# How do you graph #3 < -2y+6x#?

Graph

Standard form of a two-variable linear function: -2y + 6x - 3 > 0 First, graph Line: - 2y + 6x - 3 = 0 by its intercepts. Make x = 0 -> y = 1/2. Make y = 0 -> x = - 3/2. Substitute x = 0 and y = 0 into the inequality, we get -3 > 0. This is not true. Then, the solution set area does not contain the origin O; the solution set is the area below the line. Color this answer. Graph graph{- 2y + 6x - 3 = 0 [-10, 10, -5, 5]} (x = 0).

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To graph the inequality 3 < -2y + 6x, you first need to rewrite it in slope-intercept form (y = mx + b), where m is the slope and b is the y-intercept.

-2y + 6x > 3 Subtract 6x from both sides: -2y > -6x + 3 Divide both sides by -2, remembering to reverse the inequality sign when dividing by a negative number: y < 3x - 3/2

Now you can graph the line y = 3x - 3/2 as a dashed line (since it's not part of the solution) and determine which side of the line to shade.

To graph y = 3x - 3/2, you start by plotting the y-intercept at (0, -3/2), and then use the slope of 3 to find additional points. From the y-intercept, go up 3 units and over 1 unit to the right to find another point. Connect these points with a dashed line.

Since the inequality is y < 3x - 3/2, you shade the region below the dashed line.

So, the graph represents all points below the dashed line y = 3x - 3/2.

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