Question

find the sum of absolute values of x+y+z where x,y,z are integers such that x²z+y²z+4xy=40 and x²+y²+xyz=20​

Answer 1

Step-by-step explanation:

Correct option is

C

6

Squaring both sides

(x+y+z)

2

=x

2

+y

2

+z

2

+2(xy+yz+zx)

we can substitute given values

(x+y+z)

2

=18+(2×9)

(x+y+z)

2

=36

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Answer 2

44

z(x² + y²) + 4xy = 40 ------(i)

2(x² + y²) + 2xyz = 40 ------(ii)

after subracting eqn (ii) from (i) we get,

z(x² + y²) + 4xy - 2(x² + y²) - 2xyz = 0

=> (x²+y²)(z-2) - 2xy(z-2) = 0

=> (z-2)(x² + y² - 2xy) = 0

=> (z-2)(x-y)² = 0

therefore,

z= 2 and x= y

Case 1

putting , z=2 in eqn (i)

2x²+ 2y² + 4xy = 40

=> x² + y² + 2xy = 20

=> (x+y)² = 20

=> x+y = √20

(not possible)

Case 2

putting, x=y in eqn (i)

z(2x²) + 4x² = 40

=> 2x²(z+2) = 40

=> x²(z+2) = 20

=> x²(z+2) = 4×5 or 1×20

two possible outcomes

so, x²= 4 or 1

=> x= ±2 and ±1

so, four cases are available.

After solving case 1 where x=1 , we get

(x,y,z) = (1,1,18)

After solving case 2 where x=-1 , we get

(x,y,z) = (-1,-1,18)

After solving case 3 where x=2 , we get

(x,y,z) = (2,2,3)

After solving case 4 where x=-2 , we get

(x,y,z) = (-2,-2,3)

therefore, sum of absolute values of (x,y,z)

= |1+1+18| + |-1-1+18| + |2+2+3| + |-2-2+3|

= |20| + |16| + |7| + |-1|

= 20+ 16 + 7 +1

= 44

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