Respuesta :
We are given inequality y > –2x + 3y < x – 2 .
Given options: (0,0) (0,–1) (1,1) (3,0)
Plugging first coordinate (0,0) in given ineqauality
0 > –2(0) + 3(0) < 0 – 2
0> 0< -2.
First option is not true.
Let us check (0,-1) now.
-1 > –2(0) + 3(-1) < 0 – 2
-1 > -3 < -2.
We can see -1 is greater than -3 and -3 is less than -2.
Therefore, (0,-1) make the given inequality true.
Answer:
Option D is correct.
Step-by-step explanation:
Given inequalities are y > -2x + 3 ..............(1)
and y < x - 2 ...................(2)
Option A).
Inequality (1)
LHS = y = 0
RHS = -2x + 3 = 3
LHS < RHS
So, (0,0) does not satisfy it.
Also there is not to check for other inequality.
Option B).
Inequality (1)
LHS = y = -1
RHS = -2x + 3 = 3
LHS < RHS
So, (0,-1) does not satisfy it.
Also there is not to check for other inequality.
Option C).
Inequality (1)
LHS = y = 1
RHS = -2x + 3 = -2 + 3 = 1
LHS = RHS
So, (1,1) does not satisfy it.
Also there is not to check for other inequality.
Option D).
Inequality (1)
LHS = y = 0
RHS = -2x + 3 = -6 + 3 = -3
LHS > RHS
So, (3,0) satisfy the 1st inequality.
Now, check for other inequality.
Inequality (2)
LHS = y = 0
RHS = x - 2 = 3 - 2 = 1
LHS < RHS
So, (3,0) also satisfy the 2nd inequality.
Therefore, Option D is correct.
