Question

Select the point that is a solution to the system of inequalities

Answer 1

Hello!

The answer is:

The point C. (2,6) is a solution to the system of inequalities.

Why?

To check what point is a solution to the system of inequalities, the point must satisfy both inequalities, so:

We are given the inequalities:

$y<x^{2}+6\\y>x^{2}-4$

Checking, we have:

- Substituting A(0,8), we have:

First inequality,

$y<x^{2}+6\\8<0^{2}+6\\8<6$

Now, since 8 is not less than 6, we know that A (0,8) is not a solution to the system of inequalities since it does not satisfy both inequalities.

- Substituting B(4,2), we have:

First inequality,

$y<x^{2}+6\\2<4^{2}+6\\2<16+6\\2<22$

Second inequality,

$y>x^{2}-4\\2>4^{2}-4\\2>16-4\\2>12$

Now, since 2 is not greater than 12, we know that B(4,2) is not a solution to the system of inequalities since it does not satisfy both inequalities.

- Substituting C(2,6), we have:

First inequality,

$y<x^{2}+6\\6<2^{2}+6\\6<4+6\\6<10$

Second inequality,

$y>x^{2}-4\\6>2^{2}-4\\6>4-4\\6>0$

Now, as we can se, both inequalities are satisfied since 6 is less than 10 and greater than 0, so C(2,6) is a solution to the system.

- Substituting D(-2,-4)

$y<x^{2}+6\\-4<-2^{2}+6\\-4<4+6\\-4<10$

Second inequality,

$y>x^{2}-4\\-4>2^{2}-4\\-4>4-4\\-4>0$

Now, since -4 is not greater than 0, we know that D(-2,-4) is not a solution to the system of inequalities since it does not satisfy both inequalities.

Hence,

Only the point C. (2,6) is a solution to the system of inequalities.

Have a nice day!

Answer 2

Answer:

The correct answer option is C. (2, 6).

Step-by-step explanation:

We are given the following inequalities so we will check each point if it satisfies them:

$y<x^2+6$

$y>x^2-4$

A. (0, 8):

$y<x^2+6$ ---> $8<0^2+6 = 8<6$ - False

$y>x^2-4$ ---> $8>0^2-4 = 8 > -4$ - True

B. (4, 2):

$y<x^2+6$ ---> $2<4^2+6 = 2<22$ - True

$y>x^2-4$ ---> $2>4^2-4 = 2> 12$ - False

C. (2, 6):

$y<x^2+6$ ---> $6<2^2+6 = 6<10$ - True

$y>x^2-4$ ---> $6>2^2-4 = 6> 0$ - True

D. (-2, -4):

$y<x^2+6$ ---> $-4<(-2)^2+6 = -4<10$ - True

$y>x^2-4$ ---> $-4>(-2)^2-4 = -4> 0$ - False

Therefore, the point which is the solution to this system of inequalities is C. (2, 6).