Question

Which ordered pair makes both inequalities true? y > –2x + 3 y < x – 2 (0,0) (0,–1) (1,1) (3,0)

Answer 1

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 2

Option D is correct.

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.