Question

Solve the equation pair of linear equations by cross multiplication method.x-3y-7=0 and 3x-3y-15=0​

Answer 1

Answer:

x = 4 and 6y = 1

Refer to the above attatchment

Answer 2

$\bf \: Given \: eq.$

$: \implies \sf \: x - 3y - 7 = 0$

$: \implies \sf \: 3x - 3y - 15 = 0$

$\bf now$

$: \implies \sf \: a_{1} = 1,b_{1} = - 3 \: and \: c_{1} = - 7 \\ : \implies \sf \: a_{2} = 3,b_{2} = - 3\: and \: c_{2} = - 15$

$\bf \: Formula$

$\sf : \implies \: x = \dfrac{b _1c_2 - b_2c_1}{a_1b_2 - a_2b_1} ,y = \dfrac{c_1a_2 - c_2a_1}{a_1b_2 - a_2b_1}$

$\bf \: Put \: the \: value \: on \: formula$

$\sf : \implies \: x = \dfrac{( - 3)( - 15) - ( - 3)( - 7)}{(1)( - 3) - ( 3)( - 3)} ,y = \dfrac{( - 7)(3) - ( - 15)(1)}{(1)( - 3) - ( 3)( - 3)}$

$\sf : \implies x = \dfrac{45 - 21}{ - 3 + 9} ,y = \dfrac{ - 21 + 15}{ - 3 + 9}$

$\sf : \implies \: x = \dfrac{24}{6} ,y = \dfrac{ - 6}{6}$

$\sf : \implies \: x = \dfrac{24}{6} ,y = - 1$

$\bf \: Answer$

$\sf : \implies \: x = 4 ,y = - 1$