Question

If(x+ y )²-z²=4 , (y+ z)²-x²= 9,( z+x )²-y²=36, then( x+ y + z) =? A)0 B)± 1 C)± 3 D)± 7

Answer 1

Answer:

$x+y+z=7,-7$

option (d) is correct

Step-by-step explanation:

Given:

$(x+y)^2-z^2=4$...........(1)

$(y+z)^2-x^2=9$...........(2)

$(z+x)^2-y^2=36$...........(2)

$(1)\implies\:x^2+y^2+2xy-z^2=4$......(4)

$(2)\implies\:y^2+z^2+2yz-x^2=9$......(5)

$(3)\implies\:z^2+x^2+2zx-y^2=36$......(6)

Adding (4), (5) and (6)

$x^2+y^2+2xy-z^2+y^2+z^2+2yz-x^2+z^2+x^2+2zx-y^2=4+9+36$

$x^2+y^2++z^2+2xy+2yz+2zx=49$

$(x+y+z)^2=49$

$\implies\:x+y+z=\sqrt{49}$

$\implies\:x+y+z=7,-7$