Q1(v):

Solve the following pair of linear equations by the substitution method. (v) $\sqrt{2}x + \sqrt{3}y = 0$; $\sqrt{3}x - \sqrt{8}y = 0$

Given: A pair of linear equations in two variables:

(i) $\sqrt{2}x + \sqrt{3}y = 0$

(ii) $\sqrt{3}x - \sqrt{8}y = 0$

To find: The values of $x$ and $y$ that satisfy both equations simultaneously using the substitution method.

Step 1: Expressing one variable in terms of the other from Equation (i)

Let the given equations be:

$\sqrt{2}x + \sqrt{3}y = 0$ --- (Equation 1)

$\sqrt{3}x - \sqrt{8}y = 0$ --- (Equation 2)

From Equation (1), isolate $x$:

$\sqrt{2}x = -\sqrt{3}y$

$x = -\frac{\sqrt{3}}{\sqrt{2}}y$ --- (Equation 3)

Step 2: Substituting the expression for $x$ into Equation (2)

Substitute $x = -\frac{\sqrt{3}}{\sqrt{2}}y$ into Equation (2):

$\sqrt{3}\left(-\frac{\sqrt{3}}{\sqrt{2}}y\right) - \sqrt{8}y = 0$

[Distributing $\sqrt{3}$ into the parenthesis]:

$-\frac{3}{\sqrt{2}}y - \sqrt{8}y = 0$

[Note: $\sqrt{8} = \sqrt{4 \times 2} = 2\sqrt{2}$]:

$-\frac{3}{\sqrt{2}}y - 2\sqrt{2}y = 0$

Step 3: Solving for $y$

Factor out $y$ from the expression:

$y \left( -\frac{3}{\sqrt{2}} - 2\sqrt{2} \right) = 0$

[To combine the terms inside the parenthesis, find a common denominator of $\sqrt{2}$]:

$y \left( \frac{-3 - 2\sqrt{2}(\sqrt{2})}{\sqrt{2}} \right) = 0$

$y \left( \frac{-3 - 2(2)}{\sqrt{2}} \right) = 0$

$y \left( \frac{-3 - 4}{\sqrt{2}} \right) = 0$

$y \left( \frac{-7}{\sqrt{2}} \right) = 0$

[Since $\frac{-7}{\sqrt{2}} \neq 0$, we must have $y = 0$]:

$y = 0$

Step 4: Solving for $x$

Substitute the value $y = 0$ back into Equation (3):

$x = -\frac{\sqrt{3}}{\sqrt{2}}(0)$

$x = 0$

Verification:

Substitute $x=0, y=0$ into Equation (1): $\sqrt{2}(0) + \sqrt{3}(0) = 0 + 0 = 0$. (Correct)

Substitute $x=0, y=0$ into Equation (2): $\sqrt{3}(0) - \sqrt{8}(0) = 0 - 0 = 0$. (Correct)

Final Answer: The solution to the system of equations is $x = 0$ and $y = 0$.