Question

solve by substitution method root 2x +root 3y=0 and root 3x–root 8y=0

Answer 1

√2x+√3y=0 => x=-√3y/√2. ....(1)

√3x-√8y=0. ....(2)

$putting \: x \: in \: (2) \: we \: get \\ \frac{ - \sqrt{3×3}y }{ \sqrt{2} } - \sqrt{8} y = 0 \\ \frac{ - 3y - \sqrt{16}y }{ \sqrt{2} } = 0 \\ - 3y - 4y = 0 \\ -7y = 0 \\ y = 0 \\ \\ putting \: in \: (1) \: we \: get \\ x = 0$

Answer 2

x = 0 , y = 0 for √2 x  + √3 y  = 0  & √3x - √8 y  = 0

Step-by-step explanation:

√2 x  + √3 y  = 0

√3x - √8 y  = 0

√2 x  + √3 y  = 0

=> x = - √3 y /√2

Putting in√3x - √8 y  = 0

=√3 ( - √3 y /√2)  - √8 y  = 0

=> - 3y  - 4y = 0

=> -7y = 0

=> y = 0

=> x = 0

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