Question 132405
Let x= unknown number



To find out what's going on, we have to translate this sentence into an algebraic expression:


First let's square the unknown number to get {{{x^2}}}


Now subtract 1 to get {{{x^2-1}}}


Divide the result by that number minus 1 (ie {{{X-1}}}) to get {{{(x^2-1)/(x-1)}}}



Now subtract the original number x to get {{{(x^2-1)/(x-1)-x}}}





So the sentence translates to {{{(x^2-1)/(x-1)-x}}}. Now let's simplify:




{{{(x^2-1)/(x-1)-x}}} Start with the given expression



{{{(x+1)(x-1)/(x-1)-x}}} Factor {{{x^2-1}}} into {{{(x+1)(x-1)}}} by using the difference of squares




{{{(x+1)cross((x-1))/cross((x-1))-x}}} Cancel like terms



{{{x+1-x}}} Simplify



{{{x-x+1}}} Group like terms



{{{1}}} Combine like terms



So {{{(x^2-1)/(x-1)-x}}} simplifies to {{{1}}}. In other words, {{{(x^2-1)/(x-1)-x=1}}} for any x but {{{x<>1}}}. Can you see why?




I'll let you come up with your own number game