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\(\text{If $3a + b + c = -3, a+3b+c = 9, a+b+3c = 19$, then find $abc$. }\)

 Jul 18, 2019
 #1
avatar+792 
+4

this is my geuss

a=-1

b=3

c=19/3

 

Travisio

 Jul 18, 2019
 #2
avatar
0

Nope, it's incorrect. I have one more try, though.

Guest Jul 18, 2019
 #3
avatar+106535 
+2

3a + b + c  =  -3    (1)

a  + 3b + c  = 9       (2)

a + b +  3c  =  19     (3)

 

Multiply (1) through by -3   and we get   -9a - 3b - 3c  = 9

Add this to  (2)   and we get that  -8a - 2c  = 18    (4)

 

Multiply  (3)  through by -3  and we get that -3a - 3b -9c  = -57

Add this to (2)  and we get that  -2a - 8c  = -48   (5)

 

Add (4)  and (5)  and we get that

-10a - 10c   = -30      divide through by -10

a + c  = 3 

Sub this into (2)  to find b

3b + 3  = 9

3b = 6

b = 2

 

And since  a + c  = 3      then c = 3 - a

Sub this into (3) to find a

a + 2 + 3(3 - a)  = 19

a + 2 + 9 - 3a  = 19

-2a + 11  = 19

-2a   = 8

a  = - 4

 

And using (1)

3(-4) + 2 + c = -3

-10 + c  = -3

c = 7

 

So  {a, b, c}  =  { -4, 2, 7 }

 

So   abc   =  (-4)(2)(7)  =  -56

 

 

cool cool cool

 Jul 18, 2019

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