Question

State if the point (-3, -2) is a solution to the system of equations below.
x^2 + y^2 + x - 10y - 3=0
x+y= -4

A) Since (-3,-2) does satisfy both equations, it is a solution to the system.

B) Since (-3,-2) does not satisfy both equations, it is not a solution to the
system.

C) Since (-3,-2) does satisfy both equations, it is not a solution to the system.

D) Since (-3,-2) does not satisfy both equations, it is a solution to the system.​

Answer 1

Answer:

B) Since (-3,-2) does not satisfy both equations, it is not a solution to the

system.

Step-by-step explanation:

The given system of equations is:

${x}^{2} + {y}^{2} + x - 10y - 3 = 0$

$x + y = - 4$

If (-3,-2) is a solution, then it must satisfy both equations:

Let us substitute into the first equation to get:

${( - 3)}^{2} + {( - 2)}^{2} + ( - 3) - 10( - 2) - 3 = 0$

Evaluate the exponents;

$9 + 4 - 3 + 20- 3 = 0$

$27 = 0$

This is not true

Also when we substitute into the scond equation, we get;

$- 3 + - 2 = - 4 \\ - 5 = - 4$

This is also false.

Therefore the point is not a solution.