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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.
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