If we have 3 unknowns, there should be 3 simultaneous equations available, so I’m going to treat m as a constant. Subtracting equation 2 with double the equation 1 will obtain => -5x = 3m-3 so x = (3m-3)/5. Then rearranging to y (using equation 1), we get =>y = (m+1-3x)/2. Substituting in x we worked out will give y = (-8m+10)/2 => y = (5-4m)/2