If (x + y )^2-z^2 = 8, (y + z) ^2 -x^2 = 16, (z + x)^2 -y^2= 42, then find the value of (x +y +z)

SSC CGL Model Paper 7

Question : 64 of 100

Marks: +1, -0

(x + y )^2-z^2 = 8, (y + z) ^2 -x^2 = 16

(z + x)^2 -y^2= 42,

then find the value of (x +y +z)

Solution:

ATQ,

(x + y)^2 - z^2 = 8

(x + y + z) (x + y -z ) =8

...(i)
⇒

(y + z)^2 - x^2 = 16

(x +y + z) (y + z - x) = 16

...(ii)

( z - x )^2 -y^2 = 42

(x +y +z) (z + x - y) = 42

..(iii)
→ adding all three equations, we get

(x + y +z) (x + y -z + y +z

-x + z + x -y) = 64

⇒

(x + y +z)^2 = (8)^2

⇒ x+ y +z = ±8

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