Question

\(\sqrt{3}x + \sqrt{2}y = 2\sqrt{2}\) ; \(\sqrt{2}x - \sqrt{3}y = 3\sqrt{3}\) implies \(x=,y=\)

• A. 1,2
• B. \(\sqrt{6},-1\)
• C. 2,1
• D. \(\sqrt{2},\sqrt{3}\)

Answer 1

The Correct Option is B

We are asked to solve the following system of linear equations for $x$ and $y$:
Equation 1: $\sqrt{3}x + \sqrt{2}y = 2\sqrt{2}$ 
Equation 2: $\sqrt{2}x - \sqrt{3}y = 3\sqrt{3}$

We can use the method of elimination. Let's eliminate $y$. 
Multiply Equation 1 by $\sqrt{3}$: 
$\sqrt{3}(\sqrt{3}x + \sqrt{2}y) = \sqrt{3}(2\sqrt{2})$ 
$3x + \sqrt{6}y = 2\sqrt{6}$ --- (Equation 3) 

Multiply Equation 2 by $\sqrt{2}$: 
$\sqrt{2}(\sqrt{2}x - \sqrt{3}y) = \sqrt{2}(3\sqrt{3})$ 
$2x - \sqrt{6}y = 3\sqrt{6}$ --- (Equation 4)

Now, add Equation 3 and Equation 4: 
$(3x + \sqrt{6}y) + (2x - \sqrt{6}y) = 2\sqrt{6} + 3\sqrt{6}$ 
$3x + 2x + \sqrt{6}y - \sqrt{6}y = 5\sqrt{6}$ 
$5x = 5\sqrt{6}$ 
Divide both sides by 5: 
$x = \sqrt{6}$

Substitute the value $x = \sqrt{6}$ back into Equation 1: 
$\sqrt{3}x + \sqrt{2}y = 2\sqrt{2}$ 
$\sqrt{3}(\sqrt{6}) + \sqrt{2}y = 2\sqrt{2}$ 
$\sqrt{18} + \sqrt{2}y = 2\sqrt{2}$ 
Simplify $\sqrt{18}$: $\sqrt{18} = \sqrt{9 \times 2} = 3\sqrt{2}$. 
$3\sqrt{2} + \sqrt{2}y = 2\sqrt{2}$ 
Subtract $3\sqrt{2}$ from both sides: 
$\sqrt{2}y = 2\sqrt{2} - 3\sqrt{2}$ 
$\sqrt{2}y = -\sqrt{2}$ 
Divide both sides by $\sqrt{2}$: 
$y = -1$

So, the correct option is (B): \(\sqrt{6},-1\)