Question

Solve the following system of the linear equations by using the method of elimination by eqating 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 coefficient equal, we multiply the equation (1) by $\sqrt{3}$ and equation (2) by $\sqrt{2}$ to get
$3x-\sqrt{6}y=3$ .... (3) (0.5 Mark)
$\sqrt{10}x+\sqrt{6}y=2$ .... (4) (0.5 Mark)
Adding equation (3) and equation (4), we get
$3x+\sqrt{10}x=5\Rightarrow (3+\sqrt{10})x=5$ (1 Mark)

$\Rightarrow$ $x=\frac{5}{3+\sqrt{10}}$= $\left(\frac{5}{\sqrt{10}+3}\right)\times \left(\frac{\sqrt{10}-3}{\sqrt{10}-3}\right)$
=$\frac{5(\sqrt{10}-3)}{10-9}=5(\sqrt{10}-3)$
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 $\sqrt{10}-3)$ and y = 5$\sqrt{15}-8\sqrt{6}$ (1 Mark)