An algebra problem by Armun Alam

Using the normal procedure the equation reaches a weird situation, which I think you've already encountered. For all we know, 6 can't just equal 4. The solution, however, has to do with the step: $\frac{2x+5}{x^2+5x+6}$ = $\frac{2x+5}{x^2+5x+4}$ . The numerator is the same on both sides, i.e. 2x + 5. Hence, if 2x + 5 equalled zero, the equation would be balanced, regardless of whatever happens to the denominator (the value of x when 2x + 5 = 0 doesn't force the denominator to be zero in this case). If it wasn't for this, the equation would have been meaningless, just like the equation x = x + 1. Therefore, 2x + 5 = 0 has the solution x = – $\frac{5}{2}$ or –2.5, and so it is also the solution for the original equation.