Question

root3x+root5y=0,root5x-root8y=0

Answer 1

root3x + root5y = 0             --------------(i)
root5x + root8y = 0             --------------(ii)
multiply (i) with root5 and (ii) with root3
     root15 + 5y         = 0
 (-)root15 +(-)root8y = 0
=>y(5-root8) = 0
=> y = 0

put y=0 in (i)
root3x + 0 = 0
root3x = 0
x = 0

Answer 2

Answer: x=0 y=0

Step-by-step explanation

Root3x + root5y = 0. ....... (1)

Root5x-root8y =0...... (2)

Eq (1)= root3x + root5y = 0

Root3x = -root5y

X= -root5y/root3..... (3)

Substitute in Eq (2).

Root5×-Root5y/root3 -root8y=0

-5y/root3 -root8y =0

-5y - root24y=0

-5y=root24y ......... (square on both sides)

25y^2=24y^2

25y^2 - 24y^2 =0

y^2=0

Y=0

Substitute in (3)

X=-root5×0/root3

X=0/root3

X=0

Hence Y=0 and X=0