Is (0,−3)

a viable solution for the following system?

y<−5x+2

3y>3x−6

(1 point)
Responses

(0,−3)
is a viable solution since it makes one inequality true.
left parenthesis 0 comma negative 3 right parenthesis is a viable solution since it makes one inequality true.

(0,−3)
is a nonviable solution since it does not make both inequalities true.
left parenthesis 0 comma negative 3 right parenthesis is a nonviable solution since it does not make both inequalities true.

(0,−3)
is a nonviable solution since it makes both inequalities true.
left parenthesis 0 comma negative 3 right parenthesis is a nonviable solution since it makes both inequalities true.

(0,−3)
is a viable solution since it makes both inequalities true.

1 answer

To determine if the point (0, -3) is a viable solution for the given system of inequalities:

  1. Check the first inequality:
    \( y < -5x + 2 \)
    Substitute \( x = 0 \) and \( y = -3 \):
    \(-3 < -5(0) + 2\)
    This simplifies to:
    \(-3 < 2\)
    This inequality is true.

  2. Check the second inequality:
    \( 3y > 3x - 6 \)
    Substitute \( x = 0 \) and \( y = -3 \):
    \( 3(-3) > 3(0) - 6\)
    This simplifies to:
    \(-9 > -6\)
    This inequality is false.

Since (0, -3) makes the first inequality true but the second inequality false, the point (0, -3) does not satisfy both inequalities.

Therefore, the correct conclusion is:

(0,−3) is a nonviable solution since it does not make both inequalities true.