Question

2x + 3y = 0

3x + 4y = 5

( By substitution method )

Answer 1

$2x + 3y = 0 ............. \: eq.1\\ 3x + 4y = 5............. \: eq.2$
Taking the LCM of 2 and 3 to eliminate x :-

3(2x + 3y = 0)
2(3x +4y = 5)

Or

6x + 9y = 0.......... eq.3
-2(3x+4y =5).... eq.4

Subtracting 4th eq from the 3rd gives :-

y = - 10

Now substituting y in the 1st eq :-

2x + 3(-10) = 0
2x - 30 = 0
x = 30/2
Or x = 15

Therefore, the values of x and y are 15 and - 10 respectively.

P. S. - I just realized it has to be done using substitution. Apologies! (You've got both the methods, you lucky pal!)

Answer 2

Hey!!


2x + 3y = 0 ---------- ( 1 )

3x + 4y = 5 -----------( 2 )

From eq'n ( 2 )

3x + 4y = 5

=> 4y = 5 - 3x

=> y = ( 5 - 3x / 4 ) -------- ( 3 )

Substitute eq'n ( 3 ) in eq'n ( 1 )

2x + 3y = 0

=> 2x + 3 ( 5 - 3x / 4 ) = 0

=> 8x + 15 - 9x = 0

=> - x = - 15

=> x = 15

From eq'n ( 3 )

y = 5 - 3 ( 15 ) / 4

=> y = 5 - 45 / 4

=> y = - 40 / 4

=> y = - 10

Therefore the value of x = 15 and y = - 10



Hope it helps