Question

3 x minus 5 y is equals to 19 ,7 x minus 3 y is equals to 11 find this by cross method

Answer 1

$\huge\sf\pink{Answer}$

$\bf\purple{Formed\:equations\:are :}$

$\mathtt\blue{3x - 5y = 19}$------ equation 1

$\mathtt\blue{7x - 3y = 11}$ ------- equation 2

Comparing with ,

$\implies\sf{a_1x + b_1y + c_1 = 0}$

$\implies\sf{a_2x + b_2y + c_2 = 0}$

Therefore ,

$\implies\sf{a_1 = 3 , b_1 = -5 , C_1 = -19 }$

$\\ \\ \implies\sf{a_2 = 7 , b_2 = -3 , C_2 = -11 }$

$\\ \\ \implies\sf{\dfrac{x}{b_1c_2 - b_2c_1} = \dfrac{y}{c_1a_2-c_2a_1} = \dfrac{1}{a_1b_2-a_2b_1}}$

$\\ \\ \implies\sf{\dfrac{x}{(-5\times -11)-(-3\times -19)} = \dfrac{y}{(-19\times 7)-(-11\times 3)} = \dfrac{1}{(3\times -3) -(7\times -5)}}$

$\\ \\ \implies\sf{\dfrac{x}{55)-(57)} = \dfrac{y}{(-133)-(-33)} = \dfrac{1}{(-9) -(-35)}}$

$\\ \\ \implies\sf{\dfrac{x}{-2} = \dfrac{y}{(-133)+(33)} = \dfrac{1}{(-9) + (35)}}$

$\\ \\ \implies\sf{\dfrac{x}{-2} = \dfrac{y}{(-100} = \dfrac{1}{26}}$

let $\dfrac{x}{-2}$ be I and $\dfrac{y}{-100}$ be II and $\dfrac{1}{26}$ be 3

Taking I and III

[tex]\implies\sf