graph the inequality on a plane

3x+4y is less then or equal to 12

Check your 8:30 pm posting.

To graph the inequality 3x + 4y ≤ 12, we can start by transforming it into the slope-intercept form, which is y ≤ mx + b where "m" is the slope and "b" is the y-intercept.

Let's rearrange the given inequality:
3x + 4y ≤ 12

Subtract 3x from both sides:
4y ≤ -3x + 12

Divide both sides by 4 to isolate y:
y ≤ (-3/4)x + 3

Now we have the inequality in slope-intercept form. The slope is -3/4, and the y-intercept is 3.

To graph this, we'll start by plotting the y-intercept, which is the point (0, 3). This is where the line intersects the y-axis.

Next, using the slope, we can find another point on the line. The slope -3/4 means for every 4 units we move to the right, we move down 3 units. So, from the y-intercept, we can move right 4 units and down 3 units to get another point. This gives us the point (4, 0).

Now, draw a dashed line through these two points. Since the inequality is "less than or equal to," we use a dashed line instead of a solid line to indicate that points on the line are not included in the solution.

Finally, shade the region below the line to represent the solution set. Since the inequality includes "y ≤," all points below the line (including the line itself) satisfy the inequality.

Therefore, the graph of the inequality 3x + 4y ≤ 12 is a dashed line with a slope of -3/4 passing through the points (0, 3) and (4, 0), and shading below the line.