When calculating with numbers, the handling of brackets is easy: you can always calculate the value inside the brackets: [br][br]   [math] 2\cdot (5+4)=2\cdot 9=18.[/math][br] [br]Sometimes the numerical value is unknown. However, excluding the brackets may help in simplifying the expression.[br][br][u]Task 1[/u]. How could you solve the previous example without solving the value insde the brackets?[br][br][u]Task 2[/u]. How about in 5 - (4 - 2)?[br][br][u]Task 3[/u]. Or in (2 + 3)(5 - 2)? [br][br]Like noticed in examples, a multiplier or the minus-sign before the brackets must be handled when excluding the brackets. "Formulas" are given below:[br][br]  [math]\large \begin{eqnarray}[br]\textcolor{blue}{a+(b+c)}&=&\textcolor{blue}{a+b+c}\\[br]\textcolor{blue}{a-(b+c)}&=&\textcolor{blue}{a\textcolor{red}{-}b\textcolor{red}{-}c}\\[br]\textcolor{blue}{a(b+c)}&=&\textcolor{blue}{ab+ac}\\[br]\textcolor{blue}{(a+b)(c+d)}&=&\textcolor{blue}{ac+ad+bc+bd}\\[br]\end{eqnarray}[/math][br][br][br]When brackets are excluded and only single terms are left, the like terms can be combined. The terms are like terms, if they are exactly the same but for constant part.[br][br]