Question

solve the following pair of linear equation by the substitution method
√2x+√3y=0
√3x-√8y=0 ​

Answer 1

Step-by-step explanation:

(root 2x+ root 3=0)2

divide the brackets by 2

to make them same,

therefore it will be root 4x+root 6y=0

like this of root (3x-root8y=0)2

After that, u will be here

$\sqrt{4x} + \sqrt{6y} = 0$

$\sqrt{6x} - \sqrt{16y} = 0$

then after solve it

Answer 2

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

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.