⎩ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎨ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎧ a x y = b z c = b z + c x = b y + c z = c z + a x = a x + b y = a z + b x
What is the value of ( a + b + c ) ( x + y + z ) = 0 ?
This section requires Javascript.
You are seeing this because something didn't load right. We suggest you, (a) try
refreshing the page, (b) enabling javascript if it is disabled on your browser and,
finally, (c)
loading the
non-javascript version of this page
. We're sorry about the hassle.
Problem Loading...
Note Loading...
Set Loading...
First, by adding equations 1, 3, 5, we have
( a + b + c ) = ( b + c + a ) z + ( c + a + b ) x
so that we know z = 1 − x
From this, we solve for x in equations 1, 3, and then solve for c from the two
c = 2 1 ( a + b − 3 − a 2 + 2 a b − b 2 )
so that c is real only if a = b . This is the crux step. Immediately from this, we see that a = b = c = y
From this, we then solve for x and a from equations 2, 4, and find that x = a = 2 1 . Hence
( a + b + c ) ( x + y + z ) = 4 9
Another possible solution would be
( a , b , c , x , y , z ) = ( − 1 , − 1 , − 1 , 2 1 , − 1 , 2 1 ) which would have yielded 0