Question

Use deductive reasoning to show that the following procedure always produces the number 6. Procedure: Pick a number. Add 4 to the number and multiply the sum by 3. Subtract 6 and then decrease this difference by the triple of the original number.

Answer 1

Answer:

procedure always produces 6

Step-by-step explanation:

Let 'n' be the unknown number

Add 4 to the number : n+4

multiply the sum by 3.

multiply the sum n+4 by 3

$3(n+4) is 3n+12$

Now subtract 6, so we subtract 6 from 3n+12

$3n+12-6=3n+6$

finally decrease the difference by the tripe of the original number

triple of original number is 3n

$3n+6-3n= 6$

so the procedure always produces 6