Question

3x-4/y=1 , 2x-3/y=2. find it by elimination method​

Answer 1

$\large\underline{\sf{Solution-}}$

Given pair of equations is

$\rm :\longmapsto\:3x - \dfrac{4}{y} = 1 - - - (1)$

and

$\rm :\longmapsto\:2x - \dfrac{3}{y} = 2 - - - (2)$

On multiply equation (1) by 2 and equation (2) by 3, we get

$\rm :\longmapsto\:6x - \dfrac{8}{y} = 2 - - - (3)$

and

$\rm :\longmapsto\:6x - \dfrac{9}{y} = 6 - - - (4)$

On Subtracting equation (4) from equation (3), we get

$\rm :\longmapsto\:\dfrac{1}{y} = - 4$

$\bf\implies \:\boxed{ \tt{ \: y \: = \: - \: \frac{1}{4} \: }}$

On substituting y = - 1/4, in equation (1), we get

$\rm :\longmapsto\:3x - 4( - 4) = 1$

$\rm :\longmapsto\:3x + 16 = 1$

$\rm :\longmapsto\:3x = 1 - 16$

$\rm :\longmapsto\:3x = - 15$

$\bf\implies \:\boxed{ \tt{ \: x \: = \: - \: 5 \: }}$

Verification :-

Consider equation (1),

$\rm :\longmapsto\:3x - \dfrac{4}{y} = 1$

On substituting the values of x and y, we get

$\rm :\longmapsto\:3( - 5) - 4( - 4) = 1$

$\rm :\longmapsto\: - 15 + 16= 1$

$\bf\implies \:1 = 1$

Hence, Verified

Consider Equation (2)

$\rm :\longmapsto\:2x - \dfrac{3}{y} = 2$

On substituting the values of x and y, we get

$\rm :\longmapsto\:2( - 5) - 3( - 4) = 1$

$\rm :\longmapsto\: -10+ 12 = 2$

$\bf\implies \:2 = 2$

Hence, Verified

Answer 2

Answer:

By hit and trial method

$x = - 5$

$y = - \frac{1}{4}$