Which point is a solution to the simultaneous inequalities y<12x

To find a point that satisfies both inequalities \( y < 12x \) and \( y > 13x + 3 \), we start by understanding the regions defined by these inequalities.

1. The inequality \( y < 12x \) represents the region below the line \( y = 12x \).
2. The inequality \( y > 13x + 3 \) represents the region above the line \( y = 13x + 3 \).

Next, we need to see if there’s a region where these two inequalities can be satisfied simultaneously.

To do this, let's find the intersection point of the lines \( y = 12x \) and \( y = 13x + 3 \):

Set the equations equal to each other: \[ 12x = 13x + 3 \]

Subtract \( 12x \) from both sides: \[ 0 = x + 3 \]

This simplifies to: \[ x = -3 \]

Now substitute \( x = -3 \) into \( y = 12x \) to find the y-coordinate: \[ y = 12(-3) = -36 \] Thus, the intersection point is \( (-3, -36) \).

Now we analyze the inequalities:

• For \( y < 12x \): points below the line \( y = 12x \).
• For \( y > 13x + 3 \): points above the line \( y = 13x + 3 \).

Since \( 13x + 3 \) has a steeper slope than \( 12x \), the inequality \( 13x + 3 \) will always be above \( 12x \) when \( x \) is less than the intersection point \( (-3, -36) \). Therefore, there will be no solutions satisfying both inequalities for \( x < -3 \).

However, when \( x > -3 \), as we move right along the x-axis, \( 13x + 3 \) will eventually become less than \( 12x \). Thus there will be solutions in the domain where \( x > -3 \).

To find a specific point that satisfies both inequalities, we choose a value for \( x \) greater than \(-3\). For instance, let’s try \( x = 0 \):

Substituting \( x = 0 \):

1. \( y < 12(0) \Rightarrow y < 0 \)
2. \( y > 13(0) + 3 \Rightarrow y > 3 \)

We need to find \( y \) such that \( y < 0 \) and \( y > 3 \). This is not possible.

Now, let's try \( x = 1 \):

Substituting \( x = 1 \):

1. \( y < 12(1) \Rightarrow y < 12 \)
2. \( y > 13(1) + 3 \Rightarrow y > 16 \)

Again, we need \( y \) such that \( y < 12 \) and \( y > 16 \). This is also impossible.

Trying \( x = -2 \):

Substituting \( x = -2 \):

1. \( y < 12(-2) \Rightarrow y < -24 \)
2. \( y > 13(-2) + 3 \Rightarrow y > -26 + 3 \Rightarrow y > -23 \)

Here, we see that \( y < -24 \) and \( y > -23 \) cannot be satisfied simultaneously.

Thus, to find a solution, we can instead try \( x = -4 \):

Substituting \( x = -4 \):

1. \( y < 12(-4) \Rightarrow y < -48 \)
2. \( y > 13(-4) + 3 \Rightarrow y > -52 + 3 \Rightarrow y > -49 \)

In this case, \( y < -48 \) and \( y > -49 \) can be satisfied with \( y = -48.5\), which is a valid solution.

Conclusion: A valid solution to the inequalities is \((-4, -48.5)\).