Mathematics — Standard (041) — CBSE Class 10 board question

From equation 1: $\sqrt{2}\,x + \sqrt{3}\,y = 5$ → $x = \dfrac{5 - \sqrt{3}\,y}{\sqrt{2}}$

Substituting in equation 2: $\sqrt{3} \cdot \dfrac{5 - \sqrt{3}\,y}{\sqrt{2}} - \sqrt{8}\,y = -\sqrt{6}$

$$\frac{5\sqrt{3} - 3y}{\sqrt{2}} - 2\sqrt{2}\,y = -\sqrt{6}$$

Multiply throughout by $\sqrt{2}$:

$$5\sqrt{3} - 3y - 4y = -\sqrt{12}$$

$$5\sqrt{3} - 7y = -2\sqrt{3}$$

$$7y = 7\sqrt{3} \implies \boxed{y = \sqrt{3}}$$

Substituting back: $x = \dfrac{5 - \sqrt{3} \cdot \sqrt{3}}{\sqrt{2}} = \dfrac{5-3}{\sqrt{2}} = \dfrac{2}{\sqrt{2}} = \boxed{\sqrt{2}}$

∴ $x = \sqrt{2}$, $y = \sqrt{3}$

Source: Chapter 3, Section 3.3.1 Substitution Method

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• Use substitution: express $x$ from the simpler first equation, substitute into the second.
• Multiply by $\sqrt{2}$ to clear the denominator cleanly — this is the key step examiners look for.
• Always verify: $\sqrt{2}\cdot\sqrt{2} + \sqrt{3}\cdot\sqrt{3} = 2+3 = 5$ ✓ and $\sqrt{3}\cdot\sqrt{2} - \sqrt{8}\cdot\sqrt{3} = \sqrt{6} - 2\sqrt{6} = -\sqrt{6}$ ✓
• Show all algebraic steps clearly; even 1 mark is awarded for correct method even if arithmetic slips.