Solve[-2 < x + 3 < 5, x][br]
a := x + 3[br][br]Solve[-2 < a < 5, x]
[br]Solve[2 < (3x - 1)/2 < 5, x][br][br][br]Then define:[br][br]ineq := (3x - 1)/2[br][br][br]And solve piece by piece:[br][br]Solve[ineq > 2, x][br][br]Solve[ineq < 5, x]
[b]Does solving each part individually give the same result as the combined version?[/b][br]