Question

√2x+√3y=0
√3x-√8y=0
by substitution method​

Answer 1

Answer:

Step-by-step explanation:

Keep only one variable in LHS

√2x+√3y=0

√2x=-√3y

x=-√3y/√2 ... equation 1

√3x+√8y=0 ... equation 2

Substitute eq.1 in eq.2

√3x+√8y=0

√3(-√3y/√2)+√8y=0

-3y/√2+(√8y)√2/√2=0

-3y/√2+√16y/√2=0

-3y/√2+4y/√2=0

-3y+4y/√2=0

y/√2=0

y=0

Substitute y=0 in equation 1

x=-√3y/√2

x=(-√3)0/√2

x=0/√2

x=0

Answer 2

$\huge{ \boxed{ \red{ \sf{Answer:-}}}}$

We have,

√2x + √3y = 0

√3x - √8y = 0

From Equation (1), We deduce:

⇒ √2x = -√3y

⇒ x = -√3y / √2

Plugging in equation (2),

⇒ √3(-√3y/√2) - √8y = 0

⇒ -3y/√2 - 2√2y = 0

Taking y in common,

⇒ y ( -3/√2 - 2√2) = 0

⇒ y = 0

Then,

⇒ x = -√3 × 0 / √2

⇒ x = 0

Hence,

The required value of x and y:

$\huge{ \boxed{ \red{ \sf{0 \:and \: 0}}}}$

Explore more!!

For solving the above, you can also try using the other ways of solving like:

• Graphical method
• Elimination method
• Cross multiplication method.