[u]Task 1.[/u] The depth of water beside a dock is 80 cm. One third of a vertical pole sank onto bottom mud and two fifth was above water level. How high is the pole?[br][br][u]Task 2. [/u]There were benches for guests on a summer event. All rows had equal number of benches. If there were 12 persons in one row, 32 guests had no seat. If there were 13 persons in a row, only 9 persons were without a seat. How many guests were there in this event? [br][br]Both of these tasks require an unknown in an equation in order to find a solution. Different kind of equations are needed in engineering tasks all the time.

[br][color=#0000ff]Equation means equality for two expressions.[/color] [br][br]For example, consider the following equation:[br][br][math]2x+5=3x+4[/math][br][br]In the middle, we have the equals sign ([math]=[/math]). The expression [math]2x+5[/math], which is located on the left side of the equals sign, is called the [b]left-hand side[/b] of the equation. Similarly, the expression [math]3x+4[/math], which is located on the right side of the equals sign, is called the [b]right-hand side[/b] of the equation. The symbol [math]x[/math] is called a [b]variable[/b] (or [b]unknown[/b]).[br][br]Our objective is to [b]solve[/b] the equation, that is, find a value or expression for the variable so that the left-hand side and the right-hand side are equal. For example, in the above equation, we notice that the value of 1 for [math]x[/math] (we denote this [math]x=1[/math]) is the [b]solution[/b] of the equation: for the left-hand side, we would get [math]2\cdot 1+5 = 7[/math], and similarly for the right-hand side, we would get [math]3 \cdot 1+4 = 7[/math]. Thus, we say that the solution [math]x=1[/math] [b]satisfies[/b] the equation.[br][br]An equation is [color=#0000ff]equivalent with the original equation (meaning the equations have identical solutions), if[/color][br] [list][*][color=#0000ff]the same number is added[/color] for [b]both[/b] sides of the equation,[/*][*]the same number is [color=#0000ff]subtracted[/color] from [b]both[/b] sides of the equation,[/*][*][b]both[/b] sides are [color=#0000ff]multiplied or divided[/color] with the same number which is [color=#0000ff]NOT zero[/color].[/*][*]the left-hand side and the right-hand side are swapped.[br][/*][/list] [br]The demonstrate these techniques for solving the above equation, we subtract the number 5 from both sides of the equation:[br][math]2x+5-5 = 3x+4-5,[/math][br]which leads to[br][math]2x = 3x-1.[/math][br]Now we can subtract [math]3x[/math] from both sides of the equation:[br][math]2x-3x = 3x-1-3x,[/math][br]so[br][math]-x = -1.[/math][br]Finally, we can multiply (or divide) both sides by [math]-1[/math] to obtain[br][math]x = 1,[/math][br]which is the solution of the equation.[br][br]Alternatively, we could solve the previous equation by subtracting [math]2x[/math] from both sides:[br][math]2x+5-2x = 3x+4-2x[/math][br][math]5 = x + 4[/math][br]Now we can subtract the number 4 from both sides to obtain[br][math]5-4 = x+4-4[/math][br][math]1 = x[/math][br]Here we can swap the left-hand side and the right-hand side to obtain the solution,[br][math]x = 1.[/math][br][br]As we can see, both steps lead to the same solution.