12
so x + y + z = 4 + 6 + 12 = 22 choice (D)
Edwin
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