Question

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.

Answer 1

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.**