Question

Solve the following system of linear equations by using the method of elimination by equating the coefficients
$\sqrt{3}x-\sqrt{2}y=\sqrt{3};\sqrt{5}x+\sqrt{3}y=\sqrt{2}$

Answer 1

The given equations are
$\sqrt{3}x-\sqrt{2}y=\sqrt{3}$ ....(1)
$\sqrt{5}x+\sqrt{3}y=\sqrt{2}$ ...(2)
Let us eliminate y. To make the coefficients equal, we multiply the equation (1) by $\sqrt{3}$ and equation (2) by $\sqrt{2}$ to get
$3x-\sqrt{6}y=3$ ....(3)
$\sqrt{10}x+\sqrt{6}y=2$ ....(4)
Adding equation (3) and equation (4), we get
$3x+\sqrt{10}x=5\Rightarrow (3+\sqrt{10})x=5$

$\Rightarrow x=\frac{5}{3+\sqrt{10}}=\left(\frac{5}{\sqrt{10}+3}\right)\times \left(\frac{\sqrt{10}-3}{\sqrt{10}-3}\right)$

$=\frac{5(\sqrt{10}-3)}{10-9}=5(\sqrt{10}-3)$
Putting $x=5(\sqrt{10}-3)$ in (1) we get

$\sqrt{3}\times 5(\sqrt{10}-3)-\sqrt{2}y=\sqrt{3}$

$\Rightarrow 5\sqrt{30}-15\sqrt{3}-\sqrt{2}y=\sqrt{3}$

$\Rightarrow \sqrt{2}y=5\sqrt{30}-15\sqrt{3}-\sqrt{3}$

$\Rightarrow \sqrt{2}y=5\sqrt{30}-16\sqrt{3}$

$\Rightarrow y=\frac{5\sqrt{30}}{\sqrt{2}}-\frac{16\sqrt{3}}{\sqrt{2}}=5\sqrt{15}-8\sqrt{6}$

Hence, the solution is $x=5(\sqrt{10}-3)andy=5\sqrt{15}-8\sqrt{6}$