The given system of equation is:
2x + 3y = 0 .........(i)
3x + 4y = 5 ........(ii)
On multiplying (i) by 4 and (ii) by 3, we get:
8x + 12y = 0 .......(iii)
9x + 12y = 15 ......(iv)
On subtracting (iii) from (iv) we get:
x = 15
On substituting the value of x = 15 in (i), we get:
30 + 3y = 0
⇒ 3y = −30
⇒ y = −10
Hence, the solution is x = 15 and y = −10.