The given equations are:x/3 + y/4 =11...(i)(5x)/6 − y/3 = −7 .....(ii)From (i) we get(4x + 3y)/12 = 11=> 4x + 3y = 132 ....(iii)From (ii), we get(5x + 2y)/6 = −7=> 5x - 2y = -42 ....(iv)Let us eliminate y from the given equations. The coefficients of y in the equations(iii) and (iv) are 3 and 2 respectively. The L.C.M of 3 and 2 is 6. So, we make the coefficient of y equal to 6 in the two equations.Multiplying (iii) by 2 and (iv) by 3, we get8x + 6y = 264 ...(v)15x - 6x = -126 ...(vi)Adding (v) and (vi), we get8x + 15x = 264 - 126=> 23x = 138=> x = 138/23 = 6Substituting x = 6 in (iii), we get4 x 6 + 3y = 132=> 3y = 132 - 243y = 108=> y = 108/3 = 36Hence, the solution of the given system of equations is x = 6, y = 36.