Question

√2x+√3y=√3 ,√2x-√3y=√2
Plz solve this linear equation with 2 variables with method.

Answer 1

Answer :

x = (√3 + √2)/2√2

y = (√3 - √2)/2√3

Solution :

Here ,

The given equations are ;

√2x + √3y = √3 -------(1)

√2x - √3y = √2 -------(2)

Now ,

Adding eq-(1) and (2) , we get ;

=> √2x + √3y + √2x - √3y = √3 + √2

=> 2√2x = √3 + √2

=> x = (√3 + √2)/2√2

Now ,

Subtracting eq-(2) from (1) , we get ;

=> (√2x + √3y) - (√2x - √3y) = √3 - √2

=> √2x + √3y - √2x + √3y = √3 - √2

=> 2√3y = √3 - √2

=> y = (√3 - √2)/2√3

Hence ,

x = (√3 + √2)/2√2

y = (√3 - √2)/2√3

Answer 2

Solution:-

√2x + √3y = √3 ..........(i) eq

√2x - √3y = √2 ...........(ii) eq

By using elimination method , we get

Add (i) eq and (ii) eq , we get

√2x + √3y + √2x - √3y = √3 + √2

2√2x = √3 + √2

$\bf \: x = \frac{ \sqrt{3} + \sqrt{2} }{2 \sqrt{2} }$

Now put the value of x on (i) eq, we get

$\bf\sqrt{2} \times \frac{ \sqrt{3} + \sqrt{2} }{2 \sqrt{2} } + \sqrt{3} y = \sqrt{3}$

$\frac{ \sqrt{3} + \sqrt{2} }{2 } + \sqrt{3} y = \sqrt{3}$

Now taking lcm

$\bf\frac{ \sqrt{3} + \sqrt{2} + 2\sqrt{3}y }{2 } = \sqrt{3}$

$\bf \: \sqrt{3} + \sqrt{2} + 2 \sqrt{3} y = 2 \sqrt{3}$

$\bf \: \sqrt{2} + 2 \sqrt{3} y = \sqrt{3}$

$\bf \: 2 \sqrt{3} y = \sqrt{3} - \sqrt{2}$

$\bf \:y = \frac{ \sqrt{3} - \sqrt{2} }{2 \sqrt{2} }$

$\bf \: { \green{value \: of \: \bf \: x = \frac{ \sqrt{3} + \sqrt{2} }{2 \sqrt{2} } \: and \: \bf \:y = \frac{ \sqrt{3} - \sqrt{2} }{2 \sqrt{2} } }}$