a number n is more than 9 units from 3. write an inequality and solve it.
The number can be 9 units smaller than 3 and also 9 units larger than 3. To account for both these possibilities, we use the absolute value of the difference between the number and 3 greater than 9.
Let n = the number
|n - 3| > 9
Now solve for both possibilities for n.
Suppose n - 3 is positive.
n - 3 > 9 ⇒ n > 9+3 ⇒ n > 12
Suppose n - 3 is negative.
-(n - 3) = - n + 3 = 3 - n
3 - n > 9 ⇒ - n > 9 - 3 ⇒ -n > 6 ⇒ n < -6 (we have to flip the inequality sign when we multiply or divide by a negative)
So there are 2 possibilities n< -6 or n >12.