Question 123465
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?



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As for your own number game, you could try...


Take any number, square it, and add 6 multiplied by the original number. After that add 5. Now divide this by the original number plus one. Now subtract your original number to get 5. 



This long sentence translates to ( tell me if you need help with the translation)


{{{(x^2+6x+5)/(x+1)-x}}} 



{{{(x+5)(x+1)/(x+1)-x}}} Factor {{{x^2+6x+5}}} to get {{{(x+5)(x+1)}}}



{{{(x+5)cross((x+1))/cross((x+1))-x}}} Cancel like terms



{{{x+5-x}}} Simplify



{{{x-x+5}}} Group like terms



{{{5}}} Combine like terms



So {{{(x^2+6x+5)/(x+1)-x}}} simplifies to {{{5}}}. In other words, {{{(x^2+6x+5)/(x+1)-x=5}}} for any x but {{{x<>-1}}}. Can you see why?