Show algebraically why the following Number Trick always works?

Choose any number; add 5; double the result; subtract 4;divide by 2;subtract the original number. the result is always 3.

Show algebraically why the following Number Trick always works?

Choose any number; add 5; double the result; subtract 4;divide by 2;subtract the original number. the result is always 3.

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Saugus, MA

let me explain what is actually happening

let x be the number you choose

add 5, which gives you x+5

double the result which gives you 2(x+5)=2x+10

subtract 4 which gives you 2x+10-4=2x+6

no matter what number you choose you always end up with 2 times your number *
+ 6*

now when you divide by 2 you get *your number(x)+3:*(2x+6)/2=x+3

your number(x)+3 minus the original number(x) gives you 3 all the time

examples:choose 4

4+5=9

2*9=18

18-4=14 which is 8+6=2(4)+6; see! you get 2 times your number +6

divide by 2

[2(4)+6]/2=4+3;your number + 3

now subtract your number and you get 3

choose 5

5+5=10

2*10=20

20-4=16 which is 10+6=2(5)+6

divide by 2 to get 5+3

subtract 5 to get 3 again

choose 6

6+5=11

2*11=22

22-4=18 which is 2(6)+6

divide by 2 to get 6+3 (divide each term by 2 !)

subtract 6 and you end up with 3 again

choose 15

15+5=20

2*20=40

40-4=36=2(15)+6

divide by 2 to get 15+3

subtract 15 to get 3 again and again and again...

Middletown, CT

Hi Nancy;

{{[2(x+5)]-4}/2} -x

[(2x+10-4)/2]-x

x+5-2-x

3

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