which of the points on the graph is a solution for the simultanious inequalities y > 4x - 2 and y >= -1/3 x + 2? enter the x-value followed by the y-value

Answer 1

To find a point that satisfies both inequalities \(y > 4x - 2\) and \(y \geq -\frac{1}{3}x + 2\), we can analyze both inequalities graphically.

1. Line for \(y = 4x - 2\): This line has a slope of 4 and a y-intercept of -2. The area above this line (not including the line itself) is where \(y > 4x - 2\).

2. Line for \(y = -\frac{1}{3}x + 2\): This line has a slope of \(-\frac{1}{3}\) and a y-intercept of 2. The area above this line (including the line itself) is where \(y \geq -\frac{1}{3}x + 2\).

Now we look for points that lie in the region satisfying both conditions.

One possible way to find a solution is to pick a specific value for \(x\) and calculate the corresponding \(y\) values:

For \(x = 0\):

• From \(y > 4(0) - 2\): \(y > -2\)
• From \(y \geq -\frac{1}{3}(0) + 2\): \(y \geq 2\)

Here, any \(y\) value that is greater than -2 and greater than or equal to 2 will satisfy both inequalities. So, \(y\) can be any value greater than or equal to 2 (like \(y=3\)).

Thus, one solution that meets both conditions is the point \((0, 3)\).

Answer: 0, 3