Answer:
(1 , 7) is a solution of y > IxI + 5 ⇒ answer B
Step-by-step explanation:
* Lets revise the absolute value
- IxI = positive value
- IxI can not give negative value
- The value of x could be positive or negative
* Lets solve the problem
∵ y > IxI + 5
∴ y > x + 5 <em>OR</em> y > -x + 5
- Lets check the answers
∵ y > 0 + 5 ⇒ y > 5
- But y = 5, and 5 it is not greater than 5 and there is no difference
between the two cases because zero has no sign
∴ (0 , 5) not a solution
∵ y > 1 + 5 ⇒ y > 6
- Its true y = 7 and 7 is greater than 6
∵ y > -1 + 5 ⇒ y > 4
- Its true y = 7 and 7 is greater than 4
∴ (1 , 7) is a solution
∵ y > 7 + 5 ⇒ y > 12
- But y = 1 and 1 is not greater than 12
∵ y > -7 + 5 ⇒ y > -2
- Its true y = 1 and 1 is greater than -2
* we can not take this point as a solution because it is wrong
with one of the two cases
∴ (7 , 1) is not a solution
