Question

Solve the following pair of simultaneous equations:
$3\sqrt{2}x-5\sqrt{3}y+\sqrt{5}=0$ and $2\sqrt{3}x+7\sqrt{2}y-2\sqrt{5}=0$

• A. $x=\frac{18\sqrt{1}+6\sqrt{4}}{56};y=\frac{12\sqrt{6}+2\sqrt{2}}{52}$
• B. $x=\frac{11\sqrt{15}+7\sqrt{16}}{109};y=\frac{1\sqrt{13}-7\sqrt{11}}{2}$
• C. $x=\frac{3\sqrt{12}-5\sqrt{14}}{23};y=\frac{8\sqrt{13}+6\sqrt{11}}{56}$
• D. $x=\frac{10\sqrt{15}-7\sqrt{10}}{72};y=\frac{2\sqrt{15}+6\sqrt{10}}{72}$

Answer 1

The correct option is D $x=\frac{10\sqrt{15}-7\sqrt{10}}{72};y=\frac{2\sqrt{15}+6\sqrt{10}}{72}$

$a_{1}x+b_{1}y+c_1=0$
$a_{2}x+b_{2}y+c_2=0$
$\Rightarrow \frac{x}{b_1{c}_2-b_2{c}_1}=\frac{y}{c_1{a}_2-c_2{a}_1}=\frac{1}{a_1{b}_2-a_2{b}_1}$

Solving by cross multiplication
$\frac{x}{10\sqrt{15}-7\sqrt{10}}=\frac{y}{2\sqrt{15}+6\sqrt{10}}=\frac{1}{42+30}$
So, $\frac{x}{10\sqrt{15}-7\sqrt{10}}=\frac{1}{72}\Rightarrow x=\frac{10\sqrt{15}-7\sqrt{10}}{72}$
and
$\frac{y}{2\sqrt{15}+6\sqrt{10}}=\frac{1}{42+30}\Rightarrow y=\frac{2\sqrt{15}+6\sqrt{10}}{72}$
Hence
$x=\frac{10\sqrt{15}-7\sqrt{10}}{72};y=\frac{2\sqrt{15}+6\sqrt{10}}{72}$