# How do you solve the following system?

## ${a}_{1} {x}_{1} + {a}_{2} {x}_{2} + {a}_{3} = 0$, ${a}_{4} {x}_{1} + {a}_{5} {x}_{2} + {a}_{6} = 0$

Jun 21, 2018

$\left({x}_{1} , {x}_{2}\right) = \left(\frac{{a}_{2} {a}_{6} - {a}_{3} {a}_{5}}{{a}_{1} {a}_{5} - {a}_{2} {a}_{4}} , \frac{{a}_{3} {a}_{4} - {a}_{1} {a}_{6}}{{a}_{1} {a}_{5} - {a}_{2} {a}_{4}}\right)$

#### Explanation:

Here,

${a}_{1} {x}_{1} + {a}_{2} {x}_{2} = - {a}_{3.} . . \to \left(1\right) , w h e r e , {a}_{1} , {a}_{2} , {a}_{3} \in \mathbb{R}$

${a}_{4} {x}_{1} + {a}_{5} {x}_{2} = - {a}_{6.} . . \to \left(2\right) , w h e r e , {a}_{4} , {a}_{5} , {a}_{6} \in \mathbb{R}$

$\text{using "color(blue)"Cramer's Rule}$ $\text{to solve the system :}$

First we find determinants : $D , {D}_{x} \mathmr{and} {D}_{y}$

$\text{Cramer's Rule :}$

${x}_{1} = \frac{{D}_{x}}{D} = \frac{{a}_{2} {a}_{6} - {a}_{3} {a}_{5}}{{a}_{1} {a}_{5} - {a}_{2} {a}_{4}} , w h e r e , {a}_{1} {a}_{5} - {a}_{2} {a}_{4} \ne 0$

${x}_{2} = \frac{{D}_{y}}{D} = \frac{{a}_{3} {a}_{4} - {a}_{1} {a}_{6}}{{a}_{1} {a}_{5} - {a}_{2} {a}_{4}} , w h e r e , {a}_{1} {a}_{5} - {a}_{2} {a}_{4} \ne 0$

Hence, the solution of system is :

$\left({x}_{1} , {x}_{2}\right) = \left(\frac{{a}_{2} {a}_{6} - {a}_{3} {a}_{5}}{{a}_{1} {a}_{5} - {a}_{2} {a}_{4}} , \frac{{a}_{3} {a}_{4} - {a}_{1} {a}_{6}}{{a}_{1} {a}_{5} - {a}_{2} {a}_{4}}\right) ,$

$w h e r e , {a}_{1} {a}_{5} - {a}_{2} {a}_{4} \ne 0$