This applet illustrates some misconceptions in solving inequalities. [br]When solving inequalities we replace the original inequality with another, simpler inequality that has the same solutions as the original one (is equivalent to the original inequality). Some operations, like addition and subtraction, always produce equivalent inequalities. Multiplication and division can produce a very different inequality, depending on the expression you are dealing with; whether the expression is a constant or involves variables, whether it changes its sign or not. [br][*]Enter the original inequality in the first input box. Hit "Enter".[br][/*][*]In the second input box enter the next step in your solution to see whether the operation will produce an equivalent inequality. Hit "Enter".[br]The green segments show the solutions of the original inequality. The brown segments show the solutions of the inequality entered in the second input box. They are drawn on a little distance for clarity.[br]Some examples and possible mistakes are listed below the applet.[/*]

[b]Example 1:[/b] [math]\;\;\frac{1}{6}x+\frac{1}{3}>2[/math][br] First step: Multiply by 6: [math]6⋅\frac{1}{6}x+\frac{1}{3}>2⋅6[/math]. [br][br][b]Example 2:[/b] [math]\;\;−3x>6[/math][br] First step: Divide by [math](-3)[/math] and don't change the direction of the inequality: [math]x>−2[/math][br][br][b]Example 3:[/b] [math]\;\;\frac{x+1}{x−2}>1[/math][br] First step: Multiply both sides by the denominator: [math](x+1)>(x–2)[/math] This will produce extra solutions. Notice that the denominator changes its sign.[br][br][b]Example 4:[/b] [math]\;\;\frac{ x−2}{x+1}>1[/math][br] First step: Multiply both sides by the denominator: [math](x−2)>(x+1)[/math] The second inequality has no solutions while the first one has.[br][br][b]Example 5:[/b] [math]\;\;\frac{x^2(x−2)}{x}>1[/math][br] First step: Multiply both sides by the denominator: [math]x^2(x−2)>x[/math]. The second inequality misses some of the solutions of the original one and in the same time introduces some new solutions.[br][br][b]Example 6:[/b] [math]\;\;\frac{x+1}{x^2}<1[/math][br] First step: Multiply both sides by the denominator: [math](x+1)<x^2[/math]. The solution set is the same as the solution of the original inequality because [math]x^2[/math] is always positive.[br][b][br]Example 7:[/b] [math]\;\;\frac{x+1}{x^3}<1[/math][br]First step: Multiply both sides by the denominator: [math](x+1)<x^3[/math]. After changing the power of the denominator from 2 to 3, the same operation will not give an equivalent inequality. The second inequality will omit some of the solutions of the first one. [br][br][b]Example 8:[/b] [math]\;\;x(x−2)(x+3)<2x[/math][br]First step: Divide both sides by [math]x[/math]: [math](x−2)(x+3)<2[/math] The second inequality misses some of the solutions and in the same time introduces some new solutions.