Question

question 4 of 10
the point (0, 0) is a solution to which of these inequalities?

a. ( y - 4 < 3x - 5 )

b. ( y + 5 < 3x - 4 )

c. ( y - 5 < 3x - 4 )

d. ( y + 5 < 3x + 4 )

Explanation:

Step1: Substitute (0,0) into A

Substitute \(x = 0\), \(y = 0\) into \(y - 4<3x - 5\):
\(0 - 4<3(0)-5\) → \(- 4<-5\)? False.

Step2: Substitute (0,0) into B

Substitute \(x = 0\), \(y = 0\) into \(y + 5<3x - 4\):
\(0 + 5<3(0)-4\) → \(5<-4\)? False.

Step3: Substitute (0,0) into C

Substitute \(x = 0\), \(y = 0\) into \(y - 5<3x - 4\):
\(0 - 5<3(0)-4\) → \(-5<-4\)? True? Wait, no: \(-5 < -4\) is true? Wait, no, wait: \(-5\) is less than \(-4\)? Wait, no, \(-5\) is more negative, so \(-5 < -4\) is true? Wait, no, wait, let's check again. Wait, no, let's do D first.

Step4: Substitute (0,0) into D

Substitute \(x = 0\), \(y = 0\) into \(y + 5<3x + 4\):
\(0 + 5<3(0)+4\) → \(5<4\)? False. Wait, that can't be. Wait, I must have made a mistake. Wait, let's recheck each option:

• Option A: \(y - 4<3x - 5\). Substitute (0,0): \(0 - 4=-4\); \(3(0)-5=-5\). Is \(-4 < -5\)? No, because \(-4\) is greater than \(-5\) (since it's to the right on the number line).

• Option B: \(y + 5<3x - 4\). Substitute (0,0): \(0 + 5 = 5\); \(3(0)-4=-4\). Is \(5 < -4\)? No.

• Option C: \(y - 5<3x - 4\). Substitute (0,0): \(0 - 5=-5\); \(3(0)-4=-4\). Is \(-5 < -4\)? Yes, because \(-5\) is less than \(-4\) (since \(-5\) is more negative). Wait, but earlier when I checked D, I thought D was false. Wait, let's check D again: \(y + 5<3x + 4\). Substitute (0,0): \(0 + 5 = 5\); \(3(0)+4 = 4\). Is \(5 < 4\)? No. Wait, but the original problem—maybe I misread the options. Wait, let's check the options again.

Wait, the options are:

A. \(y - 4<3x - 5\)

B. \(y + 5<3x - 4\)

C. \(y - 5<3x - 4\)

D. \(y + 5<3x + 4\)

Wait, when we substitute (0,0) into C: \(0 - 5 = -5\); \(3(0)-4 = -4\). So \(-5 < -4\) is true. Wait, but let's confirm: on the number line, \(-5\) is to the left of \(-4\), so \(-5 < -4\) is true. So C is true? Wait, but let's check again. Wait, maybe I made a mistake with D. Wait, D: \(y + 5<3x + 4\). \(0 + 5 = 5\); \(3(0)+4 = 4\). \(5 < 4\) is false. So C is the correct one? Wait, but let's check again. Wait, maybe the problem was written incorrectly? Wait, no, let's re-express each inequality with (0,0):

• A: \(-4 < -5\) → False.

• B: \(5 < -4\) → False.

• C: \(-5 < -4\) → True.

• D: \(5 < 4\) → False.

So the correct answer is C. Wait, but let me confirm once more. Yes, \(-5\) is less than \(-4\) because it is further to the left on the number line (more negative). So C is the solution.

Answer:

C. \(y - 5<3x - 4\)